Month: March 2023

Jharkhand Board JAC Class 10 Science Solutions Chapter 12 Electricity Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 12 Electricity

Jharkhand Board Class 10 Science Electricity Textbook Questions and Answers

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio \(\frac { R }{ R }\) is
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25
Hint Here, the resistance of a piece of wire is R. This piece of wire is cut into five equal parts, so the resistance of each part would be \(\frac { R }{ 5 }\).
Now, these five parts each having resistance of \(\frac { R }{ 5 }\) are connected in parallel.
The equivalent resistance R’ of this combination is given by
\(\frac{1}{R^{\prime}}=\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}+\frac{1}{\left(\frac{R}{5}\right)}\)
∴ \(\frac{1}{R^{\prime}}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}=\frac{25}{R}\)
∴ \(\frac { R }{ R’ }\) = 25

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) \(\frac{V^2}{R}\)
Answer:
(b) IR²
Hint: Electric power P = VI = (IR) I = I²R
Thus, the three expressions given in (a), (c) and (d) represent power while expression given in (b) does not.

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be ……………..
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W
Hint: Here, the power rating of the electric bulb is P = 100 W and the voltage rating is V = 220 V
So, the resistance of the filament of the bulb is
R = \(\frac{V^2}{P}\)
= \(\frac{(220)^2}{100}\)
= 484 Ω

Now, the power consumed by the bulb when it is operated at 110 V is given by
P = \(\frac{V^{\prime 2}}{R}\)
= \(\frac{(110)^2}{484}=\frac{110 \times 110}{484}\)
= 25 W

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be-
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(c) 1 : 4
Hint: As both the wires are made of the same material and are of equal lengths and equal diameters, they have the same resistance R.
1. When they are connected in series their equivalent resistance Rs would be –
R_s = R + R = 2R
and when they are connected in parallel their equivalent resistance R_p is given by
\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}\)
∴ R_p = \(\frac { R }{ 2 }\)

2. Now, heat produced in time t is H = \(\frac{V^2 t}{R}\). As here source voltage V is same,
When wires are connected in series,
H_s = \(\frac{V^2 t}{R_{\mathrm{s}}}=\frac{V^2 t}{2 R}\)
When wires are connected in parallel
∴ H_p = \(\frac{V^2 t}{R_{\mathrm{p}}}=\frac{V^2 t}{\left(\frac{R}{2}\right)}=\frac{2 V^2 t}{R}\)
The ratio of heat produced,
\(\frac{H_{\mathrm{s}}}{H_{\mathrm{p}}}=\frac{V^2 t}{2 R} \times \frac{R}{2 V^2 t}\) = \(\frac { 1 }{ 4 }\)
∴ H_s : H_p = 1 : 4

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is (always) connected in parallel across the points in the circuit between which the potential difference is to be measured.

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10^-8 Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Solution:
We are given,
the diameter of the wire d = 0.5 mm = 0.5 x 10^-3 m = 5 x 10^-4 m
Resistivity of copper ρ = 1.6 x 10^-8 Ω m
Required resistance R= 10 Ω
Length l = ?

R ∝ \(\frac { 1 }{ d² }\) (If there is no change in p and L)
So, when diameter d is doubled, then resistance R becomes one-fourth of its original value.

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

I (amperes)	0.5	1.0	2.0	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
The graph between V and I plotted by using the given data is shown below:

Calculation of the resistance of given resistor :

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Solution:
Here, V = 12 V; I = 2.5 mA = 2.5 m A = 2.5 x 10^-3 A; R = ?
The resistance of the resistor R = \(\frac { V }{ I }\)
= \(\frac { 1 }{ 2 }\)
= 4800 Ω
= 4.8 k Ω

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω respectively. How much current would flow through the 12 Ω resistor?
Solution:
Since all the given resistors are connected in series, their equivalent resistance R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω The current through the circuit,
I = \(\frac{V}{R_{\mathrm{s}}}=\frac{9}{13.4}\) = 0.67 A
In a series combination, the same current I flows through all the resistors, so the current flowing through 12 Ω resistor = 0.67 A.

Question 10.
How many 176 Ω resistors (in parallel) are required to carry 5A on a 220V line?
Solution :
Here, I = 5 A; V = 220 V
So, the total resistance of the given circuit is
R_total = \(\frac { V }{ I }\) = \(\frac { 220 }{ 5 }\) = 44 Ω
i.e., when 44 Ω resistance is connected with 220 V line, 5 A current would flow through the given circuit.
Now, suppose ‘n’ resistors, each of resistance R, are required to be connected in parallel, so that the total resistance R_total becomes 44 Ω.
Hence, \(\frac{1}{R_{\text {total }}}=\frac{1}{R}+\frac{1}{R}\) + … n times
= \(\frac{1+1+\ldots n \text { times }}{R}=\frac{n}{R}\)
∴ R_total = \(\frac { R }{ n }\)
Now, R_total = 44 Ω and R = 176 Ω
So, 44 = \(\frac { 176 }{ 44 }\)
∴ n = \(\frac { 176 }{ 44 }\) = 4
Thus, 4 resistors each of 176 Ω connected in parallel will result in total resistance of 44 Ω causing a current of 5 A to flow when connected to 220 V line.

Question 11.
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Solution:
(i) In order to get a resistance of 9 Ω from three resistors, each of resistance 6 Ω, we connect two 6 Ω resistors in parallel and this parallel combination is connected in series with the third 6 Ω resistor as shown in the following figure:

(ii) In order to get a resistance of 4 Ω from three resistors, each of resistance 6 Ω, we connect two 6 Ω resistors in series and this series combination is connected in parallel to the third 6 Ω resistor as shown in the following figure:

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line are rated 10W. How many bulbs can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Solution:
Here, the voltage rating of each bulb is 220 V and the power rating of each bulb is 10W.
So, the resistance of each bulb
R = \(\frac{V^2}{\rho}=\frac{220^2}{10}=\frac{220 \times 220}{10}\) = 4840 Ω
Now, here, V = 220 V and I = 5 A given.
So, the total resistance of the given circuit is
R_total = \(\frac { V }{ I }\) = \(\frac { 220 }{ 5 }\) = 44 Ω
i.e., When 44 Ω resistance is connected with 220 V line, 5 A current would flow through the given circuit.
Now, when ‘n’ bulbs each of resistance R, are connected in parallel, their equivalent
resistance P_total = \(\frac { R }{ n }\)
Hence, 44 = \(\frac { 4840 }{ 44 }\)
∴ n = \(\frac { 4840 }{ 44 }\) = 110
Thus, 110 bulbs each of resistance 4840 Ω connected in parallel will result in total resistance of 44 Ω causing a current of 5 A to flow, when connected to the 220 V line.

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Solution:
Here, the potential difference V = 220 V
The resistance of each coil R_A = R_s = 24 Ω
(1) When each of the coils A or B is connected separately, the current through each coil is
I = \(\frac{V}{R_{\mathrm{A}}}\) or \(\frac{V}{R_{\mathrm{B}}}\)
= \(\frac { 220 }{ 24 }\)
= 9.166 A

(2) When the coils A and B are connected in series, the equivalent resistance of the circuit R_s = R_A + R_B = 24 + 24 = 48 Ω
So, the current through the series combination
I_s = \(\frac{V}{R_{\mathrm{s}}}\)
= \(\frac { 220 }{ 48 }\)
= 4.58 A ≈ 4.6 A

(3) When the coils A and B are connected in parallel, the equivalent resistance R_p of the circuit is given by
\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R_{\mathrm{A}}}+\frac{1}{R_{\mathrm{B}}}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}\)
∴ R_p = 12 Ω
So, the current through the parallel combination
I_p = \(\frac{V}{R_{\mathrm{p}}}=\frac{220}{12}\) = 18.33 A

Question 14.
Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Solution:
(i) As 1 Ω resistor and 2 Ω resistor are connected in series, the equivalent resistance R_s = 1 + 2 = 3 Ω
Now, the voltage of the battery V = 6 V
So, the current flowing through the circuit,
I_s = \(\frac{V}{R_{\mathrm{s}}}=\frac{6}{3}\) = 2 A
In a series combination the same current 2 A flows through each resistor. Hence the current flowing through 2 Ω resistor is also 2 A.
∴ Power used in 2Ω resistor,
P₁ = I²_sR
= (2)² x 2 = 8 W

(ii) As 12 Ω resistor and 2 Ω resistor are connected in parallel and 4 V battery is connected in parallel with this parallel combination of resistors,
The p.d. across 2 Ω resistor will also be 4 V.
∴ Power used in 2 Ω resistor,
P₂ = \(\frac{V^2}{R}=\frac{4^2}{2}=\frac{16}{2}\) = 8 W
In order to compare the power used in 2 Ω resistor in two different circuits, find the ratio of P₁ and P₂.
So, \(\frac{P_1}{P_2}=\frac{8}{8}\) = 1
∴ P₁ = P₂
Hence, 2 Ω resistor uses equal power, i.e., 8 W in both the circuits.

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Solution :
The resistance of 100 W lamp,
R₁ = \(\frac{V^2}{P_1}=\frac{220^2}{100}=\frac{220 \times 220}{100}=\frac{4840}{10}\)Ω = 484 Ω
The resistance of 60 W lamp,
R₂ = \(\frac{V^2}{P_2}=\frac{220^2}{60}=\frac{220 \times 220}{60}=\frac{4840}{6}\)Ω = 806.7 Ω
When, these lamps are connected in parallel, their equivalent resistance R_p is given by
\(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_2+R_1}{R_1 R_2}\)
∴ R_p = \(\frac{R_1 R_2}{R_1+R_2}\)
= \(\frac{484 \times 806.7}{484+806.7}=\frac{390442.8}{1290.7}\) = 302. 6 Ω
The current drawn from the line,
I = \(\frac{V}{R_{\mathrm{p}}}=\frac{220}{302.6}\) = 0.7270 ≈ 0.73 A

Question 16.
Which uses more energy, a 250 W TV set in 1 h, or a 1200 W toaster in 10 minutes?
Solution:
For TV set:
Power P = 250 W = 250 \(\frac { J }{ s }\)
Time t = 1 h = 3600 s
Electric energy (used by TV set)
= P x t
= 250 \(\frac { J }{ 2 }\) x 3600 s
= 900000
= 900 kJ
For Toaster:
Power P = 1200 W = 1200\(\frac { J }{ s }\)
Time t = 10 minute = 10 x 60 = 600 s
Electric energy (used by toaster)
= P x t
= 1200 \(\frac { J }{ s }\) x 600 s = 720000 = 720 kJ s
From above calculations, it is clear that a 250 W TV set in 1 h consumes more electric energy than a 1200 W toaster in 10 minute.

Question 17.
An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution :
Here, I = 15 A; R = 8 Ω; t = 2 h
The rate at which heat is developed in the
heater means its electric power,
P = I² R
= (15)² x 8 = 225 x 8 = 1800 W = 1800 \(\frac { J }{ s }\)

Question 18.
Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
(a) Because tungsten has a high melting point (3380 °C), It does not melt at high temperature. It retains as much of heat generated, so that it becomes very hot and emits light without melting away.

Moreover, tungsten has high flexibility and low rate of evaporation at high temperature. That is the reason why tungsten is used as filament of electric lamps.

(b) The coils of electrical heating devices such as electric toasters and electric irons are made of an alloy, e.g., Nichrome rather than a pure metal because :

• the resistivity of an alloy e.g., Nichrome is much higher than that of its constituent metals, c
• alloys do not oxidise (i.e., burn) readily at high temperature (i.e., when it is red hot at 800 °C)
  alloy has a high melting point.

(c) (1) In a series circuit the current is constant throughout the electric circuit. So it s is obviously impracticable to connect an electric
bulb and an electric heater in series because they need currents of widely different values to operate properly.

(2) In a series circuit when one component (or electrical appliance) fails due to some defect, the circuit is broken and none of the components (or electrical appliances) works.

(3) In a series circuit all the electrical appliances have only one switch due to which they cannot be turned ON or OFF separately.

(4) In a series circuit electrical appliances of different power ratings do not get the same voltage (220V) as that of the power supply line because the voltage is shared by all the appliances. The appliances get less voltage and hence do not work properly.

(d) The resistance of a wire is inversly proportional to its area of cross-section
i.e., R ∝ \(\frac { 1 }{ A }\)
Thus, if the wire is thick (large area of cross-section), then its resistance is less. If the wire is thin (less area of cross-section), then its resistance is large.

(e) Because:

• Copper and aluminium have low electric resistivity, so they conduct electric current without heavy heat losses (due to which they are known as very good conductors of electricity).
• Copper and aluminium are much more easily and cheaply available as compared to metal like silver.
  They can be easily made into wires due to high malleability.

Therefore, copper and aluminium wires are usually employed for electricity transmission.

Jharkhand Board Class 10 Science Electricity InText Questions and Answers

Question 1.
What does an electric circuit mean?
Answer:
An electric circuit is a continuous and closed path of an electric current.
OR
A continuous and closed path consisting of conducting wires and other electrical components along which an electric current flows is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
The SI unit of current is called an ampere (A).
If 1 coulomb charge flows through any cross-section of a conductor in 1 second, then the electric current flowing through the conductor is
said to be 1 ampere.
i.e., 1 A = \(\frac { 1C }{ 1s }\) = 1C s^-1

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Solution :
We know that one electron possesses a negative charge of 1.6 x 10^-19 C.
i.e., charge on 1 electron = – 1.6 x 10^-19 C

∴ No. of electron constituting – 1 C of charge
= \(\frac{-1 C \times 1}{-1.6 \times 10^{-19} \mathrm{C}}\)
= \(\frac { 10 }{ 1.6 }\) x 10¹⁸ = 6.25 x 10¹⁸
Thus, 6.25 x 10¹⁸ electrons taken together constitute 1 coulomb of charge.
In other words, the SI unit of electric charge coulomb (C) is equivalent to the charge contained in 6.25 x 10¹⁸ electrons.

Question 4.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
An electric cell or battery is a device that helps to maintain a potential difference across a conductor.

Question 5.
What is meant by saying that the potential difference between two points is 1 V?
Answer:
The potential difference between two points (in a electric field) is said to be 1 volt if 1 J of work is done to move a charge of 1 C from one point to another point.
1 V = \(\frac { 1 J}{ 1 C }\)

Question 6.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Solution:
Here the term, ‘each coulomb’ means ‘every 1 coulomb’. So Q = 1 coulomb, potential difference V = 6 volt, Energy = work done, W = ?
Now, W = VQ
= 6V x 1C
= 6 J
Since the work done on each coulomb of charge is 6 J, 6 J energy is given to each coulomb of charge passing through a 6V battery.

Question 7.
On what factors does the resistance of a conductor depend?
Answer:
The resistance R of a conductor depends on :

• its length l as R ∝ l
• its area of cross-section (i.e., thickness of the conductor) as R ∝ \(\frac { 1 }{ A }\)
• the nature of the material of the conductor (i.e., resistivity of material of the conductor)
• its temperature

Question 8.
Will current can flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
The current will flow more easily through a thick wire than through a thin wire of the s same material and having the same length when connected to the same source.

• Because, the resistance of a wire is inversly proportional to its area of cross-section, A thick wire has more area of cross-section and hence less resistance compared to a thin wire provided the two wires have the same length.
• Hence, a current can flow more easily through a thick wire compared to a thin wire when connected to the same source.

Question 9.
Let the resistance of an electrical component remain constant while the protential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
According to Ohm’s law, I = \(\frac { V }{ R }\)
Here, as R = constant, I ∝ V.
So, when the potential difference across the two ends of the electrical component decreases to half of its former value, the current through it will also decreases to half of its former value.

Question 10.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
The coils of electrical heating devices such as electric toasters and electric irons are made of an alloy, e.g., Nichrome rather than a pure metal because :

• the resistivity of an alloy e.g., Nichrome is much higher than that of its constituent metals.
• alloys do not oxidise (i.e., burn) readily at high temperature (i.e., when it is red hot at 800 °C)
• alloy has a high melting point.

Question 11.
Use the data in Table 2 to answer the following:
(a) Which is a better conductor, iron or mercury?
(b) Which material is the best conductor?
Answer:
(a) The electrical resistivity of iron is 10.0 x 10^-8 Ω m whereas that of mercury is 94.0 x 10″8fim. As the resistivity of iron is less than that of mercury, iron (Fe) is a better conductor than mercury (Hg).

(b) Silver metal has the lowest electrical resistivity of 1.60 x 10^-8 Ω m, therefore silver metal is the best conductor of electricity.

Question 12.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:
A schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series is shown in figure 12.12

Question 13.
Redraw the circuit of putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would he the readings in the ammeter and the voltmeter?
Answer:
The circuit diagram of above with an ammeter A and a voltmeter V across 12 Ω resistor is shown in figure 12.13.

Calculation of current flowing in the circuit:
Equivalent resistance of the circuit,
R_s = 5 Ω + 8 Ω + 12 Ω = 25 Ω
Potential difference V = 6 volt
In a series combination, the current flowing through each resistor is the same and equal to the total current flowing through the circuit.
∴ Current through the resistors,
I = \(\frac { V }{ R }\)
= \(\frac { 6 }{ 25 }\)
= 0.24 A
Since the current in the circuit and the current through each resistor is the same, the ammeter will show a reading of 0.24 A.
Calculation of potential difference across 12 Ω resistor:
As total current 0.24 A flows in the circuit, the same current 0.24 A would also flow through the 12 Ω resistor which is connected in series.
∴ The potential difference across the 12 Ω resistor,
V = IR
= 0.24 x 12 = 2.88 V
Thus, the p.d. across 12 Ω resistor is 2.88 volt.
So, the voltmeter will show a reading of 2.88 V.

Question 14.
Judge the eΩuivalent resistance when the following are connected in parallel:
(a) 1 Ω and 10⁶ Ω
(b) 1 Ω, 10³ Ω and 10⁶Ω
Answer:
(a) Less than 1 Ω (but approximately 1 Ω)
(b) less than 1 Ω (but approximately 1 Ω)
Explanation: When resistors are connected in parallel, the equivalent resistance is less than the least resistance (here in both the cases it is 1 Ω) connected in the combination.
or
(a) Here, R₁ = 1 Ω and R₂ = 10⁶ Ω
The equivalent resistance R_p of the parallel combination is given by,
\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R_1}+\frac{1}{R_2}\)
∴ R_p = \(\frac{R_1 \times R_2}{R_1+R_2}\)
= \(\frac{1 \times 10^6}{1+10^6}\)
= \(\frac { 1000000 }{ 1000001 }\) ≈ 1 Ω

(b) Here, R₁ = 1 Ω, R₂ = 10³ Ω and R₃ = 10⁶
The equivalent resistance Rp of the parallel combination is given by,

Question 15.
An electric lamp of 100 Ω, a toaster of resistance 50 Cl, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current through it?
Solution :
Here, Resistance of an electric lamp R₁ = 100 Ω
Resistance of a toaster R₂ = 50 Ω
Resistance of a water filter R₃ = 500 Ω.
Equivalent resistance R_p of three resistors R₁, R₂ and R₃ connected in parallel is given by,

Total current through the circuit (i.e., through all the three appliances)
I = \(\frac{V}{R_p}\)
= \(\frac { 220 }{ 31.25 }\)
= 7.04 A
Since the electric iron connected to the same source (i.e., 220 V) takes as much current as taken by all the three appliances (i.e., f = 7.04A), its resistance must be equal to R_p.
So, the resistance of the electric iron = 31.25 Ω
and the current through the electric iron = 7.04 A

Question 16.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
The advantages of connecting electrical devices in parallel with the battery instead of connecting them in series are as follows :
(1) A parallel combination / circuit is helpful when each device has different resistance and requires different current for its operation, as in this case the total resistance of the circuit is reduced due to which the current from the battery is high and hence each electrical device can draw the required amount of current.

This is not so in a series combination / circuit because total resistance increases too much in series circuit due to which the current from the battery is low and the same low current flows through all the devices, irrespective of their resistances and hence devices cannot work properly.

(2) In a parallel combination / circuit each electrical device gets the same p.d. (potential difference) across it as that of the battery due to which all the devices work properly.

While in case of a series combination / circuit the devices do not get the same p.d. as that of the battery because the p.d. is shared by all the devices connected in series.

(3) In a parallel combination / circuit if one electrical device stops working due to some defect, then other devices are not affected. They continue to work without any problem.

On the other hand, in a series combination/ circuit if one electrical device stops working due to some defect, then all other devices also stop working as the whole circuit is broken.

(4) In a parallel combination / circuit each electrical device has its own switch due to which it can be turned ‘ON’ or turned ‘OFF’ independently without affecting other devices.

But in a series combination / circuit all the electrical devices have only one switch due to which they cannot be turned ‘OFF’ or turned ‘ON’ independently.

Question 17.
How can three resistors of resistances 2 Ω, 3 Ω and G Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Answer:
(a) In order to obtain a total resistance of 4 Ω from three resistors of 2 Ω, 3Ω and 6Ω…
1. First connect the two resistors of 3 Ω and 6 Ω in parallel to get a total resistance of 2 Ω. This is because in parallel combination,

2. Now, above parallel combination of 3 Ω and 6 Ω resistors is connected in series with the remaining 2 O resistor to get total resistance of 4 Cl. This is because in series combination.
R_s = R_p + R₃
= 2 + 2
= 4 Ω
Hence, the arrangement of three resistors 2 Ω, 3 Ω and 6 Ω which gives total resistance 4 Ω can be represented as follows :

(b) In order to obtain a total resistance of 1 Ω from three resistors of 2 Ω, 3 Ω and 6 Ω, all the three resistors should be connected in parallel. This is because in a parallel combination,

Hence, the arrangement of three resistors 2 Ω, 3 Ω and 6 Ω which gives total resistance 1 Ω can be represented as follows :

Question 18.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω,
8 Ω, 12 Ω, 24 Ω?
Solution:
(a) The highest resistance can be secured (or obtained) by connecting all the four coils in series. In this case,
R_s = R₁+R₂ + R₃ + R₄
= 4 + 8 + 12 + 24 = 48 Ω
Thus, the highest resistance which can be secured is 48 Ω.

(b) The lowest resistance can be secured by connecting all the four coils in parallel. In this cae,

Thus, the lowest resistance which can be secured is 2 Ω.

Question 19.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
The heating element of an electric heater is made of an alloy (such as Nichrome) which has high resistance (partly due to high resistivity) whereas the cord is made of copper metal which has very low resistance (partly due to low resistivity).

• Now, the heating element of an electric heater made of Nichrome glows because it becomes red-hot due to the large amount of heat (according to H = I²Rt) is produced on passing current.
• On the other hand, the connecting cord of the electric heater made of copper does not glow because relatively very less (negligible) heat (according to H = I²Rt) is produced in it by passing the same current.

Question 20.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Solution:
Here, charge, Q = 96000 C,
time t = 1 h = 60 x 60 = 3600 s,
potential difference V = 50 V
Now,
Heat generated H = VIt
= V(\(\frac { Q }{ t }\))t (∵ I = \(\frac { Q }{ t }\))
= 50 x 96000
= 4800000 J
= 4.8 x 10⁶ J
or
= 4.8 x 10³ x 10³ J
= 4.8 x 10³ kJ
= 4800 kJ

Question 21.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat devloped in 30 s.
Solution :
Here, current I = 5 A,
resistance R = 20 Ω, time t = 30 s
Now,
Heat produced H = I²Rt
= (5)² x 20 x 30
= 25 x 20 x 30
= 15000 J
= 15 kJ

Question 22.
What determines the rate at which energy is delivered by a current?
Answer:
The electric power of the source determines the rate at which energy is delivered by the current to the load / appliance.
[Whereas the electric power of an appliance determines the rate at which energy delivered by a current Is consumed by appliance.]

Question 23.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2h.
Solution:
Here, I = 5A, V = 220V,
t = 2h = (2 x 60 x 60)s = 7200s
Power P = VI = 220 x 5 = 1100W = 1100J/s
Now,
Energy consumed W = Pt = 1100J/s x 7200s
= 7920000
= 7.92 x 106J
OR
Energy consumed W = Pt = 1100 w x 2 h
= 2200 Wh
= 2.2 x 10³ Wh
= 2.2 k Wh

Activity 12.1 [T. B. Pg. 203]

To verify Ohm’s law.
OR
To find the relationship between potential difference (V) and current (I).

Procedure:
1. Set up a circuit as shown in figure 12.3, consisting of a +Nichrome wire XY of length say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each.

2. First use only one cell as the source in the circuit.
Note the reading of ammeter I, for the current and reading of the voltmeter V for the potential difference across the Nichrome wire XY in the circuit.
Tabulate them in the Table given.

3. Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the Nichrome wire and potential difference across the Nichrome wire.

4. Repeat the above steps using three cells and then four cells in the circuit separately.
Calculate the ratio V to I for each pair of potential difference V and current I.

Observation table :

Sr. No.	Number of cells used in the circuit	Potential difference across the Nichrome wire V (volt)	Current through the Nichrome wire I (ampere)	\(\frac { V }{ I }\) (volt / ampere)
1.	1	1.5	0.1	15
2.	2	3	0.2	15
3.	3	4.5	0.3	15
4.	4	6	0.4	15

Plot a graph between V and I and observe the nature of the graph.
(Take the voltage (V) on X-axis and the current (J) on Y-axis. Draw the graph of I-V taking proper scale.)

Observation:
When V increases I also increases linearly, i.e., I ∝ V. The ratio V/I is found to be (approximately) the same, i.e., 15V/A.
The graph between V and 1 is a straight line passing through the origin O.

Conclusion:
The electric current flowing through a metallic wire is directly proportional to the potential difference across its ends (J ∝ V) and V/I is a constant ratio in particular case.

Activity 12.2 [T. B. Pg. 205]

To show that the strength of an electric current in a circuit depends on resistance used in it.

Procedure:
1. Take a Nichrome wire, a torch bulb, a low bulb (10W) and an ammeter (0-5A range), a plug key and some connecting wires.

2. Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in figure 12.5.

3. Complete the circuit by connecting the Nichrome wire in the gap XY. Plug the key.
Note down the ammeter reading. Take out the key from the plug after measuring the current through the circuit.
[Note: Always take out the key from the plug after measuring the current through the circuit.]

4. Replace the Nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.

5. Now repeat the above step with the 10 W bulb in the gap XY.

• Are the ammeter readings different for different s components connected in the gap XY?
• What do the above observations indicate?

6. You may repeat this Activity by keeping any material component in the gap.
Observe the ammeter readings in each case. Analyse the observations.

Observation :

• Yes, the ammeter readings are different for different components (Nichrome wire, a torch bulb and a low bulb) when connected in the gap XY.
• If we take any other material component in the gap XY, the reading of the ammeter may be different.
• This is because certain components offer an easy path for the flow of electric current while the others resist the flow.
• i.e., the current through an electric component depends on its resistance.

Conclusion :
The strength of an electric current in a circuit depends on the resistance used in it.

• A component of a given size that offers low resistance is called a good conductor.
• A component of the same size which offers appreciable resistance is called a resistor.
• A component of the same size that offers a higher resistance is called a poor conductor or a bad conductor.
• A component of the same size that offers very high resistance is called an insulator.

Activity 12.3 [T. B. Pg. 206]

To study the factors on which resistance of conducting wire depends.

Procedure:

• Complete an electric circuit consisting of a cell, an ammeter, a Nichrome wire of length l [say, marked (1)] and a plug key, as shown in figure 12.6.
  
• Now, plug the key. Note the current in the ammeter.
• Replace the Nichrome wire by another Nichrome wire of same thickness but twice the length, that is 21 [marked (2) in the figure],
• Note the ammeter reading.
• Now replace the wire by a thicker Nichrome wire, of the same length l [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
• Instead of taking a Nichrome wire, connect a copper wire [marked (4) in the figure] in the circuit.
• Let the wire be of the same length and same area of cross-section as that of the first Nichrome wire [marked (1)]. Note the value of the current.
• Notice the difference in the current in all cases,
• Does the current depend on the length of the conductor ?
• Does the current depend on the area of s cross-section of the wire used ?

Observations :

• Ammeter shows the current I flows in the Nichrome wire marked (1).
• When the length of Nichrome wire marked (2) is doubled keeping its same thickness (i.e., when the Nichrome wire marked (1) is replaced by the Nichrome wire marked (2) the ammeter reading decreases to half, i.e. current becomes 1/2.
• When the thickness of Nichrome wire marked (3) is increased keeping its length same i.e., when the Nichrome wire marked (3) is S connected in the circuit), the ammeter reading increases, i.e., current I increases.
• When Nichrome wire is replaced by a copper s wire marked (4) of same length and same cross-sectional area as marked (1), the ammeter reading increases, i.e., current I increases.

Conclusion :

• Resistance of the conducter depends on its s length. Here resistance of wire R is directly proportional to length l, i.e., R ∝ l.
• Resistance of the conductor depends on its area of cross-section. Here resistance of wire R is inversely proportional to area of cross-section A, i.e., R ∝ \(\frac { 1 }{ 2 }\)
• Copper is a better conductor than Nichrome. Resistance of wire depends on the nature of S its material.

Activity 12.4 [T. B. Pg. 210]

To study the series combination of resistors.

Procedure :

• Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in figure 12.9.
• You may use the resistors of values like 1 Ω, 2 Ω, 3 Ω, etc. and a battery of 6 V for performing this Activity.
• Plug the key. Note the ammeter reading. [part (1)]
• Change the position of the ammeter to anywhere inbetween the resistors. Note the ammeter reading each time, [part (2)]
• Do you find any change in the value of current through the ammeter?
  

Observation :

• The ammeter reading is nearly 1 A.
• When the position of the ammeter is changed to anywhere inbetween the resistors, its reading remains the same, i.e., the current flowing through it and through each part of the circuit, i.e., through each resistor is the same.
• No, current flowing through the ammeter does not change.

Conclusion :

• The value of the current in the ammeter is the same, independent of its position in the electric circuit.
• In a series combination of resistors the current is the same in every part of the circuit, i.e., the same current flows through each resistor. This is due to the reason that, the current has only one path to flow.

Important note:
(1) In discussion of Activity 12.4, ammeter and voltmeter are considered as ideal meters. (2) Ideally, an ammeter should have zero resistance and a voltmeter should have infinite resistance. But in practice, these conditions cannot be realized and hence practically the voltmeter draws some current from the main branch and so the ammeter reading in part (2) is slightly less than that in part (1).

Activity 12.5 [T. B. Pg. 211]

To show that in a series combination of resistors, the total potential difference across the combination divides itself across the individual resistors.

Procedure:

• In Activity 12.4, insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in figure 12.7.
• Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V.
• Now measure the potential difference across the two terminals of the battery. Compare the two values.
• Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown
  
• Plug the key and measure the potential difference across the first resistor. Let it be V₁
• Similarly, measure the potential difference across the other two resistors, separately. Let these values be V₂ and V₃, respectively.
• Deduce a relationship between V, V₁, V₂ and V₃.

Observation:

• The p.d. across the series combination is equal to the p.d. across the two terminals of the battery.
• p.d. across the first resistor R₁ is found out to be V₁ using voltmeter.
• Similarly with the help of voltmeter, p.d. across resistors R₂ and R₃ separately are found out to be V₂ and V₃ respectively.
• It is found out that V = V₁ + V₂ + V₃.

Conclusion:

• In a series combination of different resistors, the potential difference across different resistors is different, i.e., If R₁ ≠ R₂ ≠ R₃, V₁ ≠ V₂ ≠ V₃.
• The total p.d. across the combination is equal to the sum of p.d. across each of the resistors, i.e., total p.d. across the combination divides itself across the individual resistors, i.e., V is equal to V₁ + V₂ + V₃.

Important note:
In Activity 12.5 it is assumed that, R (ammeter) = zero and R (voltmeter) = ∞, so V₁ + V₂ + V₃, In practice, these conditions cannot be realized, hence practically V is (very nearly) equal to V₁ + V₂ + V₃.

Activity 12.6 [T. B. Pg. 213]

To study the parallel combination of resistors.
OR
To study the relationship between potential difference and respective resistance in parallel circuit as well as the relationship between s current and respective resistance in parallel circuit.

Procedure:
1. Make a parallel combination XY of three resistors having resistances R₁, R₂ and R₃, respectively. Connect it with a battery, a plug key and an ammeter, as shown in figure 12.14. Also connect a voltmeter in parallel with the combination of resistors.

2. Plug the key and note the ammeter reading. (Let the current be I).

3. Also take the voltmeter reading. (It gives the potential difference V, across the combination).

4. Connect a voltmeter across the resistor with resistance R₁ and note the voltmeter reading V₁ (see figure 12.14).

5. Similarly connect a voltmeter across the resistors with resistances R₂ and R₃ separately and note s the corresponding voltmeter readings V₂ and V₃.

6. Take out the plug from the key. Remove the ammeter and voltmeter from the circuit.

7. Insert the ammeter in series with the resistor s with resistance R₁, as shown in figure 12.15. Note the ammeter reading, I₁.

8. Similarly, measure the current through the resistors with resistances R₂ and R₃. Let these be I₂ and I₃, respectively.
What is the relationship between I, I₁, I₂ and I₃?

Observations :

• Ammeter showed the reading I when connected as shown in figure 12.14.
• Voltmeter showed reading V when connected as shown in figure 12.14.
• It is found that the p.d. across each resistor is the same (i.e., V), as across the combination of resistors, i.e., V₁ = V₂ = V₃ = V
• Ammeter showed reading I₁, when connected with the resistor R₁ as shown in figure 12.15.
• Similarly ammeter showed reading I₂ and I₃ measured separately through R₂ and R₃ respectively.
• The relationship between I, I₁, I₂ and I₃ is:
  I = I₁ + I₂ + I₃
  Where, I = Total current through the combination of resistors in parallel
  I₁, I₂, I₃ = Currents through R₁, R₂ and R₃ respectively.

Conclusion:

• In parallel combination of resistors, the potential difference across different resistances will be same.
• Total current I in the given parallel circuit is divided amongst different resistors.

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 10 वृत्त Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 10 वृत्त

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
दी गई आकृति में, दो वृत्त परस्पर बिन्दु C पर स्पर्श करते हैं। सिद्ध कीजिए कि C पर खींची गई उभयनिष्ठ स्पर्श रेखा, P तथा Q पह खींची गई उभयनिष्ठ स्पर्श रेखा का समद्विभाजन करती हैं।

हल:
दिया है : दो वृत्त जिनके केन्द्र A तथा B हैं परस्पर बिन्दु C पर स्पर्श करते हैं।
सिद्ध करना है: C पर खींची गई स्पर्श रेखा, P तथा Q पर खींची गई स्पर्श रेखा का समद्विभाजन करती है।
उपपत्ति: हम जानते हैं कि किसी बाह्य बिन्दु से वृत्त पर खींची गई स्पर्श रेखाओं की लम्बाइयाँ बराबर होती हैं।
PR = RC …..(1)
(बिन्दु R से केन्द्र A वाले वृत्त पर खींची गई स्पर्श रेखाएँ)
तथा RQ = RC …..(2)
(बिन्दु R से केन्द्र B वाले वृत्त पर खींची गई स्पर्श रेखाएँ)
समी (1) व (2) से
PR = RQ
अत: C पर खींची गई स्पर्श रेखा, P तथा Q पर खींची गई स्पर्श रेखा का समद्विभाजन करती है।

प्रश्न 2.
5 सेमी त्रिज्या वाले एक वृत्त के बिन्दु P पर स्पर्श रेखा PQ केन्द्र 0 से जाने वाली एक रेखा के बिन्दु Q पर इस प्रकार मिलती है कि OQ = 13 सेमी. तो PQ की लम्बाई ज्ञात कीजिए।
हल:
∵ PQ वृत्त पर एक स्पर्श रेखा है। OP वृत्त की त्रिज्या है।

∴ PQ ⊥ OP अर्थात् ∠OPQ = 90°
समकोण त्रिभुज OPQ में, पाइथागोरस प्रमेय से
OQ² = OP² + PQ²
⇒ 13² = 5² + PQ²
⇒ PQ² = 13² – 5²
⇒ PQ² = (13 + 5) (13 – 5)
⇒ PQ² = 18 × 8
⇒ PQ = \(\sqrt{144}\)
⇒ PQ = 12 सेमी.
अतः PQ की लम्बाई = 12 सेमी.।

प्रश्न 3.
सिद्ध कीजिए कि दो संकेन्द्रीय वृत्तों में बड़े वृत्त की जीवा, जो कि छोटे को स्पर्श करती है, स्पर्श बिन्दु पर समद्विभाजित होती है।
हल:
दिया है : माना दो संकेन्द्रीय वृत्त जिनके केन्द्र O और त्रिज्या r और r’ हैं, r > r’
माना AB बड़े वृत्त की जीवा है, जो छोटे वृत्त को C पर स्पर्श करती है।

सिद्ध करना है : AC = CB.
रचना: ∵ OC को मिलाया।
उपपत्ति: OC छोटे वृत्त की त्रिज्या है।
और जीवा AB को बिन्दु C पर स्पर्श करती है।
∴ AB, बिन्दु C पर छोटे वृत्त की स्पर्श रेखा है।
∠OCB = 90° (प्रमेय 10.1 से )
अत: AB बड़े वृत्त की जीवा है और OC ⊥ AB,
AC = CB
[∵ वृत्त के केन्द्र से जीवा पर डाला गया लम्ब जीवा को समद्विभाजित करता है]

प्रश्न 4.
O केन्द्र वाले वृत्त के बाह्य बिन्दु P से वृत्त पर दो स्पर्श रेखाएँ PQ और PR खींची गई हैं। सिद्ध कीजिए कि QORP एक चक्रीय चतुर्भुज है।
हल:
हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।

∴ ∠PRO = 90° तथा ∠PQO = 90°
∠PRO + ∠PQO = 90° + 90°
= 180° …..(1)
चतुर्भुज QORP में
∠PRO + ∠ROQ + ∠PQO + ∠QPR = 360°
⇒ ∠PRO + ∠PQO + ∠ROQ + ∠QPR = 360°
⇒ 180° + ∠ROQ + ∠QPR = 360° [समी. (1) से]
⇒ ∠ROQ + ∠QPR = 360° – 180°
⇒ ∠ROQ + ∠OPR = 180°
अतः सम्मुख कोणों का योग 180° है। अतः QORP एक चक्रीय चतुर्भुज है।

प्रश्न 5.
आकृति में, O केन्द्र वाले वृत्त की PQ एक जीवा है तथा PT एक स्पर्श रेखा है। यदि ∠QPT = 60° है, तो ∠PRQ ज्ञात कीजिए।

हल:
चित्र से,
OP ⊥ PT अर्थात् OPT = 90°
∠OPQ = ∠OPT – ∠OPT
∠OPQ = 90 – 60
= 30°
ΔOPQ मैं, (वृत्त की त्रिज्या बराबर होती है।)
∠OQP = ∠OPQ = 30°
(बराबर भुजाओं के सम्मुख कोण बराबर होते हैं)
अब,
∠QP + ∠OPQ + ∠POQ = 180°
कोण का योग = 30° + 30° + ∠POQ = 180°
∠POQ = 180° – 60° = 120°
∠POQ = 360° – 120° = 240°
हम जानते हैं, वृत्त के केन्द्र पर बना कोण वृत्त की परिधि पर बने कोण का दुगना होता है।
∠POQ = 2∠PRQ
⇒ 240° = 2∠PRQ
⇒ ∠PRQ = \(\frac{240^{\circ}}{2}\) = 120°
अतः कोण PRQ की माप 120° है।

प्रश्न 6.
सिद्ध करो कि वृत्त की किसी जीवा के सिरों पर खींची गई स्पर्श रेखाएँ, जीवा से समान कोण बनाती हैं।
हल:
माना वृत्त C(O, r) की जीवा AB के सिरे A और B पर स्पर्श रेखाएँ PA और PB खींची गई हैं जो कि बिन्दु P पर काटती हैं।
माना OP, जीवा AB को C बिन्दु पर काटती है।

सिद्ध करना है : ∠PAC = ∠PBC
उपपत्ति : ΔPCA और ΔPCB में,
PA = PB (बाह्य बिन्दु P से वृत्त पर स्पर्श रेखाएँ हैं)
PC = PC (उभयनिष्ठ भुजा)
∠APC = ∠BPC [∵ स्पर्श रेखाएँ PA व PB, OP के साथ समान कोण बनाती हैं]
S-A-S सर्वांगसमता से,
ΔPCA ≅ ΔPCB
⇒ ∠PAC = ∠PBC.

प्रश्न 7.
आकृति में, 3 सेमी. त्रिज्या वाले एक वृत्त के परिगत एक त्रिभुज ABC इस प्रकार खींचा गया है कि रेखाखण्ड BD तथा DC की लम्बाइयाँ क्रमशः 6 सेमी तथा 9 सेमी है। यदि ΔABC का क्षेत्रफल 54 वर्ग सेमी है, तो भुजाओं AB तथा AC की लम्बाइयाँ ज्ञात कीजिए।

हल:
3 सेमी त्रिज्या वाले वृत्त के परिगत एक ΔABC खींचा गया है। त्रिभुज की भुजाएँ BC, AB, AC वृत्त को क्रमश: D, E, F बिन्दुओं पर स्पर्श करती है।
∵ किसी बाह्य बिन्दु से, वृत्त पर खींची गई स्पर्श रेखाओं की लम्बाई समान होती हैं।
AE = AF = x (माना)
BD = BE = 6 सेमी
CD = CF = 9 सेमी
OF, OE, OA, OB तथा OC को मिलाया।
हम जानते हैं कि वृत्त की स्पर्श रेखा बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∴ OD ⊥ BC, OE ⊥ AB, OF ⊥ AC
ΔABC में, b = AB = (x + 6)
a = BC = (6 + 9) = 15 सेमी
c = AC = (x + 9) सेमी

ΔOCB का क्षेत्रफल = \(\frac{1}{2}\) × आधार × शीर्षलम्ब
= \(\frac{1}{2}\) × 15 × 3 = \(\frac{45}{2}\) सेमी²
ΔCOA का क्षेत्रफल = \(\frac{1}{2}\) × आधार × शीर्षलम्ब
= \(\frac{1}{2}\) × (x + 9) × 3
= \(\frac{3 x+27}{2}\) सेमी²
ΔAOB का क्षे. = \(\frac{1}{2}\) × आधार × शीर्षलम्ब
= \(\frac{1}{2}\) × (6 + x)³ = \(\frac{18+3 x}{2}\) सेमी²
ΔABC का क्षे. = ΔOCB का क्षे. + ΔCOA का क्षे + ΔAOB का क्षे.
\(\sqrt{54 x^2+810 x}=\frac{45}{2}+\frac{3 x+27}{2}+\frac{18+3 x}{2}\)
= \(2 \sqrt{54 x^2+810 x}\) = 45 + 3x + 27 + 18 + 3x
वर्ग करने पर,
4(54x² + 810x) = (6x + 90)²
54 × 4(x² + 15x) = 36(x + 15)²
6x(x + 15) = (x + 15)²
6x = x + 15
5x = 15
x = 3
AE = AF = 3 सेमी
∴ AB = AE + EB = 3 + 6 = 9 सेमी
AC = AF + FC = 3 + 9 = 12 सेमी

प्रश्न 8.
दी गई आकृति में, त्रिभुज ABC की भुजाएँ AB, BC तथा CA, केन्द्र O तथा त्रिज्या r वाले वृत्त को क्रमश: P, Q तथा R पर स्पर्श करती हैं।

सिद्ध कीजिए :
(i) AB + CQ = AC + BQ
(ii) क्षेत्रफल (ΔABC) = \(\frac{1}{2}\)(ΔABC का परिमाप) × r
हल:
(i) चूँकि हम जानते हैं कि बाह्य वृत्त से खींची गई स्पर्श रेखाएँ बराबर होती हैं।
अतः AP = AR …..(1)
(बिन्दु A से वृत्त पर खींची गई स्पर्श रेखाएँ)
BP = BQ …..(2)
(बिन्दु B से वृत्त पर खींची गई स्पर्श रेखाएँ)
तथा CQ = CR …..(3)
(बिन्दु C से वृत्त पर खींची गई स्पर्श रेखाएँ)
समी. (1), (2) तथा (3) को जोड़ने पर
AP + BP + CQ = AR + BQ + CR
⇒ (AP + BP) + CQ = (AR + CR) + BQ
⇒ AB + CQ = AC + BQ.

(ii) OR, OP, OA, OB तथा OC को मिलाया।
चूँकि हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।

अत: OP ⊥ AB, OQ ⊥ BC तथा OR ⊥ AC
अब (ΔABC) का क्षेत्रफल = त्रिभुज BOC का क्षेत्रफल + त्रिभुज AOC का क्षेत्रफल + त्रिभुज AOB का क्षेत्रफल
= \(\frac{1}{2}\)BC × OQ + \(\frac{1}{2}\) AC × OR + \(\frac{1}{2}\)AB × OP
(∵ त्रिभुज का क्षेत्रफल = \(\frac{1}{2}\)आधार × ऊँचाई)
= \(\frac{1}{2}\)(BC × r + AC × r + AB × r)
(∵ OQ, OR तथा OP वृत्त की त्रिज्याएँ हैं)
= \(\frac{1}{2}\)(BC + AC + AB) × r
= \(\frac{1}{2}\)(ΔABC का परिमाप) × r
अत: ΔABC का क्षेत्रफल = \(\frac{1}{2}\)(ΔABC का परिमाप) × r

प्रश्न 9.
यदि एक बिन्दु T से O केन्द्र वाले किसी वृत्त पर TA व TB स्पर्श रेखाएँ परस्पर 70° के कोण पर झुकी हों तो ∠AOB को ज्ञात कीजिए।
हल:
बिन्दु T से TA व TB वृत्त की दो स्पर्श रेखाएँ है। OA तथा OB वृत्त की त्रिज्याएँ है।

अत: AT ⊥ OA तथा BT ⊥ OB (प्रमेय 10.1 से)
∴ ∠OAT = 90°
तथा ∠OBT = 90°
∠AOB + ∠ATB = 180°
∠AOB + 70° = 180°
∠AOB = 180° – 70° = 110°

प्रश्न 10.
निम्न आकृति में XP तथा XQ, केन्द्र O वृत्त पर बिन्दु X से खींची गई स्पर्श रेखाएँ हैं तथा AB वृत्त के बिंदु R पर स्पर्श रेखा है।
सिद्ध कीजिए : XA + AR = XB + BR

हल:
∵ वृत्त के किसी बाह्य बिन्दु से खींची गई स्पर्श रेखाएँ बराबर होती है।
∴ XP = XQ
⇒ XA + AP = XB + BQ …..(i)
बिन्दु B से खींची गई स्पर्श रेखाएँ BQ तथा BR है।
∴ BQ = BR …..(ii)
इसी प्रकार AP = AR ……(iii)
समीकरण (i), (ii) व (iii) से
XA + AR = XB + BR

प्रश्न 11.
एक त्रिभुज ABC के अन्तर्गत एक वृत्त इस प्रकार खींचा गया है कि यह भुजाओं AB, BC तथा AC को क्रमश: P, Q तथा R पर स्पर्श करता है, यदि AB = 10 सेमी, AR = 7 सेमी तथा CR = 5 सेमी है, तो BC की लम्बाई ज्ञात कीजिए।
हल:
दिया है, AB = 10 सेमी
AR = 7 सेमी
तथा CR = 5 सेमी

∵ बिन्दु A से दो स्पर्श रेखाएँ खींची गई हैं।
∴ AP = AR = 7 सेमी
∵ AB = AP + PB
⇒ 10 = 7 + PB
⇒ PB = (10 – 7) सेमी = 3 सेमी
बिन्दु B से खींची गई स्पर्श रेखाएँ BP व BQ बराबर हैं।
∴ BQ = BP = 3 सेमी
तथा बिन्दु C से खींची गई स्पर्श रेखाएँ CQ व CR बराबर हैं।
CQ = CR = 5 सेमी
अतः BC = BQ + QC
= 3 सेमी + 5 सेमी
= 8 सेमी

प्रश्न 12.
निम्न आकृति में O केंन्द्र वाले वृत्त का व्यास AB है तथा AC इसकी एक जीवा है। ∠BAC = 30° है। यदि बिंदु C पर खींची गई स्पर्श रेखा, बढ़ाए गए व्यास AB को बिन्दु D पर प्रतिच्छेद करती है, तो दर्शाइए कि BC = BD।

हल:
अर्धवृत्त में स्थित कोण समकोण होता है,
∠ACB = 90°
ΔABC में,
∠ABC + ∠ACB + ∠OAB = 180°
⇒ ∠ABC + 90° + 30° = 180°
⇒ ∠ABC = 180° – 120° = 60°
∠ABC + CBD = 180° (रैखिक युग्म)
⇒ 60° + ∠CBD = 180°
⇒ ∠CBD = 180° – 60°- 120°
ΔAOC में
OA = OC
⇒ ∠ACO = ∠OAC = 30°
अब ∠ACB = ∠ACO + ∠OCB
⇒ 90° = 30° + ∠OCB
⇒ ∠OCB = 90° – 30° = 60°
∵ स्पर्श रेखा, त्रिज्या पर लम्ब होती है।
∴ ∠ACD = 90°
⇒ ∠OCB + ∠BCD = 90°
⇒ 60° + ∠BCD = 90°
⇒ ∠BCD = 90° – 60° = 30°
ΔCBD में,
∠BCD + ∠CBD + ∠BDC = 180°
⇒ 30° + 120° + BDC = 180°
⇒ ∠BDC = 180° – 150° = 30°
∵ ∠BCD = ∠BDC = 30°
⇒ BD = BC

वस्तुनिष्ठ प्रश्न :

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क).

1. दो वृत्त एक-दूसरे को ………………. बिन्दु पर प्रतिच्छेद करते हैं।
2. वृत्त के बाहर स्थित बिन्दु से वृत्त पर ………………. स्पर्श रेखाएँ खींची जा सकती है।
3. वह रेखा जो वृत्त को दो बिन्दुओं पर काटती है ………………. कहलाती है।
4. त्रिभुज का अन्तः वृत्त ………………. का प्रतिच्छेदक बिन्दु होता है।
5. वृत्त के बाहरी बिन्दु P से वृत्त पर स्पर्श रेखा, केन्द्र O से हमेशा OP से ………………… होती है।

उत्तर:

1. दो,
2. 2,
3. छेदक रेखा,
4. त्रिभुज के कोणों का समद्विभाजक,
5. कम ।

निम्न में सत्य / असत्य ज्ञात कीजिए :

प्रश्न (ख).

1. वृत्त की दो स्पर्श रेखाओं की लम्बाई समान होती है।
2. किसी वृत्त पर खींची गई छेदक रेखा वृत्त को दो बिन्दुओं पर प्रतिच्छेद करती है।
3. समद्विबाहु त्रिभुज ABC के परिगत वृत्त पर बिन्दु से स्पर्श रेखाएँ इस प्रकार हैं कि AB = AC जो BC के समान्तर है।
4. वृत्त के बाहर स्थित किसी बिन्दु से वृत्त पर अनेक स्पर्श रेखाएँ खींची जा सकती हैं।
5. स्पर्श रेखा और वृत्त के उभयनिष्ठ बिन्दु को स्पर्श बिन्दु कहते हैं।

उत्तर:

1. सत्य,
2. सत्य,
3. सत्य,
4. असत्य,
5. सत्य ।

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
निम्न आकृति में, यदि TP, TQ केन्द्र O वाले किसी वृत्त पर दो स्पर्श रेखाएँ इस प्रकार हैं कि ∠POQ = 115° है, तो ∠PTQ बराबर है:

(A) 115°
(B) 57.5°
(C) 55°
(D) 65°
हल:
∵ ∠PTO और ∠POQ सम्पूरक कोण हैं।
∠PTQ + ∠POQ = 180°
⇒ ∠PTQ + 115° = 180°
⇒ ∠PTQ = 180° – 115° = 650
अत: सही विकल्प (D) है।

प्रश्न 2.
निम्न आकृति में, O केन्द्र वाले वृत्त पर बिन्दु B पर स्पर्श रेखा PQ खींची गई है। यदि ∠AOB = 100° है, तो ∠ABP बराबर है :

(A) 50°
(B) 40°
(C) 60°
(D) 80°
हल:
दिया है,
∠AOB = 100°
∵ OA = OB
⇒ ∠OBA = ∠OAB = \(\frac{180^{\circ}-100^{\circ}}{2}\)
⇒ ∠OBA = ∠OAB = 40°
∵ स्पर्श रेखा त्रिज्या पर लम्ब होती है।
∴ ∠OBP = 90°
∴ ∠ABP + ∠ABO = 90°
⇒ ∠ABP + 40° = 90°
⇒ ∠ABP = 90° – 40° = 50°
अत: सही विकल्प (A) है।

प्रश्न 3.
निम्न आकृति में, O केन्द्र वाले वृत्त पर बाहय बिंदु P से दो स्पर्श रेखाएँ PQ तथा PR खींची गई हैं। वृत्त की त्रिज्या 4 सेमी है। यदि ∠QPR = 90° है, तो PQ की लम्बाई होगी:

(A) 3 सेमी
(B) 4 सेमी
(C) 2 सेमी
(D) 2\(\sqrt{2}\) सेमी
हल:
∵ त्रिज्या, स्पर्श रेखा पर लम्ब होती है,
∠OQP = ∠ORP = 90°
दिया है, ∠QPR = 90°
चतुर्भुज PQOR में,
∠PQO+ ∠QOR + ∠ORP + ∠RPQ = 360°
⇒ 90° + ∠QOR + 90° + 90° = 360°
⇒ ∠QOR = 360°- 270° = 90°
∴ PR = PQ
⇒ ∠POQ = ∠POR = \(\frac{90^{\circ}}{2}\) = 45°
समकोण ΔOQP में,
tan 45° = \(\frac{P Q}{O Q}\)
⇒ 1 = \(\frac{P Q}{4}\)
⇒ PQ = 4 सेमी
अत: सही विकल्प (B) है।

प्रश्न 4.
निम्न चित्र में 7 सेमी त्रिज्या के एक वृत्त के बाहय बिंदु P से स्पर्श रेखा PT खींची गई है कि PT = 24 सेमी है। यदि O वृत्त का केन्द्र है, तो PR की लंबाई है:

(A) 30 सेमी
(B) 28 सेमी
(C) 32 सेमी
(D) 25 सेमी
हल:
समकोण ΔPTO में,
OP² = OT² + PT²
⇒ OP² = (7)² + (24)²
⇒ Op² = 49 + 576 = 625
⇒ OP = 25 सेमी
∴ PR = PO + OR
= (25 + 7) सेमी
= 32 सेमी
अतः सही विकल्प (C) है।

प्रश्न 5.
निम्न आकृति में, O वृत्त का केन्द्र है। PQ एक जीवा है तथा PT, P पर एक स्पर्श रेखा है, जो PQ के साथ 50° का कोण बनाती है। ∠POQ का मान है:

(A) 130°
(B) 90°
(C) 100°
(D) 75°
हल:
∵ त्रिज्या स्पर्श रेखा पर लम्ब होती है।
∴ ∠OPT = 90°
∴ ∠OPQ + ∠QPT = 90°
⇒ ∠OPQ = 90° – 50° = 40°
∵ OP = OQ
⇒ ∠OQP = ∠OPQ = 40°
ΔPOQ में,
∠OQP + ∠OPQ + ∠POQ = 180°
⇒ 40° + 40° + ∠POQ = 180°
⇒ ∠POQ = 180° – 80° = 100°
अत: सही विकल्प (C) है।

प्रश्न 6.
एक वृत्त के केन्द्र से 13 सेमी दूरी पर स्थित एक बिन्दु Q से वृत्त पर खींची गई स्पर्श रेखा PQ की लम्बाई 12 सेमी है। वृत्त की त्रिज्या (सेमी. में) है:
(A) 25
(B) \(\sqrt{313}\)
(C) 5
(D) 1
हल:
बाह्य बिन्दु Q से PQ वृत्त पर खींची गई स्पर्श रेखा है तथा OP वृत्त की त्रिज्या है।

अतः ∠OPQ = 90° (प्रमेय 10.1 से)
समकोण त्रिभुज OPQ में,
OQ² = PQ² + OP² (पाइथागोरस प्रमेय से)
⇒ 13² = 12² + OP²
⇒ OP² = 13² – 12²
⇒ OP² = (13 + 12) (13 – 12)
⇒ OP² = 25
⇒ OP = \(\sqrt{25}\) = 5 सेमी.
अत: विकल्प (C) सही है।

प्रश्न 7.
दी गई आकृति में AP, AQ तथा BC वृत्त की स्पर्श रेखाएँ हैं। यदि AB = 5 सेमी., AC = 6 सेमी. तथा BC = 4 सेमी है, तो AP की लम्बाई (सेमी. में) है:

(A) 7.5
(B) 15
(C) 10
(D) 9
हल चूँकि हम जानते हैं कि बाह्य बिन्दु से वृत्त पर खींची गई स्पर्श रेखाओं की लम्बाइयाँ बराबर होती हैं।
अत: BP = BD …..(1)
(बिन्दु B से वृत्त पर स्पर्श रेखाएँ)
CQ = CD …..(2)
(बिन्दु C से वृत्त पर स्पर्श रेखाएँ)
AP = AQ …..(3)
(बिन्दु A से वृत्त पर स्पर्श रेखाएँ) अब
AP = AB + BP ⇒ AP = 5 + BD ….. (4)
AQ = AC + CQ ⇒ AQ = 6 + CD ….. (5)
समी (4) तथा (5) को जोड़ने पर,
AP + AQ = 5 + BD + 6 + CD
⇒ AP + AP = 11 + BD + CD
[समी. (3) का प्रयोग करने पर]
⇒ 2AP = 11 + BC
⇒ 2AP = 11 + 4
⇒ 2AP = 15 ⇒ AP = 7.5 सेमी.
अत: विकल्प (A) सही है।

प्रश्न 8.
यदि एक बाह्य बिन्दु P से एक O केन्द्र वाले वृत्त पर दो स्पर्श रेखाएँ PA और PB इस प्रकार खीची गई कि दोनों 80° के कोण पर झुकी है, तो ∠POA बराबर हैं:
(A) 50°
(B) 60°
(C) 70°
(D) 80°
हल:
बिन्दु P से PA तथा PB वृत्त पर स्पर्श रेखाएँ हैं तथा OA व OB वृत्त की त्रिज्याएँ हैं।
अत: AP ⊥ OA तथा PB ⊥ OB (प्रमेय 10.1 से)

∠OAP = 90°
तथा ∠OBP = 90°
अब ∠AOB + ∠APB = 180°
⇒ ∠AOB + 70° = 180°
⇒ ∠AOB = 180° – 80°
∴ ∠AOB = 100°
∵ OP रेखा, ∠AOB का समद्विभाजक है।
∠POA = \(\frac{\angle A O B}{2}=\frac{100^{\circ}}{2}\) = 50°
∴ ∠POA = 50°
अत: सही विकल्प (A) है।

प्रश्न 9.
दी गई आकृति में, एक चतुर्भुज ABCD के अन्तर्गत खींचा गया वृत्त, इसकी भुजाओं AB, BC, CD तथा AD को क्रमश: P, Q, R तथा S पर स्पर्श करता है। यदि वृत्त की त्रिज्या 10 सेमी., BC = 38 सेमी. PB = 27 सेमी. तथा AD ⊥ CD है, तो CD की लम्बाई है:

(A) 11 सेमी
(B) 20 सेमी
(C) 21 सेमी
(D) 15 सेमी
हल:
बिन्दु D से DR तथा DS वृत्त पर स्पर्श रेखाएँ हैं तथा OS व OR वृत्त की त्रिज्याएँ है।

AD⊥ OS तथा DR ⊥ OR (प्रमेय 10.1 से)
AD ⊥ CD (दिया है)
चतुर्भुज DROS में,
∠D + ∠R + ∠O + ∠S = 360°
⇒ 90° + 90° + ∠O + 90° = 360°
⇒ ∠O = 360° – 270° = 90°
इस प्रकार चतुर्भुज DROS में,
∠D = ∠R = ∠O = ∠S = 90°
तथा OS = OR [एक ही वृत्त की त्रिज्याएँ]
अत: DROS एक वर्ग होगा।
अतः SD = DR = 10 सेमी
(बिन्दु D से वृत्त पर स्पर्श रेखाएँ)
∵ बिन्दु B से BP व BQ वृत्त पर स्पर्श रेखाएँ हैं।
∴ BP = BQ = 27 सेमी (प्रमेय 10.2 से)
CQ = BC – BQ
CQ = 38 – 27 = 11 सेमी
∵ बिन्दु C से CR व CQ वृत्त पर स्पर्श रेखाएँ हैं।
∴ CR = CQ (प्रमेय 10.2 से)
⇒ CR = 11 सेमी
CD = CR + DR
⇒ CD = 11 + 19
⇒ CD = 21 सेमी
अत: विकल्प (C) सही है।

प्रश्न 10.
किसी 5 सेमी. त्रिज्या वाले वृत्त के एक व्यास AB के पर स्पर्श रेखा XAY खींची गई है। XY के समान्तर 4 से 8 सेमी की दूरी पर, जीवा CD की लम्बाई है:
(A) 4 सेमी
(B) 5 सेमी
(C) 6 सेमी
(D) 8 सेमी
हल:
चूँकि X-AY वृत्त पर स्पर्श रेखा है तथा OA व OC वृत्त की त्रिज्याएँ हैं।

अत: XY ⊥ OA अर्थात् ∠XAO = 90° (प्रमेश 10.1 से)
∵ XY || CD (दिया गया है)
∴ ∠XAO + ∠OEC = 180°
⇒ 90° + ∠OEC = 180°
⇒ ∠OEC = 180° – 90° = 90°
AE = 8 सेमी
तथा AO = 5 सेमी (वृत्त की त्रिज्या)
∴ OE = AE – AO
= 8 – 5 = 3 सेमी
समकोण त्रिभुज OEC में,
OC² = OE² + CE² (पाइथागोरस प्रमेय)
⇒ 5² = 3² + CE²
⇒ CE² = 5² – 3²
= 25 – 9 = 16
⇒ CE = \(\sqrt{16}\) = 4 सेमी
OE ⊥ CD
CE = ED
CD = 2 × CE = 2 × 4 = 8 सेमी
अतः विकल्प (D) सही है।

प्रश्न 11.
यदि 60° पर झुकी दो स्पर्श रेखाएँ 3 सेमी त्रिज्या वाले एक वृत्त पर खींची जाती हैं, तो प्रत्येक स्पर्श रेखा की लम्बाई है:
(A) 3 सेमी
(B) \(\frac{3}{2} \sqrt{3}\) सेमी
(C) 3\(\sqrt{3}\) सेमी
(D) 6 सेमी
हल:
माना बिन्दु P से PQ तथा PR वृत्त पर स्पर्श रेखाएँ हैं। OQ तथा OR वृत्त की त्रिज्याएँ हैं।

अतः PQ ⊥ OQ तथा PR ⊥ OR
अतः समकोण ΔPOQ तथा ΔPOR में
∠OQP = ∠ORP (प्रत्येक 90° है)
कर्ण PO = कर्ण PO (उभयनिष्ठ भुजा)
तथा OQ = OR (वृत्त की समान त्रिज्याएँ)
∴ ΔPOQ ≅ ΔPOR (समकोण कर्ण भुजा सर्वांगसमता गुणधर्म से)
⇒ ∠QPO = ∠RPO (CPCT)
⇒ ∠QPO = ∠RPO
= \(\frac{60^{\circ}}{2}\) = 30°
अब समकोण ΔOQP में
tan 30° = \(\frac{O Q}{P Q}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{3}{P Q}\)
⇒ PQ = 3\(\sqrt{3}\)
चूँकि बिन्दु P से PQ तथा PR वृत्त पर खींची गई स्पर्श रेखाएँ हैं और हम जानते हैं कि वृत्त पर बाह्य बिन्दु से खींची गई स्पर्श रेखाओं की लम्बाइयाँ बराबर होती है।
अत: PR = PQ
= 3\(\sqrt{3}\) सेमी.
अतः विकल्प (C) सही है।

Jharkhand Board JAC Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Additional Questions and Answers

Question 1.
Give four points of difference between the following terms / quantities :
(1) Near-sightedness and Far-sightedness
Answer:

Near-sightedness	Far- sightedness
1. The eye lens does not become thin as required but remains thick.	1. The eye lens does not become thick as required but remains thin.
2. The light rays from objects at far distances are focused short of the retina. As a result, the distant objects cannot be seen clearly.	2. The light rays from objects nearby the eyes are focused behind the retina. As a result the nearby objects cannot be seen clearly.
3. The light rays from objects nearby eyes are focused on the retina. As a result, the nearby objects are seen clearly.	3. The light rays from distant objects are focused on the retina. As a result the distant objects are seen clearly.
4. This defect can be corrected by using concave lens of appropriate focal length.	4. This defect can be corrected by using convex lens of appropriate focal length.

(2) Near point and Far point
Answer:

Near point	Far point
1. The minimum distance at which the object can be seen clearly without contraction of eye lens is called the near point of an eye.	1. The farthest distance upto which the eye can see objects clearly is called far point of an eye.
2. For young adult with normal vision, this value is 25 cm.	2. For young adult with normal vision, this value is infinite.
3. For a person having defect of far-sightedness value of the near point is at a distance more than 25 cm.	3. For a person having defect of near-sightedness value of the far point is at a distance less than infinite distance.
4. For a person having defect of far-sightedness value of the near point is obtained at a distance of 25 cm using convex lens of appropriate focal length.	4. For a person having defect of near-sightedness value of the far point is obtained at distance of infinite distance using concave lens of appropriate focal length.

Question 2.
Give scientific reasons for the following statements:
(1) To rectify the defect of near-sightedness or myopia, concave lens of suitable focal length is used as corrective lens.
Answer:
The eye lens of a person having defect of near-sightedness or myopia, does not become thin as per the requirement and so the rays after being refracted by the eye lens get focussed at a position short of retina. So distant objects cannot be seen clearly.

If a person having defect of near-sightedness, uses concave lens of suitable focal length, then light rays become slightly divergent. Hence, the image can be formed on retina.

Thus, if the image is formed on retina, then the distant object may be seen clearly.

(2) To rectify the defect of far-sightedness or hypermetropia, convex lens of suitable focal length is used as corrective lens.
Answer:
The eye lens of a person having defect of far-sightedness as hypermetropia, does not become thick as per the requirement and so the rays after being refracted by the eye lens get focussed behind retina. So nearby objects cannot be seen clearly.

If a person having defect of far-sightedness, uses convex lens of suitable focal length, then light rays become slightly convergent. Hence, the image can be formed on retina.

Thus, if the image is formed on retina, then the nearby object may be seen clearly.

(3) A rainbow is visible in the sky only after rain shower.
Answer:
In rainy season there are many clouds in the sky having tiny water droplets. When the sunlight is incident on tiny water droplets, due to these tiny water droplets, refraction, dispersion, internal reflection and at the end again refraction of sunlight take place.

Due to which a band is created which contains seven colours in the sky which is known as rainbow.

In other seasons there are no clouds in the sky. Therefore there are no tiny water droplets s in the sky. Hence, no rainbow is formed in the sky in other seasons and so it is not visible in? the sky.

(4) The sunrise is experienced two minutes early and the sunset is experienced two minutes delayed.
Answer:
As the altitude (height) from the earth’s surface gradually increases, the earth’s atmosphere becomes rarer. Hence, the refractive index of the air decreases continuously.

So a ray of light coming from the Sun towards an observer continuously passes from an optically rarer to optically denser medium and bends towards the normal. Thus, its direction of propagation changes continuously.

The actual sunrise or sunset begins when the Sun reaches the horizon.
1. In figure 11.12, S₁ is the actual position of the Sun a little below the horizon.

2. In this case, the light rays coming from S₁, continuously get refracted in the earth’s atmosphere (atmospheric refraction) and reach the observer as shown in the figure.

3. The tangent drawn to the curved path of the ray at point P passes through S₂, above the horizon.

4. S₂ is the apparent position of the Sun.

5. Thus, during the sunrise, the Sun is seen even though it is little below the horizon. Similarly, during the sunset it is seen for sometime even though it is little below the horizon.

6. Taking the refractive index of air as 1.00029 the apparent angular shift in the position of the Sun is found to be approximately (\(\frac { 1 }{ 2 }\))°.
Now, for angular displacement of 180° of the Sun, time required is 12 hours. Hence for angular displacement of (\(\frac { 1 }{ 2 }\))° of the Sun, time required is
= \(\frac{\left(\frac{1}{2}\right)^{\circ} \times 12 \text { hours }}{180^{\circ}}\)
= 0.03333 hour
= 1.9998 minutes
≈ 2 minutes
Due to advance sunrise and delayed sunset the duration of a day increases by about 4 minutes.

(5) The clear sky appears blue in colour.
Answer:

The molecules of air and other fine particles in the atmosphere are smaller in size than the wavelength of light in the visible region.

• The wavelength of red light is about 1.8 times that of blue light.
• When the sunlight passes through the atmosphere, the fine particles in the air scatter blue light more strongly than red light.
• At this time if the observer looks upwards, he finds sky blue.
• If the earth had no atmosphere, there would not have been any scattering. Then the sky would have looked dark.

Important Note:
The sky appears dark to passengers flying at very high altitudes, as scattering of light is not prominent at such heights.

(6) The danger signal lights are red in colours.
Answer:
The red light gets scattered least by fog or smoke because of its longer wavelength relative to light of any other colour. So, it can be seen even from a long distance. Therefore it is used in signals showing danger.

(7) The sun appears reddish at sunrise and sunset.
Answer:

In figure 11.14, the situation at the sunrise is shown.
Here, the white light coming from the Sun near the horizon, passes through thick layers of air and covers larger distance in the earth’s atmosphere before reaching the observer. During this, more scattering of blue light and shorter wavelengths take place. Hence, the reddish light reaches the observer and the Sun appears reddish. The same thing occurs at the sunset.
[Note : Rising and setting of the full moon from the horizon appears reddish due to this reason.]

Objective Questions and Answers

Question 1.
Answer the following questions in one word / sentence :

Question 1.
What is dispersion of white light?
Answer:
The phenomenon of a splitting of white light into its constituent colours is called the dispersion of white light.

Question 2.
What happens to the image-distance in the normal eye, when we increase the distance of an object from the eye?
Answer:
The image distance always remains constant.

Question 3.
What can be said about the focal length of the eye lens if its curvature increases?
Answer:
decreases

Question 4.
What can be said about the curvature of the eye lens if it becomes thin?
Answer:
decreases

Question 5.
For normal eye vision what is the object-distance and image-distance when the object is placed at a near point? (Take the distance between the eye lens and the retina as 2.3cm.)
Answer:
u = – 25 cm, v = + 2.3 cm

Question 6.
For normal eye vision what is the object-distance and image-distance when the object? is placed at a far point? (Take the distance between the eye lens and the retina as 2.3 cm.)
Answer:
u = – 00, a = + 2.3 cm

Question 7.
State the type of image of an object formed s on the retina.
Answer:
real, inverted and diminished

Question 8.
Write the name of the most front part of human eye.
Answer:
cornea

Question 9.
State the function of the iris.
Answer:
Iris can control the amount of light entering into the eye as well as size of pupil can be controlled by it.

Question 10.
State the function of light sensitive cells present in retina.
Answer:
Light rays falling on the retina are converted into the electrical signals.

Question 11.
Write the function of optic nerves.
Answer:
The work of optic nerves is to sent electrical signals to the brain.

Question 12.
Write use of bifocal lens.
Answer:
To rectify/remove the eye-defect known as presbyopia.

Question 13.
How much duration in second increases per day due to early sunrise and delayed sunset?
Answer:
240s

Question 14.
Due to which effect does the smoke emitted by the combustion of the engine oil in a motorcycle sometimes appears blue in colour?
Answer:
The Tyndall effect

Question 15.
Which effect is developed commercially to determine the size and density of aerosol and other colloidal particles?
Answer:
The Tyndall effect

Question 16.
Wavelength of red colour is approximately how many times the wavelength of violet colour?
Answer:
1.8

Question 2.
Fill in the blanks :

1. The type of image formed by the eye lens is ………………. and ……………….
2. A triangular glass prism has ………………. triangular bases and ………………. rectangular surfaces.
3. Light enters our eye through the ……………….
4. A person suffering from far-sightedness or hypermetropia cannot see clearly ………………. objects.
5. A ………………. corrective lens is used to rectify near-sightedness.
6. An old person suffering from near-sightedness and a far-sightedness uses ………………. to rectify his vision.
7. While passing through the prism the light ray travelling from air to glass bends towards the ……………….
8. In a glass prism ………………. light propagates with maximum speed.
9. At night stars are seen slightly at a higher position than their actual position because of the ……………….
10. For a light ray passing through the prism, the angle between the incident ray and the emergent ray is known as the ……………….
11. At the time of sunrise the sun appears ………………. in colour.
12. The fine particles in air scatter ………………. light more strongly.
13. ………………. light, while passing through a prism, does not disperse.
14. Stars behave like ………………. sources of light and planets behave like ………………. sources of light.
15. Danger signals are red in colour because red light is ……………….
16. In the spectrum of white light, ………………. and ………………. colours seen at the two ends.
17. The diameter of the human eyeball is approximately ………………. cm.
18. The distance between the eye lens and the retina is known as the ……………….
19. In the normal situation in the relaxed position of the ciliary muscles, the eye lens is ……………….
20. At night as we move up in the atmosphere of the earth, the refractive index ………………. continuously.

Answer:

1. real, inverted
2. two, three
3. cornea
4. nearby
5. concave / diverging
6. spectacles with bi-focal lenses
7. normal
8. red
9. atmospheric refraction
10. angle of deviation
11. reddish
12. blue
13. Monochromatic (It means light of single frequency, i.e., of single colour.
14. point, extended
15. scattered the least
16. violet, red
17. 2.3
18. size of the eyeball
19. thin
20. decreases

Question 3.
State whether the following statements are true or false:

1. The near point of every person is 25 cm.
2. The splitting of white light into its constituent colours is called the scattering of light.
3. Far-sightedness can be rectified by using a concave lens of suitable power.
4. In the eye of myopic person, the image of a distant object is formed behind the retina.
5. Near-sightedness arises due to more curvature of the cornea or due to the eye lens remaining thick permanently.
6. The speed of light decreases as it passes from an optically denser medium to an optically rarer medium.
7. A myopic person has the far point nearer than infinity.
8. A hypermetropic person has near point farther away from the normal near point (25 cm).
9. The construction of the human eye can be compared with that of a camera.
10. A rainbow is formed due to refraction taking place twice, one internal reflection and dispersion of sunlight by water droplets in the sky.
11. Planets twinkle.
12. When the sunlight passes through a canopy of dense forest, tiny water droplets in the mist, scatter the light. This effect is known as the Tyndall effect.

Answer:

1. False
2. False
3. False
4. False
5. True
6. False
7. True
8. True
9. True
10. True
11. False
12. True

Question 4.
Match the following:
(1)

Column I	Column II	Column III
1. Myopia	p. The focal length of the eye lens increase	a. Bifocal lens
2. Hypermetropia	q. The focal length of the eye lens decrease	b. Concave lens
3. Presbyopia	r. The power of accommodation of the eye decrease with ageing	c. Convex lens

Answer:
(1 – q – b), (2 – p – c), (3 – r – a).

(2)

Column I	Column II
1. TWinkling of stars	p. Tiny water droplets present (or suspended) in the atmosphere
2. Blue coloured sky	q. Band of colours
3. Rainbow	r. Scattering of light
4. Spectrum	s. Uneven atmosphere

Answer:
(1 – s), (2 – r), (3 – p), (4 – q).

(3)

Column I	Column II
1. Human eye or eyeball	a. It controls and regulates the amount of light entering the eye.
2. Self operated accommodation power of eye	b. Delicate membrane having large number of light sensitive cells.
3. Retina	c. Works as a photographic camera.
4. Ciliary muscles	d. Not able to see the nearby objects.
5. Myopia	e. Electrical signals related to image are sent to the brain.
6. Cataract	f. A circular muscular diaphragm which can control the size of pupil.
7. Presbyopia	g. Milky and cloudy layer is formed on the eye lens.
8. Iris	h. The capacity of eye to see objects clearly between 25 cm and infinite distance.
9. Pupil	i. Increases or decreases curvature of eye lens.
10. Optic nerves	j. Image of object at infinite distance is formed in front of retina.
11. Hypermetropia	k. Accommodation power of eye decreases with ageing.

Answer:
(1 – c), (2 – h), (3 – b), (4 – i), (5 – j), (6 – g), (7 – k), (8 – f), (9 – a), (10 – e), (11 – d).

Question 5.
Choose the correct option from those given below each question:
1. Splitting of white light into its seven constituent colours is called ……………
A. refraction
B. reflection
C. dispersion
D. interference
Answer:
C. dispersion

2. Which colour of light deviates maximum in the dispersion of white light by a prism?
A. Violet
B. Blue
C. Green
D. Red
Answer:
A. Violet

3. In the human eye, the image of an object is formed at the ……………….
A. iris
B. pupil
C. retina
D. cornea
Answer:
C. retina

4. The focal length of the eye lens is changed due to the action of the ……………….
A. pupil
B. retina
C. ciliary muscles
D. cornea
Answer:
C. ciliary muscles

5. A ………………. lens is used to correct presbyopia.
A. convex
B. concave
C. bi-focal
D. contact
Answer:
C. bi-focal

6. Out of the following, which phenomenon does not play a role in the formation of a rainbow?
A. Reflection
B. Refraction
C. Dispersion
D. Absorption
Answer:
D. Absorption

7. Where is the image formed in the eye of a person suffering from near-sightedness?
A. On the retina
B. Behind the retina
C. In front of the retina
D. On the pupil
Answer:
C. In front of the retina

8. Which phenomenon is responsible for the twinkling of stars?
A. Atmospheric reflection
B. Atmospheric refraction
C. Reflection
D. Total internal reflection
Answer:
B. Atmospheric refraction

9. The phenomenon of ………………. of light by the colloidal particles gives rise to the Tyndall effect.
A. reflection
B. refraction
C. scattering
D. dispersion
Answer:
C. scattering

10. What is the time difference between actual sunset and apparent sunset?
A. 2 seconds
B. 20 seconds
C. 2 minutes
D. 20 minutes
Answer:
C. 2 minutes

11. Which light gets scattered maximum due to atmosphere?
A. Blue
B. Yellow
C. Green
D. Red
Answer:
A. Blue

12. Which light has minimum speed in glass (prism)?
A. Red
B. Green
C. Blue
D. Violet
Answer:
D. Violet

13. When an eye is focussed on a distant object, the focal length of the eye lens is ……………….
A. maximum
B. minimum
C. half of its minimum
D. half of its maximum
Answer:
A. maximum

14. How many surfaces does a triangular prism have?
A. 3
B. 4
C. 5
D. 6
Answer:
C. 5

15. The wavelengths of violet, yellow and red light are λ_v, λ_y and λ_r respectively, then ……………………..
A. λ_v > λ_y > λ_r
B. λ_v < λ_y < λ_r
C. λ_y < λ_v < λ_r
D. λ_y < λ_r < λ_v
Answer:
B. λ_v < λ_y < λ_r

16. For normal vision, the far point is at ……………… distance.
A. 25 cm
B. 1 cm
C. 1 m
D. infinite
Answer:
D. infinite

17. For normal vision, the near point is ………………
A. 25 cm
B. 25 m
C. zero
D. infinite
Answer:

18. Which phenomenon can explain the advance sunrise and the delayed sunset?
A. Dispersion of light
B. Scattering of light
C. Tyndall effect
D. Atmospheric refraction
Answer:
D. Atmospheric refraction

19. Which of the following phenomena cannot be explained by scattering of light?
A. The red light used for signal lights for danger.
B. Blue colour of clear sky
C. White colour of clouds
D. Early sunrise
Answer:
D. Early sunrise

20. The base of an equilateral triangle ABC is BC. When it is arranged in four different situations and white light is incident on it, then in which of the following arrangements of the prism, the third colour from the top is the colour of clear sky in dispersion of light is produced?

A. (i)
B. (ii)
C. (iii)
D. (iv)
Answer:
B. (ii)
Hint: The dispersion of white light is shown in arrangement (ii) of prism

The third colour of light from the top is blue which is the colour of the clear sky.

21. The Sun appears white in afternoon. The reason is …
A. less scattering of light.
B. more scattering of all the colours of white light.
C. more scattering of blue colour.
D. more scattering of red colour.
Answer:
A. less scattering of light.
Hint: White light coming from the Sun has to travel less distance in the atmosphere before reaching the observer. So less scattering of light takes place and as a result the Sun appears white.

22. Sea water at more depth appears blue. The reason is …
A. presence of some plants in sea water.
B. the image of the sky appears in water.
C. scattering of light.
D. light is absorbed by sea water.
Answer:
C. scattering of light.

23. When the ciliary muscles are relaxed, the eye lens becomes …………….. and its focal length ……………… This enables us to see distant objects clearly.
A. thin, increases
B. thin, decreases
C. thick, increases
D. thick, decreases
Answer:
A. thin, increases

24. When the ciliary muscles contract, the eye lens becomes …………….. and its focal length ……………..
This enables us to see nearby objects clearly.
A. thick, decreases
B. thick, increases
C. thin, increases
D. thin, decreases
Answer:
A. thick, decreases

25. The rainbow on the moon ……………..
A. is not possible.
B. is rare.
C. is observed with the reverse order of colours.
D. is of two types.
Answer:
A. is not possible.

26. In dispersion of white light due to a triangular glass prism, the deviation of red colour is less compared to violet colour. The reason ………………
A. is n_v > n_r.
B. is n_r > n_v.
C. is n_v = n_r.
D. does not depend on n.
Answer:
A. is n_v > n_r.
Hint: In glass medium, speed of violet colour is less as compared to speed of red colour. So, n_v > n_r from n = \(\frac { c }{ v }\)

27. Which lens from the following, should a person suffering from near-sightedness use?
A. A convex lens
B. A concave lens
C. A cylindrical lens
D. A bi-focal lens
Answer:
B. A concave lens

28. Which lens is used by a person suffering from far-sightedness?
A. A convex lens
B. A concave lens
C. A cylindrical lens
D. A bi-focal lens
Answer:
A. A convex lens

29. Which of the following is true for near-sightedness?
A. Nearby objects cannot be seen clearly.
B. Distant objects cannot be seen clearly.
C. The eye lens cannot become thick as required.
D. This defect can be rectified using spectacles of convex lenses.
Answer:
B. Distant objects cannot be seen clearly.

30. Which of the following is true for far-sightedness?
A. Nearby objects cannot be seen clearly.
B. Distant objects cannot be seen clearly.
C. The eye lens cannot become thin as required.
D. This defect can be rectified using spectacles of concave lenses.
Answer:
A. Nearby objects cannot be seen clearly.

31. Where is the image formed in the eye of a person suffering from far-sightedness?
A. On the retina
B. Behind the retina
C. On the pupil
D. In front of the retina
Answer:
B. Behind the retina

32. A person has a defect of eye vision. His near point is 40 cm. It means …
A. he cannot clearly see objects at a distance more than 40 cm from the eye.
B. he can clearly see objects at a distance of 40 cm only.
C. he can clearly see objects at a distance equal to 40 cm or more from the eye.
D. he can clearly see objects at a distance less than 40 cm e.g., 25 cm from the eye.
Answer:
C. he can clearly see objects at a distance equal to 40 cm or more from the eye.
Hint: Here the person suffers from far-sightedness.

33. A person has a defect of vision. His far point is 1.5m. It means …
A. he cannot clearly see objects at a distance more than 1.5 m from the eye.
B. he can clearly see objects at a distance more than 1.5 m from the eye.
C. he cannot clearly see objects at a distance less than 1.5 m from the eye.
D. he suffers from far-sightedness.
Answer:
C. he cannot clearly see objects at a distance less than 1.5 m from the eye.
Hint: Here the person suffers from near-sightedness.

34. Out of the following, which light is deviated minimum in the dispersion of white light through a glass prism?
A. Green
B. Violet
C. Yellow
D. Dispersion of the given three colours is the same.
Answer:
C. Yellow

35. Which light has maximum speed in glass?
A. Violet
B. Blue
C. Green
D. Red
Answer:
D. Red

36. Which ray of light is present exactly at the middle of the spectrum obtained from white light?
A. Green
B. Yellow
C. Red
D. Violet
Answer:
A. Green

37. The lens in human eye is a …………….
A. convex mirror
B. convex lens
C. concave mirror
D. concave lens
Answer:
B. convex lens

38. For persons suffering from near-sightedness the power of the lens used in spectacles is …………….
A. positive
B. zero
C. negative
D. infinite
Answer:
C. negative

39. For a person suffering from ……………. the power of the lens used in spectacles is positive.
A. far-sightedness
B. near-sightedness
C. presbyopia
D. cataract
Answer:
A. far-sightedness

40. Which phenomenon/phenomena of light is/ are involved in the formation of a rainbow?
A. Refraction
B. Dispersion
C. Internal reflection
D. All of the given above
Answer:
D. All of the given above

41. ……………. light from the following is least scattered by fog, dust and smoke.
A. Violet
B. Blue
C. Red
D. Yellow
Answer:
C. Red

42. Which of the following controls the amount of light entering into the human eye?
A. Ciliary muscles
B. Pupil
C. Cornea
D. Iris
Answer:
D. Iris

43. The refractive index of glass is maximum for ……………. light.
A. violet
B. green
C. blue
D. red
Answer:
A. violet

Question 6.
Answer the following questions in very short as directed (Miscellaneous) :
(1) What is the power of a lens that can be used to correct the eye defect of a person who cannot see the objects distinctly kept beyond 2m?
Answer:
0.5 D
(∵ For a myopic eye the focal length of the corrective lens is equal to the far point of the myopic person, i.e., f = – 2m).

(2) Why does the Sun appear white at noon?
Answer:
The Sun appears white at noon because white light (sunlight) is least scattered by the atmosphere.

(3) Why is the eye lens not perfectly solid?
Answer:
If it becomes solid, its focal length would be fixed. Then, we would not be able to focus the objects lying at different distances on the retina. In short, accommodation of the eye would be reduced to zero.

(4) What is the focal length of plain goggles?
Answer:
Infinity

(5) What happens when elasticity of the crystalline lens is reduced to zero?
Answer:
Power of accommodation would be almost zero.

(6) Which defect of the eye occurs due to distortion of cornea?
Answer:
Astigmatism occurs due to the distortion of cornea.

(7) How the defect of astigmatism can be corrected?
Answer:
Astigmatism is corrected by the use of a cylindrical lens.

(8) What is colour blindness? How can it be cured?
Answer:
It is a defect of the eye in which a person is unable to distinguish between certain colours due to insufficient or no cone shaped cells on retina. It cannot be cured.

(9) In hypermetropia, how does the size of eyeball change?
Answer:
In hypermetropia, the eyeball becomes too small (flat).

(10) What change occurs in the focal length, when our eye lens becomes thick?
Answer:
The focal length will decrease when the eye lens becomes thick.

(11) What are rods and cones?
Answer:
Sensitive portion of retina has large number of cells, rod shaped and cone shaped cells. Rod shaped cells are sensitive to the intensity or brightness of the light whereas cone shaped cells are sensitive to colours.

(12) What is cataract?
Answer:
Sometimes, the eye lens of a person becomes hazy or even opaque resulting in reduced or total loss of vision. This is called cataract.

(13) What would have been the colour of the sky, had there been no atmosphere?
Answer:
Black

(14) Due to which phenomenon is the colour of water in deep sea blue?
Answer:
Due to scattering of light.

(15) What is the cause for presbyopia?
Answer:
The eye lens becomes less elastic or non-elastic after 40 years as the, accommodation power of eye becomes less.

(16) Give the relationship between wavelength of light and its angle of deviation, when it is passed through a prism.
Answer:
Wavelength λ = \(\frac{1}{\text { Angles of deviation } \delta}\)

(17) For which colour has the glass larger refractive index – violet or green?
Answer:
Violet

(18) Which part of human eye is also known as ‘white of the eye’?
Answer:
Sclera

(19) Why is blind spot so called?
Answer:
Blind spot is a point at which the optic nerve leaves the eye. It contains no rods or cones. So an image formed at this point, is not sent to the brain.

(20) Which liquid is filled in the space between the eye lens and the retina?
Answer:
Vitreous humour (It is a transparent jelly.)

(21) What happens to the pupil of the eye when the light is (a) very bright and (b) very dim?
Answer:
(a) In very bright light, the size of the pupil becomes very small, so that less amount of light enters into the eye.
(b) In very dim light, the size of the pupil increases, so that more amount of light enters into the eye.

(22) A man is wearing spectacles of focal length +1 m. What can be the defect in the eye?
Answer:
As the focal length of spectacles is positive, the man is using convex lenses. So, he is suffering from hypermetropia.

(23) Which portion of a bi-focal lens is
(a) a concave lens
(b) a convex lens?
Answer:
(a) Upper portion
(b) Lower portion.

(24) When sunlight enters into a room filled with dark smoke, its path becomes visible. Name the phenomenon responsible for this.
Answer:
The Tyndall effect.

(25) What is the function of the iris?
Answer:
The iris controls the size of the pupil.

(26) What are light sensitive cells?
Answer:
Rods and cones

(27) What type of signals are generated and sent to the brain by light sensitive cells in the retina?
Answer:
Electrical signals.

(28) What holds the crystalline lens in the human eye?
Answer:
Ciliary muscles.

(29) Which part of the human eye helps in changing the thickness of the eye lens?
Answer:
Ciliary muscles.

(30) What is dispersion of white light?
Answer:
The splitting of white light into its various components (i.e., 7 colours) is called dispersion of white light.

(31) Give the main difference between the lens of the human eye and the lens of a camera.
Answer:
The lens of the human eye has flexible aperture, so that its focal length can be changed, while in a camera the focal length of the lens is fixed.

(32) The image formed on the retina is inverted, but we see the object erect. Why?
Answer:
The inverted image formed on the light sensitive cells (called rods and cones of retina), generates electrical signals. These signals reach the brain via the optic nerve. It is the brain? that interprets this image and while processing the image it helps in perceiving the objects as they are.

(33) The absolute refractive index of a medium is 2.0. The speed of light in vacuum/air is 3 x 10⁸ms^-1. Find the speed of light in the medium.
Answer:
Absolute refractive index of medium is,
n = \(\frac { c }{ v }\)
∴ v = \(\frac { c }{ n }\)
= \(\frac{3 \times 10^8}{2}\)
= 1.5 x 10⁸ms^-1

(34) Match the column properly :

Column I (eye defect)	Column II (correcting lens)
1. Myopia	p. Bi-focal lens
2. Hypermetropia	q. Concave lens
	r. Convex lens

Answer:
(1 – q), (2 – r).

(35) Match the following column :

Column I (eye defect)	Column II (correcting lens)
1. Astigmatism	p. Convex lens
2. Presbyopia	q. Cylindrical lens
	r. Concave lens

Answer:
(1 – q), (2 – p).

(36) The far point of a myopic eye is 100 cm. What is the focal length of the lens required to see very distant (normal far point) objects clearly?
Answer:
As the focal length of concave lens used for correcting the myopic eye is equal to distance of far point of the myopic eye, the focal length of correcting lens required is f = – x = – 100 cm = – 1 m.

(37) The near point of a hypermetropic eye is 75 cm. What is the focal length of the lens required to see clearly an object placed at 25 cm from the eye (normal near point)?
Answer:
The focal length of convex lens used for correcting the hypermetropic eye is given by
(where x’ = defective near point = 75 cm and d = normal near point = 25 cm)
f = \(\frac{x^{\prime} d}{x^{\prime}-d}\)
∴ f = \(\frac{75 \times 25}{75-25}\)
= 37.5 cm
= 0.375 m

(38) The eye lens of human eye is a double convex lens. Agree or Disagree?
Answer:
Agree.

(39) Cone-shaped retinal cells respond to the brightness or intensity of light. Agree or Disagree?
Answer:
Disagree.

(40) Which property of vision is used in cinematography?
Answer:
Property of persistence of vision is used in cinematography.

(41) What is aqueous humour?
Answer:
Aqueous humour is a transparent viscous liquid in the space between the cornea and the eye lens.

(42) What is the maximum power of accommodation of a normal eye?
Answer:
The maximum power of accommodation of a normal eye
= \(\frac{1}{\text { near point of the normal eye (in metre) }}\)
= \(\frac { 1 }{ 0.25 m }\)
= \(\frac { 100 }{ 25 m}\)
= 4 m^-1
= 4 D

(43) What is meant by scattering of light?
Answer:
It is the phenomenon of change in the direction of light on striking a scatterer.

(44) What is the basic cause of atmospheric refraction?
Answer:
The basic cause of atmospheric refraction is variation in the refractive index (i.e., optical density) of different layers of the earth’s atmosphere with altitude.

(45)

In the above figure a narrow beam of white light is shown to pass through a triangular glass prism. After passing through the prism, it produces a spectrum XY on a screen. State the colour seen at (i) X and (ii) Y.
Answer:
The colour seen at X is violet and that seen at Y is red.

(46)

In the above figure, which angles are correctly marked?
Answer:
∠A and ∠e are correctly marked.

(47)

In the above figure (ray diagram), state the angle of incidence and the angle of deviation.
Answer:
∠PQN = i = Angle of incidence
∠P’OR = D = Angle of deviation

(48)

In the above figure (ray diagram), state angle of incidence, angle of emergence and angle of deviation.
Answer:
∠p = Angle of incidence
∠y = Angle of emergence
∠z = Angle of deviation

(49) What is the principle of the working of s the human eye?
Answer:
It is like a camera having a lens system forming an inverted, real image on the light |> sensitive screen (retina) inside the eye.

(50) On which factor does the colour of the scattered white light depend?
Answer:
Size of the particles of the medium : through which it is passing.

(51) Give the scientific names of the following parts of the eye :
(a) Carrying signals from an eye to the brain.
(b) A small opening (hole) in the middle of the iris.
Answer:
(a) Optic nerve
(b) Pupil

(52) A near-sighted person has a near point 25 cm and a far point 50 cm. Can he see clearly an object at a distance of: (i) 5 cm, (ii) 25 cm, (iii) 60 cm. Write ‘Yes’ or ‘No’ only.
Answer:

1. No
2. Yes
3. No

(53) The near point of a far-sighted person is 50 cm.
Can the person see clearly an object at a distance of:
(i) 20 cm
(ii) ∞ (infinity)
Answer:
(i) No
(ii) Yes

(54) How much is our horizontal field of view (a) with one eye open
(b) with both eyes open?
Answer:
(a) about 150°
(b) about 180°.

(55) Which of the following have a wider s field of view?
(a) Animals having two eyes on the opposite 5 sides of their head or
(b) Animals having two eyes at the front of their head.
Answer:
(a) Animals having two eyes on the opposite 5 sides of their head or

Value Based Questions With Answers

Question 1.
Maull and Vlshva are best friends and they study in 4th grade. Recently, Mauli has been facing difficulty in reading the blackboard text from the last desk/bench. Vishva wonders why Mauli avoids sitting on the last desk/bench. On observation, Vishva found that Mauli often carries junk food in her lunch. Vishva has started sharing her lunch full of green vegetables and fruits with her. Mauli is now better and has also started taking a “balanced diet”.

1. Name the eye defect Mauli is suffering from.
2. What are two possible deformities related s to her eye defect?
3. What values do you learn from Vishva and Mauli?

Answer:

1. Myopia (near-sightedness)
2. Lens defect (increased thinness i.e., excessive curvature of the eye lens) and Eyeball defect (elongation of the eyeball).
3. Friendship, concern for each other, s importance of a balanced diet.

Question 2.
An eye camp was organised by the doctors in a village. They found that the eyes of imaged people in the village have the near point receded and the far point also gets reduced. Often aged people suffer from both myopia s and hypermetropia. Doctors (opthalmologists) provide these people spectacles of bi-focal lenses to correct the defects. The people were happy and grateful to the doctors.
(1) Name the eye defect from which the people were suffering.
(2) Give any two causes of this defect.
(3) What were benefits to organise such camps > in rural areas? Give two suggestions.
Answer:
(1) Presbyopia

(2) (a) Weakening of ciliary muscles
(b) Reducing ability of the lens to change the curvature.

(3) (a) To make people aware of eye diseases
(b) To tell people to take proper and balanced diet.

Question 3.
Four friends went to a picnic. The weather was pleasant. They played various games and then had snacks. Suddenly, Raju, one of them, observed seven colours in the sky. He said to others, “wow what a rainbow” !

Ram, one of them, asked him “What is a rainbow?” Raju then explained to all about its formation. After that everyone in the group thanked Raju for the knowledge, he had given to them.
(1) When Raju was facing the rainbow, where was the Sun?
(2) Which device can be used to obtain such a phenomenon?
(3) What moral value do you learn from Raju?
Answer:
(1) The Sun was behind Raju.
(2) A small prism. (Water droplets present in the atmosphere act like small prisms.)
(3) Knowledge increases by sharing, friendship, love and affection with nature.

Question 4.
In a beautiful valley, there was a village. When trains passed from the village, the whistle and the sound of train, mixed with the sound of waterfall, seemed to be very pleasant to everyone. Hence, children of that village used to play near the railway track. Once on a very light foggy day, a group of children found that a fish plate was missing from the track. As such, all the villagers were worried.

Prashant, one of the children, suddenly put his ear to the line and tried to know whether a train is coming or not. He knew a train is coming. He asked his friends to inform the railway cabin crew and he himself put off his red shirt and started running s towards the train, waving his red shirt. The driver and cabin man got the alert signal in time and thus a major accident was averted.?
(1) Name the two physical phenomena of science used by Prashant.
(2) Why did Prashant use his red shirt instead of any other coloured shirts?
(3 ) What moral values do you learn from Prashant?
Answer:
(1) (a) Sound travels through a medium.
(b) Scattering of light

(2) The red light is least scattered by fog or smoke so it can be seen from a large distance.

(3) (a) Proper knowledge and its application.
(b) Concern for others.

Question 5.
Millions of people of the developing countries of the world are suffering from corneal blindness. They can be cured by replacing the defective cornea with the cornea of a donated eye. A charitable society of your city has organised a campaign in your neighbourhood in order to create awareness about this fact.
(1) State the objective of organising such campaigns.
(2) Write one argument which you would give s to motivate the people to donate their eyes after death.
(3) List two values which could be developed in the persons who actively participate and < contribute in such programme.
Answer:
(1) The objective of organising such campaign is to help those people who are suffering from corneal blindness and they can be cured by replacing their defective cornea with the cornea of a donated eye.

(2) Come forward to participate in this compaign because, if someone gets his vision through your eyes, it is an incredible help as the eye is one of the most valuable sense organs through which an individual can achieve so many things in his / her life.

(3) The persons who actively participate and contribute in such programme are

• strong hearted
• very much helpful for the people living in such situations.

Question 6.
Mr Bharat’s 65 year old mother was complaining about blurred vision in both the eyes due to which she could not see things clearly. Mr Bharat took his mother to an eye hospital. The doctor examined her eyes carefully and concluded that she has a medical condition which could not be corrected by using any type of spectacle lenses and it required surgery. Her eyes were operated upon and she could then see once again properly.
(1) What could be the defect in the eyes of Mr Bharat’s mother?
(2) What happens to the eye lens during this defect? What is done during surgical operation of the eyes to restore the correct vision?
(3) What values do you learn from Mr Bharat in this episode?
Answer:
(1) The defect in the eyes of Mr Bharat’s mother is known as cataract.

(2) During the development of cataract, a membrane is gradually formed over both the eye lenses making the eye lenses cloudy. This makes the vision blurred. During surgical operation, the cloudy eye lenses are removed from the eyes and suitable artificial lenses are inserted in their place.

(3)

• Awareness and knowledge about the eye defects that can be cured by eye specialist doctors.
• Desire to mitigate the suffering of others (here mother).
• Sense of responsibility.

Question 7.
Amit is a domestic help (or maid) working at ; Mr Dave’s house. One day Amit complained to Mr Dave that he had difficulty in reading the letter which he had received from his s parents. Mr Dave, realising that Amit had an eye defect, took him to an eye specialist doctor.

The doctor tested his eyes carefully and told Amit to wear spectacles containing certain type of lenses having specified power, Mr Dave bought the required spectacles for Amit. By wearing this spectacle, Amit could read and write easily. He was very happy and thanked Mr Dave.
(1) What could be the eye defect Amit was suffering from?
(2) What could be the two possible reasons responsible for his eye defect? What type of lenses do you think the doctor recommended for Amit’s spectacles?
(3) What values are displayed by Mr Dave in this episode??
Answer:
(1) Amit was suffering from an eye defect called hypermetropia (far-sightedness) in ; which a person cannot see the nearby objects clearly though he can see the distant objects clearly.

(2) (a) Low converging power of the eye lens (because of the eye lens being less convex or less thick)
(b) Eyeball being too small (flat) (because of which the distance of the retina from the eye lens is less than normal).

The doctor recommended convex lenses for the spectacles of Amit.

(3) Mr Dave displayed the values of –

• awareness, which means having s knowledge of a situation or facts.
• concern for others (to mitigate their suffering).
• kindness and generosity.

Question 8.
Rohit is a car driver working for Mr Joshi. One day Rohit complained that he had difficulty in driving the car because he could not see the distant traffic clearly though he could see the nearby objects clearly. Mr Joshi took Rohit to an eye hospital. The eye specialist doctor checked and tested his eyes with various machines and gave him the name and power of the lenses to be worn as spectacles.

Mr Joshi paid for the required spectacles for the driver. By wearing these spectacles, the driver could now see even the distant vehicles and people on the road clearly. He thanked Mr Joshi for this.
(1) Name the eye defect Rohit is suffering from.

(2) What could be the two possible reasons for his eye defect? What type of lenses do you think the doctor recommended for Rohit’s spectacles?

(3) What values (or qualities) do you learn from Mr Joshi in this episode?
Answer:
(1) Myopia (near-sightedness) in which a person cannot see distant objects clearly though he can see nearby objects clearly.

(2) (a) High converging power of the eye lens (because of the eye lens being too convex or too thick).
(b) Eyeball may be too long (elongated) (because of which the distance of the retina from the eye lens is more than normal).
The doctor recommended concave lenses for the spectacles of Rohit.

(3) (a) General awareness (that an eye defect can usually be corrected by wearing spectacles containing suitable lenses).
(b) Concern for others (because Mr Joshi wanted to mitigate or remove the suffering of driver).
(c) Kindness and generosity.

Practical Skill Based Questions With Answers

Question 1.
Dispersion is caused by refraction and not by reflection. Why?
Answer:
White light is composed of seven colours having different wavelengths. And speed of different colours is the same in vacuum / air while in different media it is different.

Now, for a given angle of incidence, the angle of reflection is the same for all the wavelengths of white light, while the angle of refraction is different for different wavelengths.

Question 2.
A beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 as shown in the diagram.

(1) The colours at positions marked 3 and 5 are similar to the colour of the sky and the colour of gold (metal), respectively. Is the above statement made by a student correct or incorrect. Justify.
(2) Which of the positions shown above correspond approximately to the colour of (a) a brinjal, (b) danger signal, (c) neel (applied to clothes), (d) orange.
Answer:
(1) No, because 3 refers to yellow and 5 to blue colour of the spectrum.

(2) (a) 7 (b) 1 (c) 6 (d) 2.

Question 3.
When a beam of white light is passed through a triangular glass prism, it gets dispersed into its component colours. Why do we get these colours? In the given figure, the colours X and Y represent the extreme components of the spectrum. Identify X and Y.

Answer:
(1) Different colours of light while passing through a prism bend through different angles with respect to the incident ray, as they travel with different speeds, this leads to dispersion of light.

(2) X – violet, Y – red

Question 4.
A narrow beam PQ of white light passes through a glass prism ABC as shown in the diagram.

Trace it on your answer sheet and show the path of the emergent beam as observed on the screen DE.
(1) Write the name and cause of the phenomenon observed.
(2) Where else in nature is this phenomenon observed?
(3) Based on this observation, state the conclusion which can be drawn about the constituents of white light.
Answer:

(1) The phenomenon of splitting of white light into its constituent colours is called dispersion of light. It occurs because different constituent colours of light travel with different speeds in a material medium (other than air /vacuum) and hence bend through different angles.

(2) In nature, this phenomenon is observed in formation of a rainbow.

(3) Based on the phenomenon of dispersion, we can conclude that –

• White light consists of seven colours.
• Violet light suffers maximum deviation and red light suffers minimum deviation.

Question 5.
(a) A narrow beam of white light is incident on three glass objects as shown below. Comment on the nature of the behaviour of the emergent beam in all three cases.

(b) There is a similarity between two of the emergent beams. Identify the two.
Answer:
(a) (1) The incident beam of light after refraction through the glass slab emerges parallel to the incident beam but is laterally shifted. No dispersion takes place in this case.

(2) The incident beam of light after refraction through the prism splits into a band of seven colours which are violet, indigo, blue, green, yellow, orange and red. These coloured rays emerge out of the prism along different directions and become distinct. Thus, in this case, dispersion of white light takes place.

(3) When the incident beam passes through the first prism, the prism splits it into a band of seven colours. These coloured rays are then incident on an identical inverted prism.

Then the recombination of the coloured rays takes place. The emergent beam of light is parallel to the incident beam but slightly shifted outward.

(b) The emergent beam in the cases (1) and (3) are similar. In both the cases, the emergent beam is parallel to the incident beam and is laterally shifted.

Memory Map:

Jharkhand Board JAC Class 10 Science Solutions Chapter 11 Human Eye and Colourful Worlds Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 11 Human Eye and Colourful World

Jharkhand Board Class 10 Science Human Eye and Colourful World Textbook Questions and Answers

Question 1.
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to ……………
(a) presbyopia.
(b) accommodation.
(c) near-sightedness.
(d) far-sightedness.
Answer:
(b) accommodation.
[Hint : Accommodation is the ability of the eye lens to focus both nearby and distant objects by adjusting its focal length.]

Question 2.
The human eye forms the image of an object at its ……………..
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.
Answer:
(d) retina.
The retina is the light sensitive surface of the eye on which the image is formed.

Question 3.
The least distance of distinct vision for a young adult with normal vision is about ……………..
(a) 25 m
(b) 2.5 m
(c) 25 m
(d) 2.5 m
Answer:
(c) 25 cm.
The minimum distance at which an object can be seen most distinctly without any strain (i.e., comfortably) is 25 cm.

Question 4.
The change in focal length of an eye lens is caused by the action of the…
(a) pupil.
(b) retina.
(c) ciliary muscles.
(d) iris.
Answer:
(c) ciliary muscles.
The ciliary muscles contract and expand, in order to change the curvature of the eye lens for focussing the image of an object at varying distances at the retina.

Question 5.
A person needs a lens of power – 5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptres. What is the focal length of the lens required for correcting (1) distant vision and (2) near vision?
Solution :
(1) For distant vision, f = ?
P = – 5.5, D = – 5.5 m^-1
Now, f = \(\frac { 1 }{ p }\)
∴ f = \(\frac{1}{-5.5 m^{-1}}\)
= – 0.182m
= – 18.2 cm (concave lens)

(2) For near vision, f = ?
P = + 1.5, D = + 1.5 m^-1
Now, f = \(\frac { 1 }{ p }\)
∴ f = \(\frac{1}{+1.5 m^{-1}}\)
= 0.667 m
= 66.7 cm (Convex lens)

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Solution:
1. The defect of an eye called myopia (short-sightedness or near-sightedness) is corrected by using spectacles containing concave lens of appropriate focal length.

2. Here the far point of myopic person is 80 cm. (while far point of normal person is infinity oo)

3. This means that this person can see the distant object (kept at infinity) clearly if the image of this distant object is formed at his own far point (which is 80 cm here).
So, in this case:
Object distance u = – ∞ (Normal far point)
Image distance v = – 80 cm (Far point of this defective eye in front of lens)
Focal length f = ?
Now,
By the lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
∴ \(\frac{1}{f}=\frac{1}{-80}-\frac{1}{-\infty}\)
∴ \(\frac{1}{f}=-\frac{1}{80}\)
∴ f = – 80 cm
= – 0.8 m (Concave lens)
Now,
Power of the lens,
P = \(\frac { 1 }{ f }\)
∴ P = \(\frac { 1 }{ -0.8m }\)
= – \(\frac { 10 }{ 8 }\)m^-1
= – 1.25 D
A concave lens of power -1.25D is required to correct the problem.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Solution:
1. The defect of an eye called hypermetropia (long-sightedness or far-sightedness) is corrected by using spectacles containing convex lenses of appropriate focal length.

[N = Near point of a hypermetropic eye and
N’ = Near point of a normal eye]
[11.16 : (a) Hypermetropic eye (b) corrected eye]

2. Here the near point of the hypermetropic eye is 1 m = 100 cm
(while near point of normal eye is 25 cm)

3. This means that this person can see the nearby object (kept at 25 cm) clearly if the image of this nearby object is formed at his own near point (which is lm = 100 cm here).
So, in this case :
Object distance u = – 25 cm (Normal near point)
Image distance u = – 1 m = – 100 cm
(Near point of this defective eye in front of lens)
Focal length f = ?
Now,

Now,
Power of the lens,
P = \(\frac { 1 }{ f }\)
= \(\frac{1}{\left(\frac{1}{3} \mathrm{~m}\right)}\)
= 3D

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
For seeing nearby objects, the ciliary muscles contract to make the eye lens thicker near the middle, so as to reduce the focal length of the eye lens.

But ciliary muscles cannot be contracted beyond a certain limit and hence we cannot see clearly the objects closer than 25 cm from the eye.

In other words, a normal eye is not able to see clearly the objects placed closer than 25 cm because all its power of accommodation has already been exhausted.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye ?
Answer:
For a normal eye, the image distance (u) in the eye is fixed = distance of the retina from the eye lens ≈ 2.3 cm
When we increase the distance of the object (u) from the eye, the focal length of eye lens is changed on account of the accommodating power of the eye so as to keep the image distance (u) constant according to the relation \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\).

Question 10.
Why do stars twinkle?
Answer:
Twinkling of stars is due to atmospheric refraction of starlight.

• Starlight, on entering the earth’s atmosphere, undergoes refraction continuously before it reaches the earth.
• As the optical density of air increases towards surface of the earth, light from the star travels from rarer to denser layers, bending every time towards the normal.
• On producing the final refracted ray backwards as shown in the following figure, we find that the apparent position (B) of a star is higher than the actual position (A) of the star as shown in figure 11.11.
  
• The star appears somewhat higher (above) than its actual position when viewed near the horizon.
  Further, this apparent position of the star is not stationary, but keeps on changing slightly, since physical conditions of the earth’s atmosphere are not stationary.
• Since the stars are very distant, they appear as point-sized sources of light.
• Due to a continuous change in the direction of propagation of light, the apparent position of the star fluctuates all the time and the amount of starlight entering the eye flickers, i.e., the brightness of the star changes continuously (the star sometimes appears brighter and at some other time fainter). This is called the twinkling s of a star.

Question 11.
Explain why the planets do not twinkle.
Answer:
Planets are much closer to the earth relative to stars. Therefore, planets appear bigger > than stars. Stars are far away from the earth. Therefore they appear smaller.

• So, stars can be considered as point-sized sources and planets can be considered as a ? collection of a large number of point-sized sources of light, i.e., extended sources of light.
• As planet is a collection of a large number of point-sized sources of light, the total variation in amount of light entering our eye from all the individual point-sized sources averages out to zero, thereby nullifying the twinkling effect. That is why planets do not twinkle.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:

• In figure 11.14, the situation at the sunrise is shown.
• Here, the white light coming from the Sun near the horizon, passes through thick layers of air and covers larger distance in the earth’s atmosphere before reaching the observer. During this, more scattering of blue light and shorter wavelengths take place. Hence, the reddish light reaches the observer and the Sun appears reddish.
• The same thing occurs at the sunset.

[Note : Rising and setting of the full moon from the horizon appears reddish due to this reason.]

Question 13.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
The sky appears dark to the astronaut because in outer space, there is no atmosphere to scatter the sunlight. Since there is no scattering of the blue component of the white sunlight which can reach the eye of an astronaut in outer space, the sky appears dark to the astrounaut, instead of blue.

Jharkhand Board Class 10 Science Human Eye and Colourful World InText Questions and Answers

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye lens to adjust its focal length, so that nearby as well as distant objects can be focussed on the retina and hence seen comfortably and distinctly is called the power of accommodation of the eye.

Question 2.
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
Answer:
A person with a myopic eye should use a concave lens of suitable focal length or power to restore proper vision.
[Here, the person with defect of myopia has the far point nearer than infinity at a distance of 1.2 m from the eye.
So, v = – 1.2m; u = – ∞; f = ?
So, from the lens formula
\(\frac{1}{f}=\frac{1}{-u}-\frac{1}{v}\)
∴ \(\frac{1}{f}=\frac{1}{-(-\infty)}+\frac{1}{-1.2}\)
∴ f = – 1.2 m
∴ P = \(\frac { 1 }{ – 1.2 }\) = – 0.83 D
∴ A concave lens of focal length 1.2 m should be used to restore proper vision.]

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
For the human eye with normal vision, f the far point is at infinity and the near point is at 25 cm from the eye.

Question 4.
A student has difficulty in reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
Answer:
As the child cannot see distant objects 5 clearly it means that he is suffering from myopia or near-sightedness. In this case the child can S see nearby objects clearly but cannot see far off objects distinctiy as their images are formed s before the retina.

To correct this defect, the child has to use spectacles with concave lens of suitable focal length.

Activity 11.1 [T. B. Pg. 192]

To study the refraction of light through a s triangular glass prism.

Procedure:
1. Fix a sheet of white paper on a drawing board ; using drawing pins.

2. Place a glass prism on it in such a way that ; it rests on its triangular base. Trace the outline of the prism using a pencil.

3. Fix two pins, say at points P and Q, on the line PE as shown in the following figure :

Refraction of light through a triangular glass prism
PE – Incident ray
EE – Refracted ray
FS – Emergent ray
∠A – Angle of the prism
∠i – Angle of incidence
∠r – Angle of refraction
∠e – Angle of emergence
∠D – Angle of deviation

4. Look for the images of the pins, fixed at P and Q, through the other face AC.

5. Fix two more pins, at points R and S, such that the pins at R and S and the images of the pins at P and Q lie on the same straight line.

6. Remove the pins and the glass prism.

7. Join the points P and Q and extend the line till it meets the nearest boundary of the prism. Similarly, join and produce the points R and S. Let these lines meet the boundary of the prism at E and F respectively. Join E and E

8. Extend PE and RS such that they meet at point G. Mark the angle of deviation (∠D) as shown in figure.

9. Draw perpendiculars to the refracting surfaces AB and AC of the prism at points E and F respectively.

10. Mark the angle of incidence (∠i), the angle of refraction (∠r) and the angle of emergence (∠e) as shown in the figure.

Compare the angle of incidence and the angle of refraction at each refracting surface of the prism.
Bending of rays PE, EF and FS are similar to the kind of bending that occurs in a glass slab.

Observation:
1. A ray of light suffers two refractions while passing through a prism.

2. The first refraction occurs at E on the surface AB. The incident ray PE, enters from air into glass E. It is refracted along EF bending towards normal NN’ on face AB at E.

3. The second refraction occurs at F on the surface AC. The initial refracted ray EF travelling in glass emerges in air at F It emerges along FS, bending away from normal MM’ on face AC at F.

4. At the first refracting surface AB, angle of refraction (r) is smaller than the angle of incidence (i).

But at the second refracting surface AC, angle of refraction (e) is larger than angle of incidence (∠EFM’).

5. Bending of rays PE, EF and FS is of the same kind of bending that occurs in a glass slab. The net deviation in a rectangular glass slab is zero and there is lateral shift. However, due to the peculiar shape of the prism, net diviation in passing through a prism is not zero and the prism makes the emergent ray bend at an angle to the direction of the incident ray. This angle is called the angle of deviation, i.e., D = ∠HGS, It is the angle through which the incident ray is deviated as it passes through the prism.

6. The angle of deviation depends on the angle of incidence, the angle of the prism and the nature of the material of the prism as it depends on the refractive index of the material.

Conclusion :
While passing through a prism, a ray of light undergoes two refractions as it does in case of rectangular glass slab.

However due to peculiar shape of the prism, net deviation through prism is never zero, though the net deviation is zero in case of rectangular glass slab.

In short, deviation suffered by a ray of light on passing through a prism is non-zero.

Activity 11.2 [T. B. Pg. 193]

To show that white light is made up of seven colours and different colours suffer different deviations on passing through a prism.

Procedure:
1. Take a thick sheet of cardboard and make a small hole or narrow slit in its middle.

2. Allow sunlight to fall on the narrow slit. This gives a narrow beam of white light.

3. Now, take a glass prism and allow the light from the slit to fall on one of its faces as shown in the following figure:

4. Turn the prism slowly until the light that comes out of it appears on a nearby screen.

• What do you observe?
• Why does this happen? OR How could white light of the Sun give us various colours of the rainbow?
• What is the sequence of colours that you observe on the screen?

Observations:

• We observe beautiful band of seven colours (VIBGYOR) on the screen. The deviation suffered by the violet light is maximum and that by the red light is minimum. Hence, the violet colour is at the lower end and the red colour is at the upper end of the screen.
• This happens because prism itself splits the incident white light into a band of colours.
• The sequence of colours seen from the lower end of the screen is violet (V), indigo (I), blue (B), green (G), yellow (Y), orange (O) and red (R).

Conclusion :
White light is made up of seven colours and different colours suffer different deviations.

Activity 11.3 [T. B. Pg. 196]

To observe scattering of light by colloidal particles.

Procedure:
1. Place a strong source (S) of white light at the focus of a convex (converging) lens (L₁) as shown in the figure 11.15 to produce a parallel beam of light.

2. Allow the light beam to pass through a transparent glass tank vessel (T) containing clean water.

3. Now, allow the beam of light to pass through the circular hole (C) made in a cardboard and obtain a sharp image of the circular hole on screen (MN) using a second convex (converging) lens (L₂) as shown in figure.

4. Dissolve 200g of sodium thiosulphate (hypo) – (Na₂S₂O₃) in about 2 L of clean water taken in the tank. Add 1 to 2 mL of concentrated sulphuric acid (H₂SO₄) in the water.
What do you observe?

Observation:
In the arrangement shown in the figure, we find that fine microscopic sulphur particles precipitate in water (from solidum thiosulphate) in a couple of minutes.

Blue colour is seen from three sides of the tank due to scattering of short wavelengths by minute sulphur particles. We see transmitted light from the fourth side of the tank facing the hole (C) of the cardboard.

We first observe orange-red colour and then bright crimson red colour on the screen (MN) because the transmitted light contains mainly longer wavelengths only.

Conclusion:
Very fine particles of liquid, mainly scatter blue light of smaller wavelength and the red light containing longer wavelength passes straight (without scattering) from the vessel.

Jharkhand Board JAC Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Additional Questions and Answers

Question 1.
Solve the following examples :
1. An object is placed at a distance of 20 cm in front of a convex mirror having focal length 0.3 m. Find the position, nature and magni-fication of the image formed by the mirror.
Answer:
a = 12 cm, virtual, erect and diminished, m = 0.6

2. The radius of curvature of a concave mirror is 12 cm. The image obtained by it is real and has magnification \(\frac { 3 }{ 2 }\). Find the object distance and image distance.
Answer:
u = – 10cm, v = – 15cm

3. A convex mirror having radius of curvature 3 m is fitted as a rear view mirror in a bus. Find the position, nature and magnification of the image of a car standing behind the bus at 6 m from the mirror.
Answer:
u = 1.2m, virtual, erect and diminished, m = 0.2

4. An object of height 2 cm is placed at 32 cm from a concave mirror. If the image formed by the mirror is real, inverted and 1.5 times the object in size, find the focal length of the mirror and the image distance.
Answer:
f = 19.2 cm, u = – 48cm

5. The refractive index of diamond relative to glass is 1.61. If the refractive index of glass is 1.5, find the absolute refractive index of diamond.
Answer:
n_da = 2.415

6. The absolute refractive index of ice is 1.31. The speed of light in vacuum is 3 x 10⁸ms^-1. Then find the speed of light in ice.
Answer:
v = 2.29 x 10⁸ ms^-1

7. The speed of light is 2 x 10⁸ms^-1 in a glass having refractive index 1.5. In a liquid the ; speed of light is 2.5 x 10⁸ms^-1. Find the absolute refractive index of the liquid.
Answer:
n_1a = 1.2

8. A ray of light incident at an angle of 45° on a rectangular glass slab having refractive index \(\sqrt{2}\) from air. Find the angle of refraction, Take the refractive index of air as 1.
Answer:
r = 30°

9. The image of a candle is formed at 10 cm from a concave lens having focal length 15 cm. At what distance is the candle placed from the mirror?
Answer:
u = – 30 cm

10. A virtual Image is obtained, at 20 cm in front of a lens, when an object is placed at 60 cm in front of the lens. Calculate the focal length s of the lens. State the type of the lens.
Answer:
f = – 30 cm, concave lens

11. The focal length of a lens is – 25 cm. Calculate the power of the lens. Mention the type of the lens.
Answer:
P = – 4 D, concave lens

12. The power of a lens is +5 D. An object is placed at a distance equal to twice the focal length of the lens. (1) Calculate the focal length of the lens. (2) Find the position of the image.
Answer:

1. f = 20cm
2. on the opposite side of the object at a distance of 40 cm from the lens

13. The focal length of a convex lens is 25 cm. The image is obtained at a distance of 75 cm on the other side of the lens. Find the position of the object. What is the nature of the image?
Answer:
u = – 37.5cm; real, inverted and magnified

14. The power of a lens is -4D. A virtual and erect image is obtained at 5 cm from it. Find the position of the object.
Answer:
u = – 6.25 cm

15. A ray of light covers 100 m in vacuum in a certain time. How much distance will it cover in glass of refractive index 1.5 in the same time?
Answer:
\(\frac { 100 m }{ 1.5 }\) = 66.67 m

16. An object is placed at a distance of 25 cm from a concave mirror of focal length 20 cm. Find the image distance and nature of the image.
Answer:
– 100 cm; real, inverted and enlarged

17. An object, 4 cm in height, is placed at 25 cm from a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be kept, so that a sharp image of the object can be obtained on the screen? Find the nature and height of the image.
Answer:
v = – 37.5 cm; size/height = 6 cm; Image is real, inverted and enlarged.

18. A doctor has prescribed a corrective lens of power +1.5 D to a patient. Find the focal length of the lens in centimetre. Is the prescribed lens diverging or converging?
Answer:
f = 67 cm, converging

19. The Sun rays are incident on a concave mirror, parallel to its principal axis. If the image of the Sun is formed at 12 cm distance from the pole of the mirror, find its radius of curvature.
Answer:
24 cm

20. An object is placed at 12 cm in front of a concave mirror. It forms a real and inverted image. Where is the image formed, if the image height is double the object height?
Answer:
In front of the concave mirror at a distance of 24 cm from it.

21. What is the position of the image when an object is placed at 10 cm in front of a concave mirror? The image is virtual and double in size as
compared to the object.
Answer:
Behind the mirror at a distance of 20 cm from it.

22. Find the size of the image formed by a concave lens of focal length 20 cm, when an object having 4 cm height is placed at 60 cm from it.
Answer:
Image size/height = 1.0 cm

23. Determine the nature and size of the image, and the type of mirror for the image formed by mirrors corresponding to magnifications + 1, -1, +0.5, -0.5, +5.0 and -5.0.
Answer:

No.	Magni-fication (m)	Nature and size of Image	Type of Mirror
1.	+ 1	Virtual, erect and of the same size as the object	Plane
2.	– 1	Real, inverted and of the same size as the object	Concave
3.	+ 0.5	Virtual, erect and diminished	Convex
4.	– 0.5	Real, inverted and diminished	Concave
5.	+ 5.0	Virtual, erect and magnified	Concave
6.	– 5.0	Real, inverted and magnified	Concave

24. An object of height 4 cm is placed at 18 cm from a concave mirror having focal length 12cm. Find the position, nature and height of the image.
Answer:
v = – 36 cm, h’ = – 8 cm. Real, inverted and enlarged

25. A convex lens forms a real and inverted image of an object at a distance of 40 cm from it. What is the distance of the object from the lens if the size of the image is the same as that of the object? Determine the power of the lens.
Answer:
u = – 40 cm, P = + 5.0 D

26. An object of height 5 cm is placed at 10 cm from a convex mirror of focal length 15 cm. Find the position, nature and size of the image.
Answer:
v = 6 cm, h’ = 3 cm. Virtual, erect and diminished

27. The refractive index of glass with respect to water is 1.12. Find the absolute refractive index of water if the absolute refractive index of glass is 1.5.
Answer:
1.34

28. Light propagating in a rectangular glass slab enters into water. The refractive index of water with respect to glass is 0.9. The angle of incidence at the surface separating the two media is 26° 48′. Find the angle of refraction. Take sin 26° 48′ = 0.45 approximately.
Answer:
30°

29. An object is placed perpendicular to the principal axis of a concave lens of focal length 30 cm. Find the position of the image when the object is at a distance of 20 cm from the lens.
Answer:
v = – 12 cm

30. The power of a convex lens is +4.0 D. At what distance from the lens should the object be placed to obtain its real and inverted image of the same size as the object on the screen?
Answer:
u = – 50 cm

Question 2.
Give four points of difference between the following terms/quantities:
(1) Concave mirror and Convex mirror
Answer:

Concave mirror	Convex mirror
1. Its inner surface is reflecting.	1. Its outer surface is reflecting.
2. It forms a real or virtual image of an object depending on the position of the object.	2. It always forms a virtual image.
3. The real image can be smaller or of the same size as the size of the object or bigger than the object. The virtual image is always bigger than the object.	3. The image is always smaller than the object.
4. This mirror is used in a torch, headlight of vehicles, search-light, solar cooker, etc.	4. This mirror is used as a rear-view (wing) mirror in vehicles.

(2) Real Image and Virtual Image
Answer:

Real Image	Virtual Image
1. In this case, the rays of light emanating from an object meet after reflection or refraction and hence, an image is formed.	1. In this case, the rays of light emanating from an object (when extended backwards) appear to meet after reflection or refraction.
2. It can be obtained on a screen.	2. It cannot be obtained on a screen.
3. It is inverted compared to an object.	3. It is erect compared to an object.
4. In case of a spherical mirror it is formed towards the reflecting surface in front of the mirror.	4. In case of a spherical mirror it is formed behind the reflecting surface of the mirror i.e., in the mirror itself.

(3) Convex lens and Concave lens
Answer:

Convex lens	Concave lens
1. It is thick at the centre and thin at the edges.	1. It is thin at the centre and thick at the edges.
2. It converges parallel rays incident on it.	2. It diverges parallel rays incident on it.
3. It forms a real or virtual image of an object depending on the position of the object.	3. It always forms a virtual image.
4. The virtual image obtained with it is always enlarged.	4. The virtual image obtained with it is always diminished.

Question 3.
Give scientific reasons for the following statements:
(1) In a headlight, torch and search-light, a concave mirror (of small aperture) is used.
Answer:
In a headlight, torch and search-light, the light source is placed at the principal focus of a concave mirror. As a result, an intense parallel beam of light comes out from a headlight, torch, search-light.

[Note: Actually a parabolic mirror = reflector is used in a headlight, torch and search-light.]

(2) The rear-view mirror (back-view mirror) of the vehicles is a convex mirror.
Answer:
The field view of a convex mirror is very large, i.e., it gives over-all view of vast area of the backside of a vehicle and it forms a virtual, erect and diminished image of a backside object. Hence, the vehicle can be driven safely by viewing backside fully.

(3) A big concave mirror is used in a solar furnace and solar cooker.
Answer:
Parallel rays of light coming from the Sun fall on the concave mirror fitted in a solar s furnace and solar cooker and after reflection, these rays meet at the principal focus F of the concave i mirror. The greater the aperture of the mirror, the greater is its power to collect light. As a result, abundant amount of heat energy is produced.

(4) For shaving as well as for make-up a concave mirror is used.
Answer:
If a concave mirror is held such that the face is between the pole and principal focus of the concave mirror, a virtual, erect and magnified image of the face is obtained behind the mirror. This is of great help in shaving as well as make-up.

[A dentist also uses a concave mirror to obtain an enlarged (magnified) image of teeth.]

(5) It is difficult to pierce a fish swimming in a river.
Answer:
The rays of light from a swimming fish emerge from river – water and enter into air. Thus, they travel from a denser medium to a rarer medium. Therefore they bend away from the normal at the interface.

Hence, the correct/true position of the fish cannot be determined by the observer from air. That is why it is difficult to pierce a fish swimming in a river.

(6) A watch repairer uses a convex lens.
Answer:
If an object is kept between optical centre O and principal focus F of a convex lens, we get a virtual, erect and enlarged image of the object on the same side of the object, beyond the distance 2f from the lens. Thus, by keeping a watch between the lens and its principal focus, the watch repairer can see minute parts of the watch clearly, so that repairing becomes easy.

Objective Questions and Answers

Question 1.
Answer the following questions in one word / sentence :
(1) Can a magnified image be formed by a convex mirror? State Yes or No.
Answer:
No

(2) What is the angle of reflection when a ray c of light falls normally on a mirror?
Answer:
0°

(3) In a ray diagram, how do you draw the normal at point A on a spherical mirror?
Answer:
By drawing a line joining point A on the spherical mirror with the centre of curvature of the mirror.

(4) State the mirror formula to find the focal length of a spherical mirror.
Answer:
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\)

(5) The focal length of a convex mirror is 15cm. What is its radius of curvature?
Answer:
R = 2f = 30 cm

(6) Which two quantities change when a ray of light
Answer:
wavelength and velocity

(7) The magnification produced by concave lens is always less than 1. Why?
Answer:
The image formed by a concave lens (for any position of an object) is always diminished.

(8) Does the refractive index of a medium depend on the frequency of light passing S through it?
Answer:
No. [From n = \(\frac{c}{v}=\frac{f \lambda}{v \lambda^{\prime}}=\frac{\lambda}{\lambda^{\prime}}\) (∵frequency f = constant)]

(9) Which mirror has large vision area?
Answer:
Convex mirror

(10) Absolute refractive index of carbon disulphide is 1.63. Then, explain this statement in context to speed of light.
Answer:
From n = \(\frac { c }{ v }\), v = \(\frac { c }{ n }\) = \(\frac{3 \times 10^8}{1.68}\)
= 1.84 x 10⁸ m s^-1
Hence speed of light in carbon disulphide would be 1.84 x 10⁸ m s^-1.

(11) Between which two points on the principal axis of a concave mirror should an object be placed so that the magnification produced by the mirror is m = – 3?
Answer:
Here, m = – 3. So, the image would be real, inverted and magnified. So, the object should be placed between F and C.

(12) What is mirror?
Answer:
Mirror means smooth and polished surface which reflects (approximate 95 %) most of the light incident on it.

(13) Which lens is used as a magnifying lens?
Answer:
Convex lens

Question 2.
Fill in the blanks :

1. The centre of a spherical mirror is called the ………………… and the centre of a lens is called the …………………
2. The power of a convex lens having focal length 10 cm is ………………… D.
3. When a ray of light travels from water to air, its speed …………………
4. Focal length of a convex lens is …………………
5. ………………… image can be obtained on a screen.
6. Magnification produced by a spherical mirror is positive. So, the type of image is …………………
7. Magnification produced by a lens is negative. So, the type of image is ………………… and …………………
8. A ray of light is incident at an angle of 30° to the surface of a plane mirror. Then the angle between the incident ray and reflected ray must be …………………
9. If a virtual, erect and magnified image of i an object is formed by a convex lens; the object must be at a distance …………………
10. In case of a mirror or lens, the ratio of the image height to object height is called …………………
11. A ………………… lens is thicker at its edges than at its centre.
12. ENT specialist doctor uses a ………………… of small aperture.
13. A ………………… is used as a magnifying glass.
14. If the power of the correcting lens used in spectacles is – 0.4 D, then the type of lens must be …………………
15. ………………… is the instrument used to measure the power of a lens.

Answer:

1. Pole, Optical centre
2. 10
3. increases
4. positive
5. Real
6. erect and virtual
7. inverted and real
8. 120°
9. less than the focal length of the lens
10. magnification
11. concave
12. concave mirror
13. convex lens
14. concave
15. Dioptre meter

Question 3.
State whether the following statements are true or false:

1. A virtual image can be obtained on a screen.
2. In case of irregular reflection, the beam of reflected light remains parallel in a specific direction.
3. Laws of reflection are applicable to any type of mirror.
4. Laws of reflection are not applicable to a rough and irregular surface.
5. The diameter of the reflecting surface of a spherical mirror is known as the aperture of the mirror.
6. At least, four rays are required to obtain the position of an image formed by a mirror.
7. If an object is kept at infinite distance from concave mirror, the image is formed at the centre of curvature of the mirror.
8. Air is optically denser than water.
9. When a ray of light enters from an optically denser medium to an optically rarer medium, its speed decreases.
10. The central part of a thin lens made of glass works as a rectangular glass slab.
11. According to the New Cartesian sign convention for every mirror and lens, Object distance is taken as negative.
12. For a mirror, m = \(\frac { v }{ u }\) and for a lens m = – \(\frac { v }{ u }\)
13. A lens has power P = 5 D. So, its focal length is f = 0.2 cm.

Answer:

1. False
2. False
3. True
4. False
5. True
6. False
7. False
8. False
9. False
10. True
11. True
12. False
13. False

Question 4.
Match the following :
(1)

Column ‘A’	Column ‘B’
1. Concave mirror	p. Negative power
2. Convex lens	q. Always m = positive
3. Convex mirror	r. Always m = negative
4. Concave lens	s. Used for shaving
	t. Simple microscope

Answer:
(1 – s), (2 – t), (3 – q), (4 – p).

(2)

Column ‘A’	Column ‘B’
1. Concave mirror	p. It is used by doctors
2. Convex lens	q. It is used as a rear-view mirror in vehicles
3. Convex mirror	r. Power is negative
4. Concave lens	s. It is used in a compound microscope

Answer:
(1 – p), (2 – s), (3 – q), (4 – r).

Question 5.
Choose the correct option from those given below each question:
1. What is the wavelength range of visible light?
A. 4 x 10^-7 m to 8 x 10^-7 m
B. 4 x 10^-9 m to 8 x 10^-9 m
C. 4 x 10^-5 m to 8 x 10^-5 m
D. 4 x 10^-6 m to 8 x 10^-6 m
Answer:
A. 4 x 10^-7 m to 8 x 10^-7 m

2. What is the relation between the radius of curvature (K) and the focal length (f) of a spherical mirror?
A. R = f/ 2
B. R = f
C. R = 2f
D. R = 3f
Answer:
C. R = 2f

3. Through which of the following points, will a ray passing through the centre of curvature (C) and reflected by a concave mirror pass?
A. Focus
B. Centre of curvature
C. Pole
D. All
Answer:
B. Centre of curvature

4. Where, in front of a concave mirror, should an object be placed to get its virtual and erect image?
A. At principal focus F
B. At centre of curvature C
C. Between the principal focus and the pole
D. Beyond the centre of curvature
Answer:
C. Between the principal focus and the pole

5. The magnification produced by a plane mirror is always
A. more than 1
B. 1
C. less than 1
D. zero
Answer:
B. 1
Hint: m = \(\frac { h’ }{ h }\) = 1 (∵ Here, h’ = h)
or
m = – \(\frac { v }{ u }\)
Here, | v | = | u | and u is negative
while v is positive.
∴ We get, m = 1

6. The focal length of a plane mirror is ……………
A. zero
B. infinity
C. equal to half the object distance f
D. equal to the object distance
Answer:
B. infinity
Hint: In case of plane mirror | v | = | u |, also according to the New Cartesian sign convention, v is positive and u is negative.
Now, from the mirror formula \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), for a plane mirror, as |u| = [u| and u is negative while v is positive, we get \(\frac { 1 }{ f }\) = 0 ∴ f = ∞ (infinity)

7. The distance between the object at 2 m from a plane mirror and its image is ………………..
A. 4 m
B. 1 m
C. 2 m
D. 3 m
Answer:
A. 4 m
Hint: The distance between the object and its image = | u | + | v |
= 2 + 2 (∵ Here, | u | = | v |)
= 4 m

8. Where should an object be placed to obtain its image as real, inverted and of the same height as the object by a convex lens?
A. At the principal focus
B. Between the principal focus and centre of curvature
C. At the centre of curvature
D. Between the optical centre and principal focus
Answer:
C. At the centre of curvature

9. Which of the following materials has maximum optical density?
A. Glass
B. Water
C. Carbon disulphide
D. Diamond
Answer:
D. Diamond
Hint: From the given substances, diamond has highest refractive index.

Question 10.
The absolute refractive index of any medium is …………………
A. 1
B. > 1
C. < 1
D. zero
Answer:
B. > 1

11. Which of the lenses with focal length 10 cm, 20cm, 25cm and 50cm has maximum power?
A. 50 cm
B. 25 cm
C. 20 cm
D. 10 cm
Answer:
D. 10 cm

Hint: From power P = \(\frac { 1 }{ f }\), the lens with less focal length f has more power P. So, the lens with focal length 10 cm has maximum power.

12. What is the focal length of a convex lens having power + 5.0 D?
A. – 10 cm
B. – 20 cm
C. + 10 cm
D. + 20 cm
Answer:
D. + 20 cm
Hint : P = \(\frac { 1 }{ f }\)
∴ f = \(\frac { 1 }{ p }\) = \(\frac { 1 }{ 5 }\) = 0.2 m = + 20 cm

13. If the absolute refractive indices of water, benzene and sapphire are 1.33, 1.50 and 1.77 respectively, then which medium has maximum relative refractive index?
A. Sapphire relative to water
B. Sapphire relative to benzene
C. Benzene relative to water
D. Water relative to benzene
Answer:
A. Sapphire relative to water
Hint:
\(n_{21}=\frac{n_2}{n_1}\)
∴ n₂₁ is maximum, when n₂ is maximum and n₁ is minimum.
Hence, it is obvious that in the present case, medium 1 must be water and medium 2 must be sapphire.
∴ Refractive index of sapphire with respect to water would be maximum.

14. Which type of image is formed by a plane mirror?
A. Real and inverted
B. Real and erect
C. Virtual and erect
D. Virtual and inverted
Answer:
C. Virtual and erect

15. If the absolute refractive indices of water and glass are \(\frac { 4 }{ 3 }\) and \(\frac { 3 }{ 2 }\) respectively, then what is the ratio of the speed of light in water to that in glass?
A. 2
B. \(\frac { 8 }{ 9 }\)
C. \(\frac { 9 }{ 8 }\)
D. \(\frac { 1 }{ 2 }\)
Answer:
C. \(\frac { 9 }{ 8 }\)
Hint: Taking water as medium 1 and glass as medium 2, speed of light in water would be v₁ and speed of light in glass would be v₂.
Here, n₁ = \(\frac { 4 }{ 3 }\) and n₂ = \(\frac { 3 }{ 2 }\)
Now, n₂₁ = \(\frac{n_2}{n_1}=\frac{v_1}{v_2}\)
∴ \(\frac{3 / 2}{4 / 3}=\frac{v_1}{v_2}\)
∴ \(\frac{v_1}{v_2}=\frac{9}{8}\)

16. The absolute refractive indices of water, glass and diamond are 1.33, 1.50 and 2.42 respectively. Which medium has maximum optical density ?
A. Water
B. Glass
C. Diamond
D. Data not enough to draw any conclusion
Answer:
C. Diamond
Hint: The greater the absolute refractive index, the greater is the optical density.

17. Which of the following always form virtual image ?
A. Concave mirror and convex lens
B. Convex mirror and concave lens
C. Convex mirror and convex lens
D. Concave mirror and concave lens
Answer:
B. Convex mirror and concave lens

18. What is the angle of refraction for a light ray incident normal to the surface of a medium?
A. 90°
B. 60°
C. 30°
D. 0°
Answer:
D. 0°
Hint: According to general form of Snell’s law n₁ sin i = n₂ sin r. As i = 0°, then r = 0° because value of n₁ and n₂ are never zero.

19. No matter how far you stand from a mirror, your image appears always erect. The mirror is likely to be ………..
A. plane only.
B. concave only.
C. convex only.
D. either plane or convex.
Answer:
D. either plane or convex.
Hint: In case of both plane and convex mirror for every position of an object its image would be always erect.

20. A virtual image of an object is formed by a mirror, then …………….
A. it would be a convex mirror.
B. it would be a concave mirror.
C. it would be a plane mirror.
D. it would be any one of the mirrors given above.
Answer:
D. it would be any one of the mirrors given above.

21. An object is kept at the centre of curvature of a concave mirror. The distance between the image of the object and pole of the mirror ……………..
A. would be equal to f.
B. would be less than 2f but more than f.
C. would be equal to 2f.
D. would be more than 2f.
Answer:
C. would be equal to 2f.

22. A ray of light travelling in air is obliquely incident on undisturbed horizontal surface of pond water. Inside the water, it…
A. will travel without deviation.
B. will bend away from the normal drawn at the point of incidence.
C. will bend towards the normal drawn at the point of incidence.
D. will turn back following the original path.
Answer:
C. will bend towards the normal drawn at the point of incidence.

23. As shown in the following figure, a light ; ray travels from medium A to medium B. Then refractive index of medium B relative? to medium A would be ……………

A. \(\frac{\sqrt{3}}{\sqrt{2}}\)
B. \(\frac{\sqrt{2}}{\sqrt{3}}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\sqrt{2}\)
Answer:
A. \(\frac{\sqrt{3}}{\sqrt{2}}\)
Hint: From Snell’s law, refractive index of medium B relative to medium A.
n_BA = \(\frac{\sin 60^{\circ}}{\sin 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\)

24. As shown in the following figure, a light ray travels from medium A to medium B. Then refractive index of medium A relative to medium B would be

Answer:
B. < 1

25. Magnification produced by a rear-view mirror fitted in vehicles ………….
A. is always less than 1.
B. is always equal to 1.
C. is always more than 1.
D. can be more than or less than 1 depending upon the position of the object in front of the mirror.
Answer:
B. is always equal to 1.
Hint: Convex mirror is used as rear-view mirror in vehicles. Image formed by it is always virtual, erect and diminished in size compared to the

26. The absolute refractive indices of three substances A, B and C are 1.33, 1.65 and 1.46 respectively. In which substance is the speed of light maximum?
A. A
B. B
C. C
D. Would be the same in A, B and C
Answer:
A. A
Hint: n_m = \(\frac { c }{ v }\)

27. An object placed in front of a convex mirror is moved from far-off distance towards the pole of the mirror. Then size of the image …
A. increases.
B. decreases.
C. remains constant.
D. first increases and then decreases.
Answer:
A. increases.

28. An object placed in front of a convex mirror is moved from far-off distance towards the pole of the mirror. Then the image ……………..
A. moves towards the pole.
B. moves away from the pole.
C. first moves away from the pole and then towards the pole.
D. first moves towards the pole and then away from the pole.
Answer:
A. moves towards the pole.

29. What is the centre of a spherical mirror called?
A. Principal focus
B. Centre of curvature
C. Principal axis
D. Pole
Answer:
D. Pole

30. What is the diameter of the reflecting circular surface of a spherical mirror called?
A. Aperture
B. Radius of curvature
C. Focal length
D. Principal axis
Answer:
A. Aperture

31. Where should the object be placed to obtain its real image of the same size as the object using a concave mirror?
A. At infinite distance from the mirror
B. At the centre of curvature of the mirror
C. At the principal focus of the mirror
D. Between the principal focus and the centre of curvature of the mirror
Answer:
B. At the centre of curvature of the mirror

32. Where should the object be placed to obtain a real and diminished image using a concave mirror?
A. Between the principal focus and the pole of the mirror
B. At the centre of curvature of the mirror
C. Between the principal focus and the centre of curvature of the mirror
D. Beyond the centre of curvature of the mirror
Answer:
D. Beyond the centre of curvature of the mirror

33. Where should the object be placed to obtain a real and magnified image using a concave mirror?
A. Between the pole and the principal focus of the mirror
B. At the principal focus of the mirror
C. Between the principal focus and the centre of curvature of the mirror
D. At the centre of curvature of the mirror
Answer:
C. Between the principal focus and the centre of curvature of the mirror

34. Which type of image cannot be obtained by a concave mirror?
A. Real and enlarged
B. Real and diminished
C. Virtual and enlarged
D. Virtual and diminished
Answer:
D. Virtual and diminished

35. Which type of image is formed by a convex mirror?
A. Real, inverted and diminished
B. Virtual, erect and diminished
C. Real, inverted and enlarged
D. Virtual, erect and enlarged
Answer:
B. Virtual, erect and diminished

36. What happens to an obliquely incident ray travelling from glass to air?
A. It bends towards the normal.
B. It bends away from the normal.
C. It does not undergo refraction.
D. It turns back following the original path.
Answer:
B. It bends away from the normal.

37. A point object emits rays in all directions. Consider one ray that the object emits which is parallel to the principal axis of a concave mirror as shown in the following figure.

Select the point in the figure given here, through which the reflected ray passes.
A. C
B. F
C. A
D. P
Answer:
B. F

38. Which type of image of an object cannot be obtained by a convex lens?
A. Real and diminished
B. Real and enlarged
C. Virtual and diminished
D. Virtual and enlarged
Answer:
C. Virtual and diminished

39. From which of the following letters, we cannot obtain its laterally inverted image?
A. N
B. O
C. P
D. Q
Answer:
B. O

40. From which of the following letters, can we obtain its laterally inverted image?
A. W
B. X
C. Y
D. Z
Answer:
D. Z

41. A boy is running with speed u towards a plane mirror, then at what speed, his image will be running towards him?
A. 2v
B. > v
C. v
D. < v
Answer:
A. 2v

42. If a spherical mirror is immersed in water, then its focal length…
A. changes.
B. does not change.
C. may or may not change
D. becomes zero.
Answer:
B. does not change.

Question 6.
Answer the following questions in very short as directed (Miscellaneous) :
(1) Write Snell’s law in mathematical form.
Answer:
If i is the angle of incidence and r is the angle of refraction, then
\(\frac { sin i }{ sin r }\) = constant
This constant is called the refractive index of the second medium with respect to the first medium.

(2) There are five different transparent media, which are in physical contact with one another. Then, write the formula for the refractive index of the fifth medium with respect to the first medium.
Answer:
n₅₁ = \(\frac{n_5}{n_1}\)

(3) If the radius of curvature of a spherical mirror is 40 cm, what is its focal length in metre?
Answer:
f = \(\frac{R}{2}=\frac{40 \mathrm{~cm}}{2}\) = 20 cm = 0.2 m

(4) A concave mirror produces a real image of an object, placed at 10 cm from the mirror. If the image is enlarged by a factor of four relative to object, find the focal length of the mirror.
Answer:
Here, u = – 10 cm
As image formed is real and four times larger than object m = – 4
Now, m = – \(\frac { v }{ u }\)
∴ – 4 = – \(\frac { v }{ (-10) }\) ∴ v = – 40 cm
Using the mirror formula
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
We get
\(\frac{1}{f}=\frac{1}{(-10)}+\frac{1}{(-40)}\)
∴ f = \(\frac{-40 \times-10}{-40-10}=\frac{400}{-50}\)
∴ f = – 8 cm

(5) The magnification of an image formed by a convex lens is – 1. Find the object distance in terms of its focal length.
Answer:
Here, m = – 1 and u = – u
So, from m = \(\frac { v }{ u }\)
– 1 = \(\frac { v }{ (-u) }\)
∴ v = + u
Now, \(\frac{1}{f}=\frac{-1}{u}+\frac{1}{v}\)
∴ \(\frac{1}{f}=\frac{-1}{(-u)}+\frac{1}{(+u)}\)
∴ \(\frac{1}{f}=\frac{2}{u}\)
∴ u = 2 f

(6) An object is kept at 40 cm from a concave lens of focal length 60 cm. Find the image distance.
Answer:
Here, u = – 40 cm; f = – 60 cm
Using the lens formula,
\(\frac{1}{f}=\frac{-1}{u}+\frac{1}{v}\)
We get
∴ \(\frac{1}{-60}=\frac{-1}{-40}+\frac{1}{v}\)
∴ \(\frac{1}{v}=\frac{-1}{40}-\frac{1}{60}\)
∴ \(\frac{1}{v}=-\left(\frac{1}{40}+\frac{1}{60}\right)\)
∴ v = \(-\left(\frac{60 \times 40}{40+60}\right)=-\frac{2400}{100}\) = – 24 cm

(7) What is diffused reflection ?
Answer:
When a parallel beam of light falls on a rough surface, the light is reflected in different directions. This type of reflection is known as diffused reflection.

(8) The magnifying power of convex lens is ……………….
A. > 1
B. < 1
C. = 1
D. cannot be determined
Answer:
A, B and C

(9) A pencil, when dipped obliquely in water in a transparent glass tumbler, appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use some other liquid like kerosene or turpentine? Support your answer with reason.
Answer:
Yes.
The angle of refraction depends on the refractive index of the medium. So, bending will vary in different liquids. It will be less for a liquid with low refractive index and more for a liquid with high refractive index.

The refractive index of kerosene as well as that of turpentine oil is more than that of water. Hence, the bending will be more for both the oils than in case of water.

(10) Draw a ray diagram showing the path of a ray of light when it enters at oblique incidence from air into water.
Answer:

(11) Draw a ray diagram showing the path of a ray of light when it enters at oblique incidence from water into air.
Answer:

(12) A student traces the path of a ray of light passing through a rectangular glass slab and marks the angle of incidence i, angle of refraction r and angle of emergence e, as shown in the figure :

Which angle is correctly marked?
Answer:
r.

(13) What is the angle of reflection, when a ray of light falls normally on the surface of a plane mirror?
Answer:
r = 0°

(14) A ray of light strikes a plane mirror at an angle of 50° to the mirror surface. What will be the angle of reflection?
Answer:
40°

(15) Name the spherical mirror which has virtual principal focus.
Answer:
Convex mirror

(16) What is the angle of incidence for a ray of light passing through the centre of curvature of a concave mirror?
Answer:
0°

(17) What sign [+ve (plus) or -ve (minus)] has been given to the following on the basis of the New Cartesian sign convention?
(a) Height of a real image and
(b) Height of a virtual image?
Answer:
(a) -ve (minus)
(b) +ve (plus)

(18) What is the nature of the image formed by a concave mirror if the magnification produced by the mirror is (a) + 3 (b) – 2 ?
Answer:
(a) Virtual
(b) Erect

(19) An object is placed at a very long distance in front of a convex mirror of radius of curvature 30 cm. State the position of its image.
Answer:
At the principlal focus; 15 cm behind the convex mirror.

(20) The refractive index of glass with respect to air is 3/2. The refractive index of air with respect to the glass is 4/6. Do you agree with this statement?
Answer:
Yes

(21) Name the following:
A point inside a lens through which the light passes undeviated.
Answer:
Optical centre.

(22) A 2 cm high object is placed at a distance of 2f from a convex lens. What is the height of the image formed?
Answer:
m = \(\frac { v }{ u }\) = \(\frac { h’ }{ h }\)
∴ h’ = h(\(\frac { v }{ u }\))
Here, u = – 2f and also v = – 2f
∴ The height of the image formed,
h’ = 2 cm\(\left(\frac{-2 f}{-2 f}\right)\) (∵ h = 2 cm)
= 2 cm

(23) Where should an object be placed in order to use a convex lens as a magnifying glass?
Answer:
At a distance less than the focal length of the convex lens.

(24) Draw the given diagram in your answer book and complete the path of the ray of light on refraction by the lens.

Answer:

(25) Copy and complete the diagram given below to show what happens to the ray of light when it passes through the concave lens :

Answer:

(26) A concave lens produces an image at 20 cm from the lens of an object placed at 30 cm from the lens. Find the focal length of the lens.
Answer:
Using the lens formula,

(27) Which causes more bending (or more refraction) of light rays passing through it: a convex lens of long focal length or a convex lens of short focal length?
Answer:
A convex lens of short focal length

(28) Which has more power : a thick convex lens or a thin convex lens made of the same material (glass)? Give reason for your choice.
Answer:
Thick convex lens because it has shorter focal length and hence more power.

(29) For a converging lens, the power is positive while for a diverging lens, the power is negative. Agree / Disagree?
Answer:
Agree

(30) The optician’s prescription for a pair of spectacles is :
Right eye : – 3.50 D, Left eye : – 4.00 D
(a) Are these lenses thinner at the middle or at the edges?
(b) Which is the weaker eye?
Answer:
(a) Thinner at the middle
(b) Left eye

(31) A person got his eyes tested by an optician. The prescription for the spectadle lenses to be made reads :
Right eye : + 2.00 D, Left eye : + 2.50 D
(a) State whether these lenses are thicker in the middle or at the edges.
(b) State whether these spectacle lenses will converge light rays or diverge light rays.
Answer:
(a) Thicker in the middle
(b) Converge light rays

(32) Match the following columns correctly:

Column I	Column II
1. Converging mirror	p. It is used as rear view mirror in vehicles
2. Diverging mirror	q. It is used in the design of solar furnaces
	r. It is used in a dressing-table

Answer:
(1 – q), (2 – p).

(33) Match the following columns correctly:

Column I	Column II
1. Converging lens	p. It Is a concave lens
2. Diverging lens	q. It acts as a magnifier
	r. It is used as a reflector

Answer:
(1 – q), (2 – p).

(34) State any two factors on which the lateral displacement of an emergent ray from a rectangular glass slab depend.
Answer:

1. Angle of incidence
2. Thickness of the slab

(35) What is the minimum distance between an object and its real image in case of a concave mirror?
Answer:
0 (zero). [When the object is at centre of curvature C of the mirror.]

(36) When a light ray passes obliquely through the atmosphere in the upward direction, how does its path generally change?
Answer:
Light will bend away from its normal direction.

(37) Explain briefly, why a ray of light passing through the centre of curvature of a concave mirror, gets reflected along the same path.
Answer:
The ray of light passing through the centre of curvature of a concave mirror is incident on the mirror along its normal. So, i = r = 0. Therefore, the ray retraces its path.

(38) A ray of light is incident on a concave mirror as shown in the adjoining figure. Redraw the diagram and complete the path of the ray after reflection from the mirror.

Answer:

(39) A ray of light is incident on a convex mirror as shown in the adjoining figure. Redraw the diagram and complete the path of the ray after reflection from the mirror.

Answer:

(40) What is the unit of refractive index?
Answer:
Refractive index as no unit as it is the ratio of two similar quantities.

(41) State the conditions for no bending of light travelling from one medium to another.
Answer:

1. Light incident normally.
2. Equal refractive index of the two different media.

(42) Why is the refractive index of atmosphere different at different altitudes?
Answer:
The refractive index of atmosphere is different at different altitudes because the air density changes with altitude.

(43) For the same angle of incidence in media P Q and R, the angles of refraction are 45°, 35° and 30° respectively. In which medium is the speed of light minimum? Give reason for your answer.
Answer:
In medium R, the speed of light is minimum.
In the usual notation, as n = \(\frac { c }{ v }\) = \(\frac { sin i }{ sin r }\).
For medium R, r is minimum and hence the speed v is minimum.

(44) Observe the following incomplete ray diagram, where the image A’B’ is formed after refraction from a convex lens :

On the basis of the above information, fill in the blanks :
(1) The position of object AB (not shown in the figure) must be
(2) The size of the object must be than the size of the image.
Answer:
(1) beyond 2F₁
(2) greater.

(45) Raju focussed the image of a candle flame on a white screen using a convex lens. He noted the position of the candle as 26.0 cm, the position of the convex lens as 50.0 cm and the position of the screen as 74.0 cm. What is the focal length of the convex lens?
Answer:
Here, u = the distance of the candle from the convex lens = – (50.0 – 26.0) = – 24.0 cm
v = the distance of the screen from the convex lens = (74.0 – 50.0) = 24.0 cm
From the lens formula,
∴ \(\frac{1}{f}=\frac{-1}{u}+\frac{1}{v}\)
∴ \(\frac{1}{f}=\frac{-1}{(-24)}+\frac{1}{24}=\frac{2}{24}\)
∴ f = 12 cm

(46) Study the given ray diagram. Name the device Y.

Answer:
Device Y is a concave mirror.

(47) Which property of a concave mirror is utilized for using it as a shaving mirror ?
Answer:
When an object is held between the pole and the principal focus of a concave mirror, a virtual, erect and magnified image of the object is formed. This property of a concave mirror is utilized for using it as a shaving mirror.

(48) How do you draw a normal to a spherical mirror at a particular point on the mirror?
Answer:
Join that point on the mirror to the centre of curvature of the mirror and extend the line if required.

(49) The magnification produced by a concave mirror is -1. What is the position of the object?
Answer:
The object must be at the centre of curvature of the concave mirror.
(The image formed is real, inverted and of the same size as the object. That is why magnification = -1).

(50) The magnification produced a spherical mirror is ±2. What kind of mirror can it be?
Answer:
The mirror must be a concave mirror. (Only then magnification can be positive or negative).

(51) Bharat claims to have obtained an image twice the size of the object with a concave lens. Is he correct? Give reason for your answer.
Answer:
Bharat’s claim is not correct.
This is because the image formed by a concave lens is always smaller in size than the object.

(52) Out of alcohol with n= 1.36 and carbon disulphide with n = 1.63 which is denser optically?
Answer:
Carbon disulphide with greater refractive index is optically denser than alcohol.

(53) What is the nature of the image formed by a concave mirror if the magnification produced by the mirror is +3?
Answer:
The image must be virtual, erect, enlarged (three times in size relative to the object) and formed behind the mirror.

(54) With respect to air, the refractive index of ice is 1.31 and that of rock salt is 1.54. Calculate the refractive index of rock salt with respect to ice.
Answer:
n_ia = 1.31 and n_ra = 1.54, n_ri = ?
Now, n_ri = \(\frac{n_{\mathrm{ra}}}{n_{\mathrm{ia}}}=\frac{1.54}{1.31}\) = 1.17

(55) What is the minimum number of rays required for locating the image formed by a concave mirror?
Answer:
At least two rays are required for locating the image formed by a concave mirror.

(56) By changing the object distance the focal length of a given spherical mirror cannot be changed but the linear magnification produced by the mirror can be changed. Agree / Disagree?
Answer:
Agree

(57) Write the correct sequence of angle of incidence, angle of emergence, angle of refraction and lateral displacement shown in the following diagram by numbers 1, 2, 3 and 4.

Answer:
2, 1, 4, 3.

(58) What is the relation between n1 and n2 in the following three ray diagrams (figures)?

Answer:
(a) n₁ < n₂
(b) n₁ = n₂
(c) n₁ > n₂

Important note: In case of a concave lens the corresponding relation between n₁ and n₂ along with, the ray diagrams (figures) is shown below:

(59) Which factors determine the focal length of a lens?
Answer:
The focal length of a lens depends on:

• the radii of curvature of the surfaces of the lens.
• the nature of the material of the lens.
• the nature of the medium in which the lens is placed.

[Note : It also depends upon the frequency of the incident light.]

Value Based Questions With Answers

Question 1.
Two fast friends named Anjalee and Shweta spend most of the time with each other. One S day, Anjalee observed that Shweta has pain in gums while eating her school-lunch. Anjalee’s S father is a dentist. So she advised Shweta to come with her to her father’s clinic.

Anjalee’s father examined Shweta’s mouth and teeth with a mirror and light and advised her not to eat too many chocolates and drink soft drinks, Then she started taking care of her mouth as she washed her mouth properly every time after having food and also started taking calcium-rich diet.

1. Which type of mirror is used by the dentist?
2. Name the phenomenon of light by which the doctor is able to examine Shweta.
3. What values are shown by Anjalee?

Answer:

1. Concave mirror
2. Reflection of light
3. Friendship
4. Concern for other
5. Helping nature

Question 2.
In a small town-fair, Amey took his friend and showed him a mirror in which his image showed the upper half of the body very fat and lower half of the body very thin. Amey’s friend got upset but Amey explained him by showing his similar image in the mirror.

1. Name the two mirrors used in this fair shop.
2. Name the mirror in which the size of image is small.
3. What values are displayed by Amey?

Answer:

1. Concave and convex mirror
2. Convex mirror
3. Compassion and empathy.

Question 3.
Prashant was going to his office in his car. While driving his car, Prashant saw a man behind him on a motorcycle through his rear-view mirror. A woman was also sitting behind the man on the motercycle. Through his rear-view mirror, Prashant noticed that the saree of the woman was almost touching the spokes of the motorcycle wheel. He signalled the motorcyclist to stop and alerted the woman. She tied her saree properly and thanked Prashant for the s alert.
(1) What type of mirror is used as a rear-view mirror?
(2) State two characteristics of the image formed by such a mirror.
(3) What values are displayed by Prashant in this incident?
Answer:
(1) Convex mirror

(2)

• Virtual and erect
• diminished (smaller than the object)

(3) The values displayed by Prashant in this incident are …

• Vigilant (because he kept careful watch for a likely mishap around him)
• Concern about others (here, he was concerned about the safety of the woman)
• Responsible citizen (because, he stopped the motorcycle and alerted the woman about an impending danger to her life).

Question 4.
On the way from Ahmedabad to Vadodara there were four friends in a car. Sanjay was driving the car. He saw from his side mirror that the car which was behind their car had met with an accident.

He suddenly applied the brakes, even after his friends asked him to leave the situation as it is. But still Sanjay did not agree and got down from the car and persuaded his friends to help the injured. All of them took the injured person to the nearest hospital. After taking first-aid from the hospital, the victim thanked them for saving his life.
(1) Name the type of mirror from which Sanjay saw the accident.
(2) Why is this type of mirror used as a side mirror in vehicles?
(3) What can you learn from Sanjay’s character?
Answer:
(1) A convex mirror

(2) Convex mirrors have a wider field of view, as they are curved outwards. Therefore, a convex mirror enables the driver to view much larger area.

(3) Sanjay is of helping nature.
We learn to help needy people.

Question 5.
At one of the blind ends of a road in a colony, lot of accidents used to take place. One day, Shalini moved into a house near ; the blind end. On a certain day, she saw an accident taking place at the blind end.

She, at once rushed to the spot and helped the victims with the first-aid. She then telephoned for an ambulance. Later on she met the president of the welfare committee of the colony and requested him to install huge convex mirrors at both the ends of the blind end. After the installation of the mirrors the accident rate suddenly dropped.

1. Why did Shalini advise the installation of huge convex mirrors at both ends of the blind end?
2. What according to you are the values displayed by Shalini?
3. What according to you are the values displayed by president of the welfare committee of the colony?

Answer:

1. A convex mirror has a wide range of view. As a result, it is capable of showing the traffic on the other side or the blind end. Since, she was aware of this fact, Shalini advised for their installation.
2. The values displayed by Shalini are concern for others and right use of knowledge.
3. The values displayed by the president of the welfare committee of the colony are understanding nature, sense of duty and concern for the safety of others.

Practical Skill Based Questions With Answers

Question 1.

(1) In the figure, u = ……………. and f = ……………..
(2) For the image of the object AB, v = ……………. and h’ = …………….
(3) Find magnification m.
Answer:
(1) – 30 cm, +15 cm
because from the given figure,
u = 2f₁ = – 2 x 15 = – 30 cm and
f = + 15 cm as the given lens is convex.

Question 2.

(1) In the figure, f = …………….. and u = ………………
(2) State the type of image.
(3) Will the value (modulus) of magnification be 1 or less than 1 or more than 1?
Answer:
(1) – 12 cm, – 24 cm
From the given figure, f = – 12cm.
(∵ the given lens is concave) and u = – 2f₁ = – 2 x 12 = – 24 cm
(∵ (Real) object distance is always negative.)

(2) Type of image : virtual, erect and diminished

(3) Less than 1

Question 3.

As shown in the figure, three different reflected rays are obtained from three different mirrors. Carefully study the given diagrams and answer the following questions:

1. Which one of the given mirrors is convex?
2. Which of the mirrors will (i) reflect light as a parallel beam of light (ii) converge the parallel beam when a parallel beam of light is incident on it?
3. Which one of the given mirrors will always produce an image equal in size to that of the object?
4. Which mirror will form an image smaller than object size?
5. Which mirror will form an image smaller or bigger than the object depending on the object distance?
6. Which mirror is used (i) in a dressing table, (ii) as a rear-view mirror in vehicles (iii) by a dentist?

Answer:

1. Mirror B
2. Mirror A will reflect light as a parallel beam of light and mirror C will converge a parallel beam of light.
3. Mirror A
4. Mirror B
5. Mirror C
6. Mirror A is used in a dressing table, mirror B is used as a rear-view mirror in vehicles and mirror C is used by a dentist.

Question 4.
For a spherical mirror, obtain the equation for magnification m in terms of (1) f and u only (2) f and v only.
Answer:
(1) Mirror formula : = \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
Multiply both the sides by – u
∴ \(\frac{-u}{f}=\frac{-u}{v}+\frac{-u}{u}\)
∴ \(\frac{-u}{f}=\frac{-u}{v}\) – 1
∴ \(\frac{-u}{v}=1-\frac{u}{f}\)
∴ \(\frac{-u}{v}=\frac{f-u}{f}\)
∴ \(\frac{-v}{u}=\frac{f}{f-u}\)
∴ m = \(\frac{f}{f-u}\) (∵ m = – \(\frac{v}{u}\))

(2) Mirror formula : \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
Multiply both the sides by – v
∴ \(\frac{-v}{f}=\frac{-v}{v}+\frac{-v}{u}\)
∴ \(\frac{-v}{f}=-1-\frac{v}{u}\)
∴ \(\frac{-v}{u}=1-\frac{v}{f}\)
∴ \(\frac{-v}{u}=\frac{f-v}{f}\)
∴ \(\frac{-v}{u}=\frac{f-v}{f}\)
∴ m = \(\frac{f-v}{f}\) (∵ m = – \(\frac{v}{u}\))

Question 5.
For a lens, obtain the equation for magnification m in terms of (1) f and u only (2) f and v only.
Answer:
(1) Lens formula : \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
Multiply by u on both the sides,
∴ \(\frac{u}{f}=\frac{u}{v}-\frac{-u}{u}\)
∴ \(\frac{u}{f}=\frac{u}{f}\) + 1
∴ \(\frac{u}{v}=\frac{u+f}{f}\)
∴ \(\frac{v}{v}=\frac{f}{f+u}\)
∴ m = \(\frac{f}{f+u}\) (∵ m = \(\frac{v}{u}\))

(2) Lens formula : \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
Multiply by v on both the sides,
∴ \(\frac{v}{f}=\frac{v}{v}-\frac{v}{u}\)
∴ \(\frac{v}{f}= 1-\frac{v}{u}\)
∴ \(\frac{v}{u}=1-\frac{v}{f}\)
∴ \(\frac{v}{u}=\frac{f-v}{f}\)
∴ m = \(\frac{f-v}{f}\) (∵ m = \(\frac{v}{u}\))

Question 6.
In case of a convex mirror, if an object is kept at a distance equal to its focal length, then the height of the image is half the height of the object. Prove this result from the formula for magnification only.
Answer:
Here u = – f
Now, m = \(\frac { f }{ f-u }\)
∴ m = \(\frac { f }{ f-(-f)}\)
m = \(\frac { f }{ 2f }\)
= \(\frac { 1 }{ 2 }\)
Now, m = \(\frac { h’ }{ h }\)
∴ \(\frac { h’ }{ h }\) = \(\frac { 1 }{ 2 }\)
∴ h’ = \(\frac { h }{ 2 }\)

Question 7.
The focal length of a convex lens is +10 cm and the focal length of a concave lens is -10 cm. If two objects of equal height are kept on the principal focus of both the lenses, then find the position of the image formed in each case.
Answer:
In case of a convex lens, f = + 10 cm,
u = – 10 cm
Now, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
∴ \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\)
\(\frac{1}{+10}+\frac{1}{-10}\)
= 0
∴ v = ∞
Hence, the image formed by the convex lens would be at infinite distance from the lens.
In case of concave lens, f = – 10cm, u = – 10 cm
Now, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
∴ \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\)
= \(\frac{1}{-10}+\frac{1}{-10}\)
= \(\frac{(-10)+(-10)}{-10 \times-10}\)
= \(\frac { -20 }{ 100 }\)
= – \(\frac { 1 }{ 5 }\)
Hence, the image formed by the concave lens would be exactly between principal focus F and optical centre O of the concave lens on the same side as that of the object.

Question 8.
If p, q and r denote the object distance, image-distance and the radius of curvature of a spherical mirror, then prove, r = \(\frac{2 p q}{p+q}\).
Answer:

Question 9.
A real, inverted image of the size of the object is to be formed by holding the object at 1 metre from a convex lens. What should be the focal length of the lens? Draw the ray diagram.
Answer:
When the size of the image = the size of the object, the distance of the object from a convex lens, u = 2f where f is focal length of a convex lens.
Here, u = 2f = 1 metre (given).
∴ f = \(\frac { 1 }{ 2 }\) m = 0.5 m = 50 cm

The course of rays in this case is as shown in the figure above.

Question 10.
AB and CD, two spherical mirrors, form parts of a hollow spherical ball with its centre at O, as shown in the diagram. If arc AB = \(\frac { 1 }{ 2 }\) arc CD,

(1) What is the ratio of their focal lengths?
(2) State which of the two mirrors will always form a virtual image of an object placed in front of it and why?
Answer:
(1) f = \(\frac { R }{ 2 }\) The radius of curvature is the same in two cases. Hence, numerically, the focal length |f| will be the same. But the focal length of a convex mirror (AB) is positive while that of a concave mirror (CD) is negative. Hence the ratio of the focal lengths will be – 1.
[Note: Ratio of numerical value of the focal lengths will be 1.]

(2) Mirror AB will always form a virtual image as mirror AB is a diverging / convex mirror.

Question 11.
What is lateral displacement? Show that a ray incident obliquely on a rectangular glass slab, emerges parallel to the incident ray.
Answer:

1. Place a glass slab on a plane sheet of paper fixed on a drawing board.
2. Draw its boundary.
3. Fix two pins A and B on one side of the slab.
4. Look through the opposite side of the slab and fix two pins C and D in such a way that all the four pins
5. A, B, C and D appear to be in a straight line.
6. CD is the emergent ray. Let it subtend an angle e with the normal.
7. In the first refraction from medium a (air) to medium b (glass), i is the angle of incidence and r is the angle of refraction.
   

Refraction through a rectangular glass slab of thickness: A lateral displacement In the second refraction at E, the angle of incidence in medium b is r and the angle of refraction in medium a is e.
n_ab = \(\frac{\sin r}{\sin e}\) … (2)
From (1) and (2), we have
n_ba x n_ab = \(\frac{\sin i}{\sin r} \times \frac{\sin r}{\sin e}\)
Butn_ba x n_ab = 1, according to the principle of reversibility.
Hence, 1 = \(\frac{\sin i}{\sin e}\)
∴ sin i = sin e
∴ i = e
Thus, the angle of emergence is equal to the angle of incidence. Therefore, incident ray produced (or extended) i.e., AOd is parallel to emergent ray CD.

Thus, the emergent ray is parallel to the incident ray, but displaced sideways as shown in the figure. This sideways displacement FE is known as lateral displacement.

Question 12.
Identify the device used as a spherical mirror or lens in the following cases, when the image formed is virtual and erect in each case.

1. An object is placed between the device and its principal focus. The image formed is enlarged and behind the device.
2. The object is placed between the principal focus and device. The image formed is enlarged and on the same side as that of the object.
3. The object is placed at finite distance from the device (i.e., between infinity and device). The image formed is diminished and lies between the principal focus and optical centre and on the same side as that of the object.
4. The object is placed at finite distance from the device (i.e., between infinity and device). The image formed is diminished and lies between the pole and principal focus as well as behind it.

Answer:

1. A concave mirror.
2. A convex lens.
3. A concave lens.
4. A convex mirror.

Question 13.
Sudha finds that a sharp image of the s windowpane of her science laboratory is formed at 15 cm from the lens. She now tries to focus the building visible to her outside the? window instead of the windowpane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of s the building? What is the approximate focal length of this lens?
Answer:
To obtain a sharp image of the building instead of the windowpane, Sudha will have to move the screen slightly towards the lens. The approximate focal length of this lens is 15 cm.

Question 14.
Under what condition in an arrangement of two plane mirrors, will an incident ray and the reflected ray be always parallel to each other? (whatever may be the angle of incidence) Show the same with a diagram.
Answer:
When two plane mirrors are placed at 90° to each other, then the incident and reflected rays will always be parallel to each other, whatever may be the angle of incidence. The course of rays is shown in Fig. for i = 45°.

[Note: Students can repeat this activity for different angles of incidence to verify the answer given above.]

Question 15.
Study the diagram given below carefully and answer the following questions :

1. Due to which phenomenon does the path of a ray of light get deviated?
2. Name the device X.
3. What is point A called ?
4. Which quantity is the inverse of the distance between device X and point A?
5. State the SI unit of the physical quantity mentioned as the answer to question (4).

Answer:

1. Refraction
2. Convex lens
3. Principal Focus
4. Power of a lens
5. D (dioptre)

Memory Map:

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 11 रचनाएँ Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 11 रचनाएँ

प्रश्न 1.
7.5 सेमी. रेखाखण्ड को 2 : 3 के अनुपात में विभाजित कीजिए। केवल चित्र बनाइए।
हल:
AP : PB = 2 : 3 होगा।

प्रश्न 2.
एक समकोण त्रिभुज ABC खींचिए, जिसमें BC = 12 सेमी., AB = 5 सेमी. और ∠B = 90° है। इस त्रिभुज के समरूप एक त्रिभुज की रचना कीजिए, जिसका स्केल गुणक \(\frac{2}{3}\) हो। क्या नया त्रिभुज भी क त्रिभुज है।
हल:
रचना के चरण :

• BC = 12 सेमी. की एक रेखा खींची।
• BC के बिन्दु B से 90° का कोण बनाती हुई BX रेखाखण्ड खींची।
• रेखाखण्ड BX से AB = 5 सेमी. काटा तथा AC को मिलाया। इस प्रकार ABC प्राप्त हुआ।
• रेखाखण्ड BC के नीचे की ओर B बिन्दु से न्यून कोण बनाती हुई BY किरण खींची।
• किरण BY के तीन बराबर भाग इस प्रकार किए कि BB₁ = B₁B₂ = B₂B₃।
• B₃ को C से मिलाया।
• B₂ से B₃C के समान्तर रेखा B₂B’ खींची जो BC को B’ पर मिलती है।
• B’ से AB के समान्तर रेखा A’B’ खींची जो AC से A’ पर मिलती है।

इस प्रकार ΔA’B’C, अभीष्ट समरूप त्रिभुज है।

औचित्य (उपपत्ति) : ΔBB’B₂ तथा ΔBCB₃ में
∠B = ∠B (उभयनिष्ठ)
∠BB₂B’ = ∠BB₃C (रचना से)
ΔBB’B₂ ~ ΔBCB₃ (AA समरूपता कसौटी से)
\(\frac{B B^{\prime}}{B C}=\frac{B B_2}{B B_3}\)
(समरूप त्रिभुज की संगत भुजाएँ समानुपाती होती है)

∠A’BC = ∠ABC = 90° (रचना से)
अतः ΔA’B’C की भुजाएँ ΔABC की संगत भुजाओं \(\frac{2}{3}\) गुनी होंगी तथा ΔA’B’ C समकोण Δ होगी।

प्रश्न 3.
3 सेमी त्रिज्या के एक वृत्त पर 5 सेमी. त्रिज्या के एक संकेन्द्रीय वृत्त के किसी विन्दु से एक स्पर्श रेखा की रचना कीजिए और उसकी लम्बाई मापिए।
हल:
दिया है : 3 सेमी त्रिज्या का एक वृत्त और 6 सेमी त्रिज्या का एक संकेन्द्रीय वृत्त जिस पर एक बिन्दु माना P दिया गया है।

रचना के चरण :

• 3 सेमी त्रिज्या लेकर केन्द्र O वाला एक वृत्त खींचा।
• केन्द्र O से 5 सेमी त्रिज्या का एक संकेन्द्रीय वृत्त खींचा और इस पर एक बिन्दु P लिया।
• रेखाखण्ड OP खींचा और इसका लम्ब समद्विभाजक खींचा जो OP को बिन्दु M पर काटता है।
• बिन्दु M को लेकर MP त्रिज्या का एक वृत खींचा जो केन्द्र O के 3 सेमी त्रिज्या वाले वृत्त को T₁ और T₂ बिन्दुओं पर काटता है।
• PT₁ और PT₂ को मिलाया जो वृत्त की अभीष्ट स्पर्श रेखाएँ हैं।

उपपत्ति: हम जानते हैं स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∠OT₁P = ∠OT₂P = 90°
OT₁ तथा OT₂ को मिलाया, OP वृत्त का व्यास है।
∠OT₁P तथा ∠OT₂P अर्द्धवृत्त के कोण हैं।
∠OT₁P = 90° तथा ∠OT₂P = 90°
OT1 ⊥ PT1 तथा OT2 ⊥ PT2
अत: PT₁ तथा PT₂ अभीष्ट स्पर्श रेखाएँ हैं।
लम्बाई मापने पर-
परिकलन : स्पर्श रेखा

प्रश्न 4.
8 सेमी. लम्बी एक रेखाखण्ड AB खींचिए। A को केन्द्र मानकर 4 सेमी त्रिज्या पर एक वृत्त बनाइये तथा बिन्दु B से इस वृत्त पर स्पर्श रेखा युग्मों की रचना कीजिए एवम् उनकी लम्बाइयाँ मापिए।
हल:
रचना के चरण :

• 8 सेमी. लम्बाई का AB रेखाखण्ड खींचा।
• AB के A बिन्दु से 4 सेमी. त्रिज्या का एक वृत्त खींचा।
• AB का समद्विभाजक खींचा तथा समद्विभाजक बिन्दु को M अंकित किया।
• M को केन्द्र मानकर MA त्रिज्या का वृत्त खींचा जो A केन्द्र वाले वृत्त को T₁ तथा T₂ पर काटता है।
• BT₁ तथा BT₂ को मिलाया BT₁ तथा BT₂ अभीष्ट रेखाएँ हैं। नापने पर स्पर्श रेखाओं की लम्बाई = 6.93 सेमी।

औचित्य (उपपत्ति) : AT₁ तथा AT₂ को मिलाया।
∵ हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर जाने वाली त्रिज्या पर लम्ब होती है।
अत: ∠AT₁B = ∠AT₂B = 90°
AB, M केन्द्र वाले वृत्त का व्यास है।
∠AT₁B = ∠AT₂B = 90°
(अर्द्धवृत्त में बनी कोण समकोण होता है)
अत: BT₁ तथा BT₂ अभीष्ट स्पर्श रेखाएँ हैं।

प्रश्न 5.
4 सेमी, 5 सेमी और 6 सेमी भुजाओं वाले एक त्रिभुज की रचना कर इसके समरूप एक अन्य त्रिभुज की रचना कीजिए जिसकी भुजाएँ दिये गये त्रिभुज की संगत भुजा की \(\frac{3}{5}\) गुनी हो।
हल:
माना ΔABC है जिसमें AB = 5 सेमी, AC = 4 सेमी, BC = 6 सेमी।

रचना के चरण :

• एक रेखाखण्ड BC 6 सेमी खींचा।
• B को केन्द्र मानकर 5 सेमी त्रिज्या का चाप लगाया।
• C को केन्द्र मानकर 4 सेमी. त्रिज्या लेकर एक चाप लगाया जो पहले चाप को A बिन्दु पर काटता है।
• AB और AC को मिलाया ΔABC वांछित त्रिभुज है।
• आधार BC के नीचे की ओर कोई न्यून कोण बनाती BX किरण खींची।
• BX किरण के पाँच बराबर भाग इस प्रकार किए कि BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
• B₅C को मिलाया। B₃ से B5C के समान्तर B₃C’ रेखा खींची जो B₃ से C’ पर मिलती है।
• C’ से AC के समान्तर A’C’ रेखा खींची जो AB को A’ पर मिलती है।

अतः A’BC’ अभीष्ट समरूप त्रिभुज होगा।

औचित्य (उपपत्ति) : ΔBB₃C तथा BB₅C में
∠B = ∠B (उभयनिष्ठ)
∠BB₃C’ = ∠BB₅C (रचना से)
ΔBB₃C’ ~ ΔBB₅C (AA समरूपता कसौटी से)
\(\frac{B C^{\prime}}{B C}=\frac{B B_3}{B B_5}\)
[समरूप त्रिभुजों की संगत भुजाएँ समानुपाती होती हैं]

A’B = \(\frac{3}{5}\)AB, A’C’ = \(\frac{3}{5}\)AC तथा BC’ = \(\frac{3}{5}\)BC
अत: ΔA’BC’ की भुजाएँ ABC की संगत भुजाओं \(\frac{3}{5}\) की हुँ गुनी होंगी।

प्रश्न 6.
त्रिज्या 5 सेमी का वृत्त खींचिए। वृत्त के केन्द्र से 13 सेमी दूरी पर स्थित किसी बिन्दु से वृत्त की स्पर्श रेखाएं खींचिए। स्पर्श रेखाओं की लम्बाई नापिए तथा गणना करो व औचित्य भी दीजिए।
हल:
रचना के चरण :

• सर्वप्रथम 5 सेमी का वृत्त खींचा जिसका केन्द्र O है।
• बिन्दु ‘O’ से 13 सेमी दूरी पर बिन्दु P लिया, OP को मिलाया।
• PO को सम द्विभाजित किया सम द्विभाजक को M अंकित है।
• M केन्द्र मानकर PM त्रिज्या का एक वृत्त खींचा जो O केन्द्र वाले वृत्त को A और B बिन्दुओं पर काटता है।
• PA और PB को मिलाया अतः PA और PB अभीष्ट स्पर्श रेखाएँ हैं। स्पर्श रेखा को नापने पर लम्बाई = 12 सेमी हैं।

गणना द्वारा लम्बाई ज्ञात करना : ΔPOA में,
∠PAO = 90° (प्रमेय 10.1 से)
अत: ΔPOA एक समकोण त्रिभुज है, अत: पाइथागोरस प्रमेय के अनुसार,
PO² = AO² + PA²
PA² = PO² + AO²
= (13)² – (5)²
= 169 – 25
PA = 144
PA = \(\sqrt{144}\)
= 12 सेमी
अतः नापने और गणना द्वारा स्पर्श रेखाओं की लम्बाई 12 सेमी।

औचित्य (उपपत्ति) :
∵ हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती हैं।
अतः ∠PAO = 90° तथा ∠PBO = 90°
∵ OA, OB की मिलाया, OP वृत्त का व्यास हैं।
∠PAO व ∠PBO अर्द्धवृत्त में बने कोण हैं।
∴ ∠PAO = 90°, ∠PBO = 90°
अत: PA व PB अभीष्ट स्पर्श रेखाएँ हैं।

प्रश्न 7.
4 सेमी. त्रिज्या का एक वृत्त खींचिए। इस पर स्पर्श रेखाओं के ऐसे युग्म की रचना कीजिए कि इनके बीच का कोण 60° हो। रचना का औचित्य भी दीजिए। वृत्त के केन्द्र और स्पर्श रेखाओं के प्रतिच्छेद बिन्दु के बीच की दूरी को मापिये।
हल:
रचना के चरण :

• O को केन्द्र मानकर 4 सेमी. त्रिज्या वाला वृत्त खींचा।
• वृत्त का व्यास AB खींचा।
• त्रिज्या AO के बिन्दु O पर 60° का कोण बनाती हुई एक रेखा OC खींची जो वृत्त को C बिन्दु पर काटती है।
• OC रेखाखण्ड के बिन्दु C से 90° का कोण बनाती हुई स्पर्श रेखा खींची।
• OB के बिन्दु B से 90° का कोण बनाती हुई दूसरी स्पर्श रेखा खींची जो पहली स्पर्श रेखा को D बिन्दु पर मिलती है।

अत: CD तथा BD वृत्त की स्पर्श रेखाएँ हैं जो एक दूसरे के साथ 60″ का कोण बनाती है। मापने पर केन्द्र और स्पर्श रेखाओं के प्रतिच्छेद बिन्दु के बीच की दूरी = 8 सेमी.।
औचित्य (उपपत्ति) :
∠AOC + ∠BOC = 180° (रैखिक समीकरण युग्म)
⇒ 60° + ∠BOC = 180°
⇒ ∠BOC = 180° – 60° = 120°
∵ OB तथा OC वृत्त की त्रिज्याएँ हैं तथा CD व BD वृत्त पर स्पर्श रेखाएँ हैं।
∴ ∠OCD = ∠OBD = 90° (प्रमेय 10.1 से)
अब ∠BOC + ∠OCD + ∠BDC + ∠OBD = 360°
⇒ 120° + 90° + ∠BDC + 90° = 360°
⇒ 300° + ∠BDC = 360°
⇒ ∠BDC = 360° – 300°
= 60°
अंतः CD व BD वृत्त की स्पर्श रेखाएँ हैं।

प्रश्न 8.
5 सेमी त्रिज्या के एक वृत्त की रचना कीजिए। वृत्त के बाहर स्थित बिन्दु P से केन्द्र का उपयोग किए बिना वृत्त की स्पर्श रेखाओं की रचना कीजिए तथा औचित्य भी दीजिए।
हल:
दिया है : एक वृत्त और इसका बाह्य बिन्द P हैं। वृत का केन्द्र अज्ञात है।
रचना के चरण :

•  वृत्त की छेदक रेखा PAB खींची।
• PB को समद्विभाजित किया और इसके मध्य-बिन्दु M से MP = MB त्रिज्या का अर्द्धवृत खींचा।
• बिन्दु A पर लम्ब AC खींचा जो अर्द्ध वृत्त को C बिन्दु पर मिलता है।
• P को केन्द्र मानकर PC त्रिज्या के चाप खींचे जो वृत्त को Q और R पर प्रतिच्छेद करते हैं।
• PR और PQ को मिलाया। अब PQ और PR अभीष्ट स्पर्श रेखाएँ हैं।

औचित्य (उपपत्ति) : PB को व्यास मानकर अर्द्धवृत PCB खींचा और PB के बिन्दु A पर AC ⊥ PB पर,
PC² = PA.PB
(∵ PC त्रिज्या के चाप R व Q हैं)
∴ PR² = PA.PB (PC = PR = PQ)
PQ² = PA.PB
अत: PR और PQ अभीष्ट स्पर्श रेखाएँ हैं।

प्रश्न 9.
4.5 सेमी त्रिज्या के एक वृत्त पर ऐसी दो स्पर्श रेखाएं खींचिए जो परस्पर 45° का कोण बनाती हों। औचित्य भी दीजिए।
हल:
रचना के चरण :

• O को केन्द्र मानकर 4.5 सेमी त्रिज्या का वृत्त खींचा।
• वृत्त का व्यास AB खींचा।
• बिन्दु O पर OA से 45° का कोण बनाती हुई एक रेखा OC खींची जो वृत्त को C बिन्दु पर काटती है।
• बिन्दु B पर OB के लम्बवत् रेखा खींची तथा विन्दु C पर OC के लम्बवत् एक रेखा खींची। दोनों रेखाएँ एक-दूसरे को P बिन्दु पर काटती हैं।

अत: PB और PC वृत्त की दो अभीष्ट स्पर्श रेखाएँ हैं जो एक-दूसरे से 45° का कोण बनाती हैं।
औचित्य (उपपत्ति) : माना वृत्त का केन्द्र O है। PB व PC स्पर्श रेखाएँ हैं।
इनके बीच का कोण BPC = 45° है।
∠BOC = 180° – 45° = 135°
∠AOC = 180° – ∠BOC (रैखिक समीकरण युग्म से)
= 180° – 135°
= 45°.

प्रश्न 10.
8.5 सेमी लम्बा एक रेखाखण्ड AB खींचिए। A को केन्द्र मानकर 5 सेमी त्रिज्या का एक वृत्त तथा B को केन्द्र मानकर 2 सेमी त्रिज्या का एक अन्य वृत्त खींचिए। प्रत्येक वृत्त पर दूसरे वृत्त के केन्द्र से स्पर्श रेखाओं की रचना कीजिए।
हल:
केन्द्र A से 5 सेमी त्रिज्या का वृत्त खींचा गया है तथा B को केन्द्र मानकर 2 सेमी त्रिज्या का वृत्त खींचा गया है।
रचना के चरण :

• सर्वप्रथम AB रेखाखण्ड 8.5 सेमी खींचा।
• A को केन्द्र मानकर 5 सेमी त्रिज्या का एक वृत्त खींचा और केन्द्र B से 2 सेमी त्रिज्या का वृत्त खींचा।
• रेखाखण्ड AB का समद्विभाजक किया जो कि AB को M बिन्दु पर काटता है।
• बिन्दु M को केन्द्र मानकर MA त्रिज्या का एक वृत्त खींचा जो 4 केन्द्र वाले वृत्त को S वT बिन्दुओं पर तथा B केन्द्र वाले वृत्त को P और Q बिन्दुओं पर काटता है।
• SB, TB, PA व QA को मिलाया।

अत: SB और TB केन्द्र A वाले वृत्त की स्पर्श रेखाएँ हैं तथा PA और QA केन्द्र B वाले वृत्त की स्पर्श रेखाएँ हैं।
औचित्य (उपपत्ति) : ∵ हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∠ASB = ∠ATB = 90°
और ∠APB = ∠AQB = 90°
M केन्द्र वाले वृत्त का व्यास AB है।
∠ASB, ∠ATB अर्द्धवृत्त में बने कोण हैं।
∴ ∠ASB = 90°, ∠ATB = 90°.
∴ SB और TB केन्द्र A वाले वृत्त की स्पर्श रेखाएँ हैं।
इसी प्रकार ∠APB और ∠AQB अर्द्धवृत्त में बने कोण हैं।
∴ ∠APB = 90°, ∠AQB = 90°,
∴ PA और QA केन्द्र B वाले वृत्त की स्पर्श रेखाएँ हैं।

प्रश्न 11.
5 सेमी भुजा वाले समबाहु त्रिभुज ABC की रचना कीजिए। फिर एक अन्य त्रिभुज की रचना कीजिए जिसकी भुजाएँ दिए हुए त्रिभुज ABC की संगत भुजाओं \(\frac{2}{3}\) की गुनी हों।
हल:
रचना के चरण :

• एक रेखाखण्ड BC = 5 सेमी खींचिए।
• बिन्दु B को केन्द्र मनाकर 5 सेमी त्रिज्या लेकर लगाइए।
• इसी प्रकार, बिन्दु C को केन्द्र मनाकर 5 सेमी त्रिज्या एक अन्य चाप लगाइए, जो बिन्दु B से लगे चाप के है। यह प्रतिच्छेदित बिन्दु A है।
• A से C को मिलाइये। अतः एक समबाहु त्रिभुज BC रचना हो गई।
• BC से शीर्ष A के दूसरी ओर न्यूनकोण बनाती किरण XY खींचिए।
• 3 बिन्दु B₁, B₂, B₃ किरण BY पर इस प्रकार कीजिए कि BB₁ = BB₂ = B₂B₃ है।
• B₃ को C से मिलाइए।
• बिन्दु B₂ से, B₂D || B₃C खींचिए।
• बिन्दु D से, DE || CA खींचिए।
  तब ΔEBD अभीष्ट त्रिभुज है, जिसकी भुजाएँ त्रिभुज की संगत भुजाओं \(\frac{2}{3}\) की गुनी है।

प्रश्न 12.
2 सेमी त्रिज्या के वृत्त पर 5 सेमी त्रिज्या का एक संकेन्द्री वृत्त खींचिए। बाह्य वृत्त पर लिए गए एक बिन्दु P में छोटे वृत्त पर दो स्पर्श रेखाओं PA तथा PB की रचना कीजिए। PA की लम्बाई मापिए।
हल:
रचना के चरण :

• एक 2 सेमी त्रिज्या का वृत्त खींचिए। अब इस O से एक 5 सेमी त्रिज्या का वृत्त खींचिए।
• बाह्य वृत्त पर बिन्दु P लेकर उसे केन्द्र O से मिलाइये।
• अब OP का लम्ब अर्द्धक खींचिए, जो OP को M पर काटता है।
• बिन्दु M को केन्द्र मानकर तथा OM त्रिज्या से वृत्त खींचिए जो अंतः वृत्त को बिन्दु A व B परकाटता है।
• बिन्दु A व B को बिन्दु P से मिलाइए। PA तथा PB अभीष्ट स्पर्श रेखाएँ हैं। PA = 4.5 सेमी।

प्रश्न 13.
एक त्रिभुज ABC की रचना कीजिए, जिसमें भुजा BC = 6 सेमी ∠B = 45° तथा ∠A = 105° हो, तब एक अन्य त्रिभुज की रचना कीजिए जिसकी भुजाएँ ΔABC की संगत भुजाओं की \(\frac{3}{4}\) गुनी हों।
हल:
रचना के चरण :
(1) एक रेखाखण्ड BC = 6 सेमी खींचिए।
(2) बिन्दु B पर ∠B 45° बनाया।
∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180°
∠C = 180° – 150°
∠C = 30°
(3) बिन्दु C पर ∠C = 30° बनाया। दोनों किरणें 4 पर काटती हैं। इस प्रकार ΔABC प्राप्त होता है।
(4) BC पर न्यून कोण बनाती एक किरण BX खींचिए।
(5) बिन्दुओं B₁, B₂, B₃ और B₄ को इस प्रकार अंकित कीजिए कि BB₁ = BB₂ = B₂B₃ = B₃B₄

(6) B₂C को मिलाइए।
(7) B₄ में होती हुई रेखा B₄C || B₃ खींचिए।
(∠BB₃C = ∠BB₄C
(8) C में होती हुई रेखा C₄‘A’ || AC खींचिए।
(∠BCA = ∠BC’A)
अत: ΔA’BC’ एक अभीष्ट त्रिभुज हैं।

प्रश्न 14.
एक त्रिभुज ABC की रचना कीजिए जिसमें भूजा BC = 7 सेमी, ∠B = 45°, ∠A = 105° हो। तब एक अन्य त्रिभुज की रचना कीजिए, जिसकी भुजाएँ ΔABC की संगत भुजाओं की \(\frac{3}{4}\) गुनी हों।
हल:
BC = 7 सेमी, ∠B = 45°, ∠A = 105°
∠C = 180° – ( ∠B + ∠A)
= 180° – (45° + 105°)
= 180° – 150°
= 30°
रचना के चरण :

• BC = 7 सेमी की रेखा खींचे।
• बिंदु B पर 45° तथा बिंदु C पर 30° का कोण बनाएँ। यह एक दूसरे को पर काटते हैं।
• बिंदु B पर एक न्यून कोण बनाएँ।
• कोण किरण को चार सामान भागों B₁, B₂, B₃ और B₄ पर विभाजित किया।
• B₄ को C पर मिलाएँ।
• बिंदु B₃ मैं रेखा B₄C के समानान्तर रेखा बनाये जो BC को C’ पर काटती है।
• C से AC के समानान्तर CA’ रेखा खींची जो AB को A’ पर काटती है।
• ΔA’BC’ अभीष्ट त्रिभुज है। जिसमें A’B = \(\frac{3}{4}\)AB.

प्रश्न 15.
आधार 5 सेमी और ऊँचाई 4 सेमी वाले एक समद्विबाहु त्रिभुज की रचना कीजिए। एक अन्य त्रिभुज की रचना कीजिए जिसकी भुजाएँ समद्विबाहु त्रिभुज की संगत भुजाओं का \(\frac{2}{3}\) गुना हो।
हल:
रचना के चरण :

• BC = 8 सेमी खींचें।
• BC रेखा का लम्ब समद्विभाजक XY खींचा जां BC को M पर काटता है।
• XM पर MA = 4 सेमी काटा, तब BA और CA को मिलाया जिसमें ΔABC प्राप्त हुआ।
• बिन्दु B पर एक न्यूनकोण बनाया तथा उस पर तीन चाप B₁, B₂ और B₃ बनाए।
• B₃C मिलाया और B₂ से B₃C के समान्तर रेखा खींची जी BC को C’ पर काटती है।
• C’ से A’C’ || AC खींची।
• अत: A’C’B अभीष्ट त्रिभुज है।

Jharkhand Board JAC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 10 Light Reflection and Refraction

Jharkhand Board Class 10 Science Light Reflection and Refraction Textbook Questions and Answers

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay
[Hint: The material of a lens must be transparent but clay is opaque.]

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length of the lens
(c) At infinity
(d) Between the optical centre of the lens and its principal focus
Answer:
(b) At twice the focal length of the lens

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of – 15 cm. The mirror and the lens are likely to be …
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer:
(a) both concave.

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be _______.
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
Answer:
(d) either plane or convex.
[Hint: Both give erect images. The size of image in plane mirror is same as the object but it is diminished in convex mirror.]

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
Answer:
(c) A convex lens of focal length 5 cm [Hint: Since a convex lens gives a magnified image of the object and the smaller the focal length, the more the magnifying power.]

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Solution:
When the object is placed between the pole and the principal focus of a concave mirror, an erect, virtual and enlarged image is formed. This image can be viewed in the mirror itself, not on the screen. In short, image is formed behind the mirror as shown in the figure 10.57.

Therefore, the range of distance of the object from the mirror must be greater than zero and less than 15 cm (focal length of the concave mirror).

Question 8.
Name the type of mirror used in the following situations:
(a) Headlights of a car
(b) Side / rear-view mirror of a vehicle
(c) Solar furnace
Support your answer with reason.
Answer:
(a) For headlights of a car, a concave mirror is used.
The light source is kept at the focus s of the mirror. On reflection, a strong? parallel beam of light emerges out.

(b) For side / rear-view mirror of a vehicle, s convex mirror is used.
This is because its field of view is very large and it forms a virtual, erect and diminished image of the object behind the vehicle, which enables a driver to see most of the traffic behind him/her.

(c) For a solar furnace, a concave mirror is used.
Light from the Sun, on reflection from t the mirror, is concentrated at the focus of s the mirror, producing heat and temperature increases sharply up to 180°C-200°C

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
Yes.
Even when one-half of a convex lens is covered with a black paper, the lens produces a complete (or full) image of the object.

Here, the intensity (brightness) of image will become one-forth as that with complete lens exposed, because less number of light rays can be passed / refracted through the lens.

The nature, size and location of the image will be the same since light from all parts of the object reach the exposed part of the lens.

It can be understood by the following two cases (Experimental verification) :
(1) When the upper half of the lens is covered :
In this case, the ray of light coming from the object will be refracted by the lower half of the lens.

These rays meet at the other side of the lens to form the image of the given object.

(2) When the lower half of the lens is covered:
In this case, the ray of light coming from the object will be refracted by the upper half of the lens.

These rays meet at the other side of the lens to form the image of the given object.

[Note: Covering one-half of the lens reduces the brightness of the image.]

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Solution:
Here, Object size h = + 5 cm
Object distance u = – 25 cm
Focal length of lens f = + 10 cm
(∵ Converging, i.e., Convex lens)
Image distance v = ?
Image height h’ = ?
As, \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\)
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) + \(\frac { 1 }{ u }\)
= \(\frac{1}{(+10)}+\frac{1}{(-25)}=\frac{5-2}{50}=\frac{3}{50}\)
= 16.67 cm
Image distance v is positive. This shows that the image formed is real and on the other side of the lens, at 16.67 cm from the lens as shown in the figure 10.60:

Now, Magnification m = \(\frac { h’ }{ h }\) = \(\frac { v }{ u }\)
∴ h’ = h\(\frac { v }{ u }\)
= \(\frac { 1 }{ f }\)
= – \(\frac { 10 }{ 3 }\) cm
= – 3.3 cm
The negative (minus) sign shows that the image is inverted, real, diminished (3.3 cm).
Figure 10.60 show the position, size and nature of the image formed.

[Note: The ray diagram can be drawn without calculating v and h’. Once, f, u and h are fixed, v and h’ have definite values.

Question 11.
A concave lens of focal length 15 cm forms an image at 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Solution:
Here, Focal length of the lens f = – 15cm (∵ Concave lens)
Image distance v = – 10 cm
(∵ In case of a concave lens the image is formed on the same side of the object)
Object distance u = ?

The negative (minus) sign shows that the object is placed on the left side of the lens.

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Solution:
Here, Object distance u = – 10 cm
Here, Object distance u = – 10 cm
Focal length of the mirror f = + 15 cm (∵ Convex mirror)
Image distance v = ?
Using the mirror formula \(\frac { 1 }{ f }\) = \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\)
We have
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) – \(\frac { 1 }{ u }\)
= \(\frac{1}{(+15)}-\frac{1}{(-10)}\)
= \(\frac{1}{15}+\frac{1}{10}\)
= \(\frac { 2+3 }{ 3 }\)
= \(\frac { 5 }{ 30 }\)
= \(\frac { 1 }{ 6 }\)
∴ v = 6 cm
The positive (plus) sign of v indicates that the image is formed behind the mirror.
Now, magnification m = – \(\frac { v }{ u }\) = – \(\frac { 6 }{ – 10 }\) = + 0.6
Positive value of magnification indicates that the image is virtual and erect.
The magnitude of magnification is 0.6, which is less than 1. This shows that the S image is diminished.

Question 13.
The magnification produced by a plane mirror is + 1. What does this mean?
Solution:
Here, m = + 1
As, m = \(\frac { h’ }{ h }\)
\(\frac { h’ }{ h }\) = + 1
∴ h’ = h
So, the size of the image is equal to the size of object.
Further, the positive (plus) sign of m indicates that the image is erect and hence virtual.
Again, m = – \(\frac { v }{ u }\) and m = + 1
∴ – \(\frac { v }{ u }\) = 1
∴ v = – u
This shows that the image is formed behind the mirror and the distance of the image from the mirror equals that of the object from the mirror.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Solution:
Here, Size of object h = + 5.0 cm
Object distance u = – 20 cm
Radius of curvature R = + 30 cm (∵ Convex mirror)
Focal length f = \(\frac { + 30 cm }{ 2 }\) = + 15 cm
Image distance v = ?
Image size h’ = ?
Mirror formula: \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)

The positive (plus) sign of u indicates that the image is behind the mirror or on the right-hand side of the mirror.

Positive value of image height indicated that the image is virtual and erect.
Moreover, h’ < h
So, the image is smaller in size than the object.

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Solution:
Here, Object size h = 7.0 cm
Object distance u = – 27 cm
Focal length f = – 18 cm
(∵ Concave mirror)
Image distance u = ?
Image size h’ = ?
From mirror formula :
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)

The negative (minus) sign of v indicates that ’ the image is formed on the same side of the object. So, screen should be held in front of the mirror at a distance of 54 cm from the mirror. The image can be obtained on the screen and hence, it is real.
Now, magnification m = \(\frac { h’ }{ h }\) = – \(\frac { v }{ u }\)
∴ h’ = – h(\(\frac { v }{ u }\))
= – 7.0 (\(\frac { -54 }{ -27 }\))
= – 14 cm
The negative (minus) sign of h’ shows that the image is inverted and hence would be real.

Question 16.
Find the focal length of a lens of power -2.0D. What type of lens is this?
Solution :
Here, Focal length of the lens f = ?
Power P = – 2.0 D = – 2.0 m^-1
Now,
As, P = \(\frac { 1 }{ f }\)
f = \(\frac { 1 }{ p }\)
= \(\frac { 1 }{ -2D }\)
= – 0.5 cm
As the power of the lens is negative (given), the lens must be concave.

Question 17.
A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Solution:
Here, Power P = + 1.5 D = + 1.5 m^-1
Focal length f = ?
From, f = \(\frac { 1 }{ p }\)
f = \(\frac { 1 }{ 1.5 D }\)
= 0.67 m
= 67 cm
As the power of the lens is positive (given), it is a converging lens, i.e., a convex lens.

Jharkhand Board Class 10 Science Light Reflection and Refraction InText Questions and Answers

Question 1.
Define the principal focus of a concave mirror.
Answer:
The point on the principal axis of a concave mirror, at which rays of light incident on the mirror in a direction parallel to the principal axis (actually) meet / intersect after reflection from the mirror is called the principal focus (F) of the concave mirror.

[Note : In general, the point on the principal axis where rays of light incident parallel to the principal axis converge to or appear to diverge from after reflection is called the principal focus of the spherical mirror.]

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Solution:
Here, R = 20 cm; f =?
As f = \(\frac { R }{ 2 }\), f = \(\frac { 20 }{ 2 }\)
= 10 cm
[Note: According to the New Cartesian sign convention, if the given spherical mirror is a convex mirror, then f = + 10cm and if the given spherical mirror is a concave mirror, then f = – 10 cm.]

Question 3.
Name a mirror that can give (form) an erect and enlarged image of an object.
Answer:
Concave mirror:
A concave mirror produces an erect and enlarged image of an object, when an object is placed between the pole and principal focus of the concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
This is because a convex mirror (always) forms an erect, virtual and diminished image of an object, wherever the object may be located.

Also, a convex mirror has a wider field of view, relative to a plane mirror, as it is curved outwards. Thus, a convex mirror fitted on the side of a vehicle enables the driver to view much larger area than that would be possible with a plane mirror, enabling the driver to see traffic behind him/her to facilitate safe driving.

Question 5.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Solution:
Here, the radius of curvature R = 32 cm
We know that, the focal length f = \(\frac { R }{ 2 }\)
∴ f = \(\frac { 32 }{ 2 }\) = 16 cm

Question 6.
A cancave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where 5 is the image located?
Solution:
Here, the linear magnification m = – 3
(negative sign for a real image, which is inverted)
The object distance u = – 10 cm
(object distance (real) is always negative)
The image distance v =?
As m = – \(\frac { v }{ u }\)
v = – mu
∴ v = – (-3) (- 10)
= – 30 cm
Thus, the real image is located at 30 cm in front of the mirror (on the same side as the object).

Question 7.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
For oblique incidence, the light ray bends towards the normal, on entering water. It happens because water is optically denser than air.

The speed of light is higher in an optically rarer medium than in an optically denser medium. So, a ray of light travelling from air to water slows down and bends towards the normal.
OR

Thus, the angle of refraction is less than the angle of incidence. It implies that the light ray bends towards the normal.

Question 8.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 x 10⁸ m s^-1.
Solution:
Speed of light in vaccum c = 3 x 10⁸ m s^-1.
Refractive index of glass n_g = 1.50
Speed of light in glass v = ?
Now, absolute refractive index of glass,
n_g = \(\frac { c }{ v }\)
∴ The speed of light in the glass,
v = \(\frac{c}{n_{\mathrm{g}}}=\frac{3 \times 10^8}{1.50}\)
= 2 x 10⁸ m s^-1.

Question 9.
Find out from Table 2 (Given with Q. 42), the medium having highest optical density. Also, find the medium with lowest optical density.
Answer:
The higher the refractive index, the higher is the optical density. Diamond has the highest optical density as it has the highest refractive index, 2.42 and air has the lowest optical density as it has the lowest refractive index, 1.0003.

Question 10.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 2.
Answer:
From Table 2,
refractive index of kerosene = 1.44
refractive index of turpentine = 1.47 and
refractive index of water = 1.33
Here, from given liquids, turpentine has the highest refractive index and water has the lowest refractive index.
Now, the lower the refractive index of a medium, the higher is the speed of light in the medium.
Hence, out of the given liquids the light travels fastest in water.

Question 11.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
Refractive index of diamond
= \(\frac{\text { Speed of light in vacuum / air }}{\text { Speed of light in diamond }}\)
∴ Speed of light in diamond
= \(\frac{\text { Speed of light in vacuum / air }}{\text { Refractive index of diamond }}\)
This expression states that the speed of light in diamond is \(\frac { 1 }{ 2.42 }\) times the speed of light in vacuum / air.
Speed of light in diamond = \(\frac{3 \times 10^8}{2.42}\)
= 1.24 x 10⁸ms^-1
In other words, it can also be said that, the ratio of the speed of light in vacuum/air to the speed of light in diamond is equal to 2.42.

Question 12.
Define 1 dioptre of power of a lens.
OR
Define the dioptre.
Answer:
1 dioptre is the power of a lens of focal length 1 metre.

Question 13.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Solution:
A convex lens forms a real, inverted image of the same size as that of the object, if the object is placed at 2F₁.
In this case v = + 50 cm and m = – 1

Thus, the needle is placed at 50 cm from the convex lens of power 4 D.

Question 14.
Find the power of a concave lens of focal length 2 m.
Solution :
Here, focal length f = – 2 m (∵ Concave lens)
∴ Power P = \(\frac { 1 }{ f(m) }\)
= \(\frac { 1 }{ -2m }\)
= – 0.5 m^-1
= – 0.5 dioptre
= – 0.5 D

Activity 10.1 [T. B. Pg. 161]

To determine the nature of the inner and outer curved surface of a spoon.

Procedure :
1. Take a large shining spoon. Try to view your face in its inner curved surface.

• Do you get the image?
• Is it smaller or larger?

2. Move the spoon slowly away from your face. Observe the image.

• How does it change?

3. Reverse the spoon and repeat the activity.

• How does the image look like now?

4. Compare the characteristics of the image on the two surfaces.

Observation :

• Yes.
• The image formed is enlarged and erect when the spoon is closer to the face.
• As the spoon is moved slowly away from the face, the image becomes large and inverted. When the face is very far off, highly diminished and inverted image of our face is seen.
• When the spoon is reversed, the image formed by the outer surface of the spoon is virtual and diminished in size.
• Now, if we move the spoon away from our face, the image moves away, however the image continues to be virtual and diminished in size.
• Finally, at last the image is diminished to almost point-size.

Conclusion:
The inner surface of the spoon acts as a concave mirror, whereas the outer surface acts as a convex mirror.

Activity 10.2 [T. B. Pg. 162]

To show the converging nature of a concave mirror and find its focal length.

Procedure:

• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
• Hold the mirror and the paper in the same position for a few minutes.
• What do you observe? Why?
  

Observation:
The paper initially turns blackish, then burns producing smoke. Eventually it catches fire. The light from the Sun is converged at a point, as a sharp bright spot by the mirror. In fact, this spot of light is the image of the Sun on the sheet of paper.

This point is the principal focus of the concave mirror. The heat produced due to concentration of sunlight ignites the paper.

Conclusion:

• The distance of this image of the Sun from the pole of the mirror gives the approximate value of the focal length of the concave mirror.
• It is also concluded that when a parallel beam of light from a far off object falls on a concave mirror, a real, inverted and point-ized image is formed at the focus of the concave mirror.

Activity 10.3 [T. B. Pg. 163]

To locate the image formed by a concave mirror for different positions of the object.

Procedure :
1. Take a concave mirror.

2. Find out its approximate focal length in the way described in Activity 10.2.

3. Note down the value of the focal length.

4. Mark a straight line on a table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that the pole of the mirror lies over the line.

5. Draw with a chalk two more straight lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror.

These lines will now correspond to the positions of the points P F and C respectively.
(Remember : For a spherical mirror of small aperture, the principal focus F lies mid-way between the pole P and centre of curvature C.)

6. Keep a bright object, say a burning candle, at a position far beyond C.
Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.

7. Observe the image carefully. Note down its nature, position and relative size with respect to the object size.

8. Repeat the activity by placing the candle –

• just beyond C
• at C
• between F and C
• at F
• between P and F

In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Where would you look to observe the image of that object?

9. Note down and tabulate your observations.

Observation:
1. Image formation by a concave mirror for different positions of the object.

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F	Highly diminished, point-sized	Real and inverted
Just beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

2. When an object (the burning candle in the present case) is placed between P and F, its image cannot be obtained on the screen.

3. Hence, to observe its image, in this case one has to look in the mirror itself because the image is virtual.

Conclusion:
The nature, position and size of the image formed by a concave mirror depends on the position of the object in relation to the points E F and C.

Activity 10.4 [T. B. Pg. 166]

To locate the image formed by a concave mirror for different positions of an object using ray diagrams.

Procedure:

• Draw a neat ray diagram for each position of the object as discussed in the Activity 10.3. You may take any two of the rays (mentioned in the previous point 10.2.2), for locating the image.
• Observe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.

Observation:
The following figures illustrate the ray diagrams for the formation of image by a concave mirror for various positions of the object:

Conclusion:
Image formation by the concave mirror for different positions of an object using ray diagrams is tabulated as follows :

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F	Highly diminished, point-sized	Real and inverted
Just beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

Activity 10.5 [T. B. Pg. 167]

To locate the image formed by a convex mirror for different positions of an object.

Procedure:

• Take a convex mirror. Hold it in one hand.
• Hold a pencil in the upright position in the other hand.
• Observe the image of the pencil in the mirror.
• Is the image erect or inverted? Is it diminished or enlarged?
• Move the pencil away from the mirror slowly.
• Does the image become smaller or larger?
• Repeat this activity carefully.
• State whether the image will move closer to or farther away from the focus as the object is moved away from the mirror.

Observation:
1. When we hold a pencil in the upright position in front of a convex mirror, the image of the pencil is observed on the back side of the mirror, i.e., in the mirror itself.
The image is erect and virtual.
The image is diminished in size relative to the object.

2. When the pencil is moved away from the mirror slowly, the image becomes smaller and smaller, moving away from the mirror.

3. On repeating this activity, we find that as the object is moved away from the mirror, the image moves closer to the focus of the mirror.

Conclusion:
This activity confirms all the characteristics of the images formed by a convex mirror for different positions of the object and it is tabulated as follows :

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus F, behind the mirror	Highly diminished, point-sized	Virtual and erect
Between infinity and the pole P of the mirror	Between P and F, behind the mirror	Diminished	Virtual and erect

Activity 10.6 [T. B. Pg. 167]

To demonstrate the wide field of view nature of a convex mirror.

Procedure:

• Observe the image of a distant object, say a distant tree in a plane mirror.
• Could you see a full-length image?
• Try with plane mirrors of different sizes.
• Did you see the entire object in the image?
• Repeat this activity with a concave mirror.
• Did the concave mirror show full-length image of the object?
• Now, try using a convex mirror.
• Did you succeed?
• Explain your observation with reason.

Observation :
No, we can not see the full-length image of an object in plane mirror.

When we try with plane mirrors of different sizes, we find that the entire object in the image is seen when the size of the plane mirror is at least half of the size of the object.

No, concave mirror did not show full-length S image of the object (here a distant tree).

When we use a small convex mirror, we succeed in seeing the full-length image of an object, wherever the object may be located.

The reasons for these observations are as follows :
(1) In a plane mirror, size of image is S always equal to the size of the object.

(2) In case of a concave mirror, when object is between P and F, then only its enlarged, virtual and erect image is seen behind the concave mirror.
(i.e., full-length image of an object can be seen).
But here our object is far away from the concave mirror, e.g., distant tree; so its full- length image can not be seen in the concave mirror itself (i.e., behind the concave mirror).

(3) While, in a convex mirror, the image is always virtual, erect and smaller than the object, wherever the object may be located.

Conclusion:
Out of plane mirror, concave mirror and convex mirror, only a convex mirror can show a full-length image of a tall object in all its positions.
OR
A convex mirror has a wide field of view and the image formed by it is always virtual, erect and shorter than the object, wherever the object may be located.

Activity 10.7 [T. B. Pg. 172]

To demonstrate refraction of light.
OR
To show that the apparent depth of the bucket filled with water is less than the real depth of the bucket.

Procedure:

• Place a coin at the bottom of a bucket filled with water.
• With your eye to a side above water, try to pick up the coin in one go.
• Did you succeed in picking up the coin?
• Repeat the Activity.
• Why did you not succeed in doing it in one go?
• Ask your friends to do this.
• Compare your experience with theirs.

Observation:
No.
When you try picking up the coin with your eye to a side above water, you do not succeed in one go.

The reason for this can be explained as follows :
As shown in the figure 10.31, a coin is at point ‘O’ at the bottom of a bucket filled with water.

When we view this coin with our eye to a side above water, we observe the image I of the coin, which is above ‘O’. When we try to pick up this coin in one go, we do not succeed in picking up the coin. This is because we move our hand up to I, where the coin is being observed, but actually the coin lies below at O, the bottom of the bucket.

When our friends try the same way, they also fail to pick up the coin.

Conclusion:
In going from water to air, rays of light bend away from the normal and the bottom of the bucket appears to be raised, i.e., the apparent depth of the bucket is less than the real depth of the bucket.

Activity 10.8 [T. B. Pg. 172]

To show that the object in water appears to be raised due to refraction.
OR
Apparent depth of an object in the large shallow bowl filled with water increases as its real depth increases.

Procedure:

• Place a large shallow bowl filled with water on a table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.
• Keep looking for the coin from your position.
• Does the coin becomes visible again from your position?
• How could this happen?

Observation:

O = Real position of the coin
I = Initial apparent position of the coin
I’ = New apparent position of the coin

As shown in the figure 10.32, in a large shallow bowl filled with water, the real position of the coin is ‘O’.
On account of refraction of light the coin appears raised from O to I. Image I of the coin appears to be raised which is original apparent position of a coin.

When we move our eye slowly away from the bowl as shown by an arrow in the figure, the coin disappears from our sight. This happens because rays starting from the coin O fail to enter our eye, after refraction at water surface AB.

When a friend pours water gently into the bowl without disturbing the coin, the coin becomes visible again as it appears raised further to position I’.

This happens because on adding water, the real- depth of the coin increases. As the apparent depth of a coin is equal to the real depth divided by the refractive index (n) of water, the apparent depth of the coin also increases. Hence, the coin appears raised from I to I’. Therefore, the coin becomes visible from the displaced position (new position) of our eye.

Conclusion:
The coin becomes visible again and slightly raised above its actual position on pouring water into the bowl. This is because of refraction of light.

In other words, apparent depth of a coin in the large shallow bowl filled with water increases as its real depth increases.

Activity 10.9 [T. B. Pg. 172]

To show that refraction does not occur for normal incidence.

Procedure:

1. Draw a thick straight line in ink, over a sheet of white paper placed on a table.

2. Place a glass slab over the line in such a way that one of its edges makes an angle with the line.

3. Look at the portion of the line under the slab from the sides.

• What do you observe?
• Does the line under the glass slab appear to be bent at the edges?

4. Next, place the glass slab such that it is normal to the line.

• What do you observe now?
• Does the part of the line under the glass slab appear bent?

5. Look at the line from the top of the glass slab.

• Does the part of the line, beneath the slab, appear to be raised?
• Why does this happen?

Observation :

• When we look at the portion of the line under the slab from sides, careful observation shows that line AB is bent at edge B and line BC at edge C. [figure 10.33 (a)]
• Yes. The line under the glass slab appears bent at edges B and C.
• On placing the glass slab normal to the line, we observe that the part of the line under the glass slab does not appear bent, [figure 10.33 (c)]
• No. The line under the glass slab does not appear bent at edges B and C.
• Yes. The part of the line, beneath the slab, appears raised while looking at the line from the top of the glass slab.
• This is due to refraction of light.

Conclusion:
The object appears raised due to refraction of light and for normal incidence, refraction of light does not occur.

Activity 10.10 [T. B. Pg. 173]

To show the refraction of light and lateral displacement through a rectangular glass slab.

Procedure:

• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil and name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge.
• Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight-line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly,
• join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
• Join O and O’. Also, produce EF up to P as shown by a dotted line in the following figure.
• At O, draw NN’ perpendicular to AB. At O’, draw MM’ perpendicular to CD. Also, draw O’L perpendicular to OP.
  

[Note : (1) In figure 10.34, face AB is air-glass interface and face CD is glass-air interface. ( 2 ) Reflection of light is not shown in the figure.]

Observation :

• At O, light ray along EF enters from air into glass. It bends towards the normal NN’. This is the first refraction.
• At O’, the light ray enters from glass into air. It bends away from the normal MM’ and travels 5 along GH. This is second refraction.
• Here, angle of emergence r₂ is equal to angle of incidence i₁, i.e., the emergent ray is parallel to the original direction of the incident ray. This is because there is identical medium, air in this case, adjacent to both edges AB and CD.
• However, the light ray is shifted sideways slightly. This is lateral displacement and is represented by O’L.

Conclusion:

• The ray of light travelling from a rarer medium (air in this case) to a denser medium (glass in this case) bends towards the normal.
• The ray of light travelling from a denser medium (glass in this case) to a rarer medium (air in this case) bends away from the normal.
• The emergent ray is parallel to the incident ray but is slightly displaced sideways.

Activity 10.11 [T. B. Pg. 177]

To show the converging nature of a convex? lens and find its focal length.

Procedure :
1. Hold a convex lens in your hand. Direct it towards the Sun.

2. Focus the light from the Sun (through lens) on a sheet of paper. Obtain a sharp bright image? of the Sun.

3. Hold the paper and the lens in the same position for a while.
Keep observing the paper.

• What happened?
• Why?
  

Observation :

• When the paper and the lens were held in the same position for sometime, the paper began to burn producing smoke. After sometime, it caught fire.
• Because, the light from the Sun constitutes parallel rays of light. These rays were converged by the lens at the sharp bright spot formed on the paper and so heat is produced by these Sun rays as they concentrate on the spot. This caused the paper to burn.

Conclusion:

• When a parallel beam of light from a far-off object like the Sun falls on a convex lens, a real, inverted, point size image of the Sun is formed at the focus of the lens.
• The distance between the position of the lens and the position of the image of the Sun gives the approximate focal length of the lens.

[Note : The distance between the lens and the image of the Sun on the paper is the approximate focal length of the lens.]

Activity 10.12 [T. B. Pg. 178]

To locate the position and examine the nature of the image formed by a convex lens for different positions of the object.

Procedure:

• Take a convex lens. Find its approximate focal length in a way described in Activity 10.11.
• Draw five parallel lines, using chalk, on a long table such that the distance between the successive lines is equal to the focal length of the lens.
• Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
• The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F₂ and 2F₂ respectively.
• Place a burnig candle, far beyond 2F₁ to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
• Note down the nature, position and relative size of the image.
• Repeat this Activity by placing the object just beyond 2F₁, at 2F₁, between F₁ and 2F₁, at F₁, between F₁ and O.
• Note down and tabulate your observations.

Observation:
Position, relative size and the nature of the image formed by a convex lens for various positions of the object.

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F2	Highly diminished, point-sized	Real and inverted
Beyond 2F1	Between F2and 2 F2	Diminished	Real and inverted
At F1	At 2F2	Same size	Real and inverted
Between F1and 2F1	Beyond 2 F2	Enlarged	Real and inverted
At focus F1	At infinity	Infinitely large or Highly enlarged	Real and inverted
Between focus F1 and optical centre O*	On the same side of the lens as the object	Enlarged	Virtual and erect

Conclusion:
The nature, position and relative size of the image formed by a convex lens for various positions of the object depends on the position of the object in front of the lens.

Activity 10.13 [T. B. Pg. 179]

To locate the position and examine the nature of the image formed by a concave lens for different positions of the object.

Procedure:

• Take a concave lens. Place it on a lens stand.
• Place a burning candle on one side of the lens.
• Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible.
• If not, observe the image directly through the lens.
• Note down the nature, relative size and approximate position of the the image.
• Move the candle away from the lens. Note the change in the size of the image.
• What happens to the size of the image when the candle is placed too far away from the lens?
• What conclusion can you draw from this Activity?

Observation:
The image formed by a concave lens, when a burning candle is kept too far away from the lens is highly diminished, i.e., point-sized.
Position, relative size and the nature of the image formed by a concave lens for various position of an object

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F1	Highly diminished, point-sized	Virtual and erect
Between infinity and optical centre O of the lens	Between focus F1 and optical centre O	Diminished	Virtual and erect

Conclusion:
The image formed by a concave lens is always virtual, erect and diminished irrespective of the position of the object.

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 5 समांतर श्रेढ़ियाँ Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 5 समांतर श्रेढ़ियाँ

लयूत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
यदि किसी A. P. के प्रथम 12 पदों का योग 468 है तथा इसका सार्वअंतर 6 है, तो 10वाँ पद ज्ञात कीजिए।
हल :
दिया है,
S_n = 468, d = 6, n = 12
a₁₀ = ?
हम जानते हैं :
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
468 = \(\frac{12}{2}\)[2a + (12 – 1)6]
468 = 6(2a + 66)
= 12a + 396
a = \(\frac{468-396}{12}=\frac{72}{12}\)
a = 6
∴ a₁₀ = a + (n – 1)d
= 6 + (10 – 1)6
= 6 + 9 × 6 = 6 + 54
a₁₀ = 60.

प्रश्न 2.
एक समान्तर श्रेणी का 14वाँ पद उसके 8वें पद का दुगना है। यदि उसका छटा पद – 8 है, तो उसके प्रथम 20 पदों का योगफल ज्ञात कीजिए।
हल :
माना समान्तर श्रेणी का प्रथम पद = a
तथा सर्वाअंतर = d
∴ समान्तर श्रेढी के nth पद
a_n = a + (n – 1)d
इस प्रकार a₁₄ = a + (14 – 1)d
= a + 13d
a₈ = a + (8 – 1)d = a + 7d
a₆ = a + (6 – 1)d = a + 5d
प्रश्नानुसार-
a₁₄ = 2a₈
⇒ a + 13d = 2(a + 7d)
⇒ a + d = 0 ………..(1)
इसी प्रकार,
a₆ = a + 5d = – 8 ………..(2)
समी (1) व (2) को हल करने पर,
a = 2, d = – 2
∴ S₂₀ = \(\frac{20}{2}\) [2 × z + (20 – 1) (- z)]
S₂₀ = – 340
अतः प्रथम 20 पदों का योगफल = – 340.

प्रश्न 3.
1 से 100 तक के मध्य की 6 से विभाजित होने वाली संख्याओं का योगफल ज्ञात कीजिए।
हल :
1 से 100 तक के मध्य की 6 से विभाज्य संख्याएँ है….
6, 12, 18, 24, …, 96
स्पष्ट है कि उपर्युक्त संख्याएँ समान्तर श्रेढी में हैं।
यहाँ a = 6, d = 12 – 6 = 6 तथा a_n = 96
अत: a_n = 96
⇒ a + (n – 1)d = 96
⇒ 6 + (n – 1) × 6 = 96
⇒ (n – 1) × 6 = 96 – 6 = 90
⇒ n – 1 = \(\frac{90}{6}\) = 15
⇒ n = 15 + 1 = 16
∵ हम जानते हैं कि
S_n = \(\frac{n}{2}\)(a + l)
⇒ S₁₆ = \(\frac{16}{2}\)(6 + 96)
⇒ S₁₆ = 8 × 102
⇒ S₁₆ = 816
अतः S₁₆ = 816

प्रश्न 4.
समीकरण हल कीजिए :
1 + 4 + 7 + 10 +……+ x = 287.
हल :
दिया गया समीकरण है :
1 + 4 + 7 + 10 +………+ x = 287
स्पष्ट है कि यह एक समान्तर श्रेढी है।
यहाँ a = 1, d = 4 – 13, a_n = x तथा S_n = 287
∵ a_n = x
⇒ a + (n – 1)d = x
⇒ 1 + (n – 1) × 3 = x
⇒ 1 + 3n – 3 = x
⇒ x = 3n – 2 ……………(i)
और S_n = 287
⇒ \(\frac{n}{2}\)(a + l) = 287
⇒ \(\frac{n}{2}\)(1 + x) = 287
⇒ \(\frac{n}{2}\)(1 + 3n – 2) = 287
[समी. (i) का प्रयोग करने पर]
⇒ n(3n – 1) = 574
⇒ 3n² – n – 574 = 0
श्रीधराचार्य सूत्र से –
n = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

∵ n एक प्राकृतिक धन पूर्णांक है। अतः n = \(\frac{-41}{3}\) को छोड़ देते हैं।
अत: n = 14
n का मान समीकरण (i) में रखने पर
x = 3 × 14 – 2
= 42 – 2 = 40
अतः x = 40.

प्रश्न 5.
A. P. 17, 15, 13 के कितने पद लिए जाएँ ताकि उनका योग 81 हो।
हल :
a = 17, d = 15 – 17 = – 2
S_n = 81
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
81 = \(\frac{n}{2}\)[2 × 17 + (n – 1) × (- 2)]
81 = \(\frac{n}{2}\)[34 – 2n + 2]
= n[17 – n + 1]
= n[18 – n]
81 = 18n – n²
n² – 18n + 81 = 0
n² – 9n – 9n + 81 = 0
n(n – 9) – 9 (n – 9) = 0
(n – 9) (n – 9) = 0
n = 9.

प्रश्न 6.
चार क्रमागत संख्याएँ जोकि समान्तर श्रेढी में हैं, का योग 32 है। प्रथम और अन्तिम संख्याओं का गुणनफल मध्य संख्याओं के गुणनफल से 7 : 15 के अनुपात में हो। वह संख्याएँ ज्ञात करो ।
हल :
माना चार क्रमागत संख्याएँ हैं-
a – 3d, a – d, a + d, a + 3d
प्रश्नानुसार,
संख्याओं का योग = 32
⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a = \(\frac{32}{4}\) = 8
तथा \(\frac{(a-3 d)(a+3 d)}{(a-d)(a+d)}\) = \(\frac{7}{15}\)
⇒ \(\frac{a^2-9 d^2}{a^2-d^2}\) = \(\frac{7}{15}\)
⇒ \(\frac{64-9 d^2}{64-d^2}\) = \(\frac{7}{15}\) [∵ a = 8]
⇒ 960 – 135d² = 448 – 7d²
⇒ 960 – 448 = 135d² – 7d²
⇒ 512 = 128d²
⇒ d² = \(\frac{512}{128}\) = 4
⇒ d = ± 2
अतः a – 3d = 8 – 3 × 2 = 2
a – d = 8 – 2 = 6
a + d = 8 + 2 = 10
a + 3d = 8 + 3 × 2 = 14
अतः अभीष्ट संख्याएँ हैं-
2, 6, 10, 14.

प्रश्न 7.
यदि एक समान्तर श्रेढी के m पदों का योग ‘n’ तथा n पदों का योग ‘m’ हो, तो सिद्ध कीजिए कि (m + n) पदों का योगफल – (m + n) होगा।
हल :
माना दी गई श्रेढी का प्रथम पद a तथा सार्वअन्तर d हैं।
दिया है S_m = n
⇒ \(\frac{m}{2}\){2a + (m – 1)d} = n
⇒ 2am + m(m-1)d = 2n ……..(1)
तथा S_n = m
⇒ \(\frac{n}{2}\){2a + (n – 1)d} = m
⇒ 2am + n(n – 1)d = 2m …….(ii)
समीकरण (i) में से समीकरण (ii) को घटाने पर,
2a(m – n) + {m(m – 1) n (n-1)} d = 2n – 2m
⇒ 2a(m – n) + {(m² – n²) – (m – n)} d
= – 2(m – n)
⇒ 2a + (m + n – 1)d = – 2 ………….(iii)
[दोनों पक्षों में (m – n) से भाग देने पर]
अब S_m+n = \(\frac{m+n}{2}\){2a + (m + n – 1)d}
⇒ \(\frac{(m+n)}{2}\)(-2) [(iii) के प्रयोग से]
⇒ S_m+n = – (m + n)
अतः S_m+n = – (m + n) इति सिद्धम् ।

प्रश्न 8.
यदि एक समान्तर श्रेढी का m वाँ पद \(\frac{1}{n}\) तथा n वाँ पद \(\frac{1}{m}\) हो, तो दर्शाइए कि mn पदों का योगफल \(\frac{1}{2}\) (mn + 1) होगा।
हल :
माना दी गई समान्तर श्रेढी का प्रथम पद a तथा सार्वअन्तर d है।
दिया है : a_m = \(\frac{1}{n}\)
⇒ a + (m – 1)d = \(\frac{1}{n}\) ……..(i)
तथा a_n = \(\frac{1}{m}\)
⇒ a + (n – 1)d = \(\frac{1}{m}\) …………….(ii)
समीकरण (i) में से समीकरण (ii) को घटाने पर,
(m – n)d = \(\frac{1}{n}-\frac{1}{m}\)
⇒ (m – n)d = \(\frac{m-n}{m n}\)
⇒ d = \(\frac{1}{mn}\)
समीकरण (i) में d = \(\frac{1}{mn}\) रखने पर,
a + (m – 1)\(\frac{1}{mn}\) = \(\frac{1}{n}\)
⇒ a + \(\frac{1}{n}\) – \(\frac{1}{mn}\) = \(\frac{1}{n}\)
⇒ a = \(\frac{1}{mn}\)
∴ अभीष्ट योगफल S_mn = \(\frac{mn}{2}\){2a + (mn – 1)d}
⇒ S_mn = \(\frac{mn}{2}\){\(\frac{2}{mn}\) + (mn – 1) × \(\frac{1}{mn}\)}
⇒ S_mn = \(\frac{1}{2}\)(mn + 1)
अत: mn पदों का योगफल
= \(\frac{1}{2}\)(mn + 1) इति सिद्धम् ।

प्रश्न 9.
एक समान्तर श्रेढी के प्रथम n पदों केयोगफल को S_n द्वारा दर्शाया जाता है। इस श्रेढी में यदि S₅ + S₇ = 167 तथा S₁₀ = 235 है, तो समान्तर श्रेणी ज्ञात कीजिए ।
हल :
माना श्रेढी का प्रथम पद = a
और सार्वअंतर = d
n पदों का योगफल
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
दिया है,
S₅ + S₇ = 167
⇒ \(\frac{5}{2}\)(2a + 4d) + \(\frac{7}{2}\)(2a + 6d) = 167
⇒ 5a + 10d + 7a + 21d = 167
⇒ 12a + 31d = 167 ……….. (1)
इसी प्रकार,
S₁₀ = 235
⇒ \(\frac{10}{2}\)(2a + 9d) = 235
⇒ 2a + 9d = 47 ……….. (2)
समी. (1) व (2) से,
a = 1 और d = 5
अत: अभीष्ट A. P. 1, 6, 11… है ।

प्रश्न 10.
दर्शाइए कि उस A. P का योग, जिसका प्रथम पद a, द्वितीय पद b और अन्तिम पद c हो \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\) के बराबर है।
हल :
दिया है,
प्रथम पद = a
द्वितीय पद = b
∴ सार्वअन्तर (d) = ba
अन्तिम पद (l) = c
माना कि समान्तर श्रेढी में पद है।
∴ a_n = c
⇒ a + (n – 1) × d = c
⇒ a + (n – 1) × (b – a) = c
⇒ (n – 1) (b – a) = c – a
⇒ (n – 1) = \(\frac{c-a}{b-a}\)
⇒ n = \(\frac{c-a}{b-a}\) + 1
⇒ n = \(\frac{c-a+b-a}{b-a}\)
⇒ n = \(\frac{b+c-2 a}{b-a}\)
अब S_n = \(\frac{n}{2}\)[a + l]
⇒ S_n = \(\frac{(b+c-2 a)}{2(b-a)}\)[a + c]
⇒ S_n = \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\) इति सिद्धम् ।

प्रश्न 11.
एक मोटर कार A स्थान से B स्थान तक 175 किमी दूरी 70 किमी / घंटा समान गति से सभी 10 हरे यातायात्रा सिग्नलों को पार करती है। भारी यातायात के कारण यह प्रथम सिग्नल पर एक मिनट, दूसरे सिग्नल पर 3 मिनट, तीसरे सिग्नल पर 5 मिनट एवम् इसी प्रकार दसवें सिग्नल पर 19 मिनट रूकती हैं। स्थान B तक पहुँचने में इसे कुल कितना समय लगेना उपयुक्त गणितीय विधि से हल कीजिए ।
हल :
यहाँ, हम देखते हैं कि 10 सिग्नलों पर लगा समय समान्तर श्रेढी में हैं।

प्रथम पद a = 1, तथा सार्वअन्तर d = 3 – 1 = 2
∴ 10 सिग्नलों को पार करने के लगा समय = 1 + 3 + 5 + 5 + 9 + 11 + 13 + 15 + 17 + 19
= \(\frac{10}{2}\)[2 × 1 + (10 – 1) × 2]
= 5 [2 + 18] = 5 × 20 = 100 मिनट
= \(\frac{10}{60}\) घंटे = \(\frac{5}{3}\)घंटे
∴ A से B तक पहुँचाने में लगा कुल समय
= (\(\frac{175}{70}+\frac{5}{3}\))घंटे = (\(\frac{35}{14}+\frac{5}{3}\))घंटे
= (\(\frac{35 \times 3+5 \times 14}{42}\))घंटे = \(\frac{175}{42}\)घंटे
= 4\(\frac{1}{6}\) घंटे = 4 घंटे 10 मिनट

प्रश्न 12.
उस A. P. के प्रथम 15 पदों का योग ज्ञात कीजिए, जिसका पाँचवा और नवाँ पद क्रमश: 26 और 42 है।
हल :
माना सं. क्षे. का प्रथम पद a तथा सार्वअन्तर d है।
स.क्षे. का पाँचवाँ पद,
a₅ = a + (n – 1)d
⇒ 26 = a + (5 – 1)d
⇒ 26 = a + 4d ………..(i)
स.क्षे. का नव पद, a₉ = a + (9 – 1)d
42 = a + 8d ………….(ii)
समीकरण (ii) में से समीकरण (i) घटाने पर
a + 8d = 42
a + 4d = 26
4d = 16
d = 4
समीकरण (i) में d = 4 रखने पर
a + 4 × 4 = 26
a = 26 – 8 = 18
स.क्षे. के 15 पदों को योग
= [2 × 18 + (15 – 1) × 4]
= [36 + 56]
= 15 × 92
= 690

प्रश्न 13.
एक कार A स्थान से B स्थान पर 260 किमी दूरी 65 किमी / घंटा समान गति से सभी 13 हरे यातायात सिग्नलों को पार करती है। भारी यातायात के कारण यह प्रथम सिग्नल पर 4 मिनट, दूसरे सिग्नल पर 7 मिनट, तीसरे सिग्नल पर 10 मिनट एवम् इसी प्रकार तेरहवें सिग्नल पर 40 मिनट रूकती है। स्थान B तक पहुँचने इसे कुल कितना समय होगा? उपयुक्त गणितीय विधि से हल कीजिए।
हल :

A से B तक पहुँचने में लगा कुल समय
= \(\frac{260}{65}\)घंटे + \(\frac{13}{2}\)[2 × 4 + (13 – 1) × 3] मिनट
= \(\frac{20}{5}\)घंटे + \(\frac{13}{2}\) [8 + 36] मिनट
= 4 घंटे + \(\frac{13}{2}\) × 44 मिनट = 4 घंटे + 286 मिनट
= 4 घंटे + 46 मिनट = 8 घंटे 46 मिनट

प्रश्न 14.
दर्शाइए कि (a – b)², (a² + b²), (a + b)² एक समान्तर श्रेणी में हैं।
हल :
माना a₁ = (a – b)², a₂ = (a² + b²), a₃ = (a + b)²
a₂ – a₁ = (a² + b²) – (a – b)²
= (a² + b²) – (a² + b² – 2ab)
= a² + b² – a² – b² + 2ab = 2ab … (i)
तथा
a₃ – a₂ = (a + b)² – (a² + b²)
= a² + b² + 2ab – a² – b²
= 2ab ……………(ii)
समीकरण (i) व (ii) से,
∵(a² + b²) – (a – b)² = (a + b)² – (a² + b²)
a₂ – a₁ = a₃ – a₂
अत: (a – b)², (a² + b²), (a + b²) एक स.क्षे. मैं हैं।

प्रश्न 15.
एक स. क्षे. के प्रथम 7 पदों का योग 63 है और इसके अगले 7 पदों का योग 161 है। स. क्षे. ज्ञात कीजिए ।
हल :
माना स.क्षे. का प्रथम पद ‘a’ तथा सार्व अन्तर ‘d’ है।
स. क्षे. के प्रथम 7 पदों का योग,
S_n = \(\frac{n}{2}\)[2a+ (n – 1)d]
S₇ = \(\frac{7}{2}\)[2a + (7 – 1)d]
63 = \(\frac{7}{2}\)(2a + 6d)
⇒ \(\frac{1}{2}\)(2a + 6d) = 9
⇒ a + 3d = 9
a = 9 – 3d ……………(i)
प्रश्नानुसार,
S₁₄ – S₇ = 161
\(\frac{14}{2}\)[2a + (14 – 1)d] – 63 = 161
\(\frac{14}{2}\)[2a + (14 – 1)d] = 161 + 63
7[2(9 – 3d) + 13d] = 224
18 – 6d + 13d = \(\frac{224}{7}\)
7d = 32 – 18
d = \(\frac{14}{7}\) = 2
समीकरण (i) से, a = 9 – 3 × 2 = 3
अतः समान्तर श्रेढी हैं :
a, a + d, a + 2d, a + 3d, ……………..
= 3, 5, 7, 9, …………

प्रश्न 16.
एक समान्तर श्रेणी के सभी 11 पदों का योगफल ज्ञात कीजिए, जिसका मध्य पद 30 है।
हल :
माना स. क्षे. का प्रथम पद a तथा सार्वअन्तर d है।
∵ स. क्षे. में 11 पद हैं।
∴ मध्य पद = (\(\frac{n+1}{2}\)) वाँ पद = (\(\frac{11+1}{2}\))वाँ पद = 5वाँ पद
a₅ = a + (5 – 1)d
30 = a + 5d ……………(i)
स. क्षे. के 11 पदों का योग = \(\frac{11}{2}\)[2a + (11 – 1)d]
= \(\frac{11}{2}\)[2a + 10d]
= \(\frac{11}{2}\) × 2(a + 5d)
= 11 × 30 = 330

प्रश्न 17.
निम्न समीकरण को हल कीजिए: 1 + 5 + 9 + 13 + ……………. + x = 1326
हल :
यहाँ, प्रथम पद, a = 1
सार्वअन्तर, d = 5 – 1 = 4
प्रश्नानुसार,
S_n = 1326
⇒ \(\frac{n}{2}\)[2 × 1 + (n – 1)4] = 1326
⇒ \(\frac{n}{2}\)(2 + 4n – 4) = 1326
⇒ \(\frac{n}{2}\)(4n – 2) = 1326
⇒ n(2n – 1) = 1326
⇒ 2n² – n – 1326 = 0
⇒ 2n² – 52n + 51n – 1326 = 0
⇒ 2n(n – 26) + 51(n – 26) = 0
⇒ (n – 26) (2n + 51) = 0
यदि n – 26 = 0 तो n = 26
और यदि 2n + 51 = 0, तो n = \(\frac{-51}{2}\) असम्भव, क्योंकि ‘n’ एक प्राकृतिक संख्या है।
अर्थात् n = 14 है।
प्रश्नानुसार,
14वाँ पद = x
⇒ a + (14 – 1)d = x
⇒ 1 + 13 × 4 = x
अतः x = 53

प्रश्न 18.
यदि संख्याएँ a, 7, b, 23, c एक समान्तर श्रेढी में है, तो a, b तथा c के मान ज्ञात कीजिए।
हल :
माना स. क्षे. का सार्व अन्तर d है।
प्रश्नानुसार,
a₂ = 7
a + (2 – 1)d = 7
a + d = 7 ……………(i)
तथा a₄ = 23
a + (4 – 1)d = 23
a + 3d = 23 ……………(ii)
समीकरण (ii) में से समीकरण (i) घटाने पर

समीकरण (i) में, d = 8 रखने पर
a + 8 = 7
⇒ a = 7 – 8 = – 1
तीसरा पद, a₃ = a + (3 – 1)d
b = – 1 + 2 × 8 = 15
पाँचवाँ पद,
a₅ = a + (5 – 1)d
c = – 1 + 4 × 8 = 31.
अतः a = – 1, b = 15, c = 31

प्रश्न 19.
समानान्तर श्रेणी
20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\), ………… का कौन-सा पद प्रथम ऋणात्मक पद है?
हल :
प्रश्नानुसार, समानान्तर श्रेणी
20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\), …….. = 20, \(\frac{77}{4}\), \(\frac{37}{2}\), \(\frac{71}{4}\), ………..
यहाँ, a = 20, d = \(\frac{77}{4}\) – 20 = \(\frac{77-80}{4}\) = \(\frac{-3}{4}\)
माना पहला ऋणात्मक पद a_n है।

अतः दी गयी समानान्तर श्रेणी का 28वाँ पद प्रथम ऋणात्मक पद होगा।

प्रश्न 20.
यदि दो समानान्तर श्रेणियों के प्रथम पदों के योगफलों का अनुपात (7+ 1) (4 + 27) है, तो उनके 9वें पदों का अनुपात ज्ञात कीजिए।
हल :
दिया है, समानान्तर श्रेणियों के प्रथम n पदों के योगफलों का अनुपात :

अतः दो समानान्तर श्रेणियों के 9वें पदों का अनुपात = 24 : 19

प्रश्न 21.
n के किस मान के लिए, दो समानान्तर श्रेणियों 63, 65, 67, ….. तथा 3, 10, 17 ……… के n वें पद समान होंगे?
हल :
पहल समानान्तर श्रेणी 63, 65, 67, …………..
a = 63, d = 65 – 63 = 2
a_n = a + (n – 1)d
= 63 + (n – 1)d
= 63 + 2n – 2 = 61 + 2n
दूसरी समानान्तर श्रेणी 3, 10, 17, ………..
a = 3, d = 10 – 3 = 7
a_n = a + (n – 1)d
= 3 + (n – 1) (7) = 3 + 7n – 7
= 7n – 4
प्रश्नानुसार
61 + 2n = 7n – 4
61 + 4 = 7n – 2n
65 = 5n
n = \(\frac{65}{5}\) = 13
अतः दोनों समानान्तर क्षेणियों का 13वाँ पद समान हैं।

प्रश्न 22.
एक समान्तर श्रेढी के प्रथम 6 पदों का योग 42 है। इसके 10वें पद तथा 30वें पद में अनुपात 1 : 3 का है। इस समान्तर श्रेढी का प्रथम पद तथा 13वाँ पद ज्ञात कीजिए।
हल :
माना स.क्षे. का प्रथम पद a तथा सार्व अन्तर d है।
स. क्षे. का 6 पदों का योग,
S₆ = \(\frac{6}{2}\)[2a + (6 – 1)d]
⇒ 42 = 3 (2a + 5d)
⇒ 2a + 5d = 14 ………(i)
10वें तथा 30वें पद में अनुपात 1 : 3

समीकरण (i) में d = 2 रखने पर
2a + 5 × 2 = 14
⇒ 2a = 14 – 10
⇒ a = \(\frac{4}{2}\) = 2
अतः स.क्षे. का 13वाँ पद = 2 + (13 – 1) × 2
= 2 + 24 = 26

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क)

1. संख्याओं के एक निश्चित नियमानुसार क्रम को ………………. कहते हैं।
2. वह श्रेणी जिसमें अगल पद, पहले पदे में एक निश्चित संख्या जोड़ने अथवा घटाने पर प्राप्त होती है, ………….. श्रेणी कहलाती है।
3. समान्तर श्रेणी का …………… धनात्मक, ऋणात्मक या शून्य हो सकता है।
4. यदि समान्तर श्रेणी में पदों की संख्या निश्चित है, तो उसे ……………… समान्तर श्रेणी कहते हैं।
5. यदि 3k – 2, 4k – 6 तथा k + 2 एक समान्तर श्रेणी के क्रमित पद हैं, तो k का मान ………….. हैं।

हल :

1. अनुक्रम
2. समान्तर
3. सार्व अन्तर,
4. परिमित,
5. 2.

निम्न में सत्य / असत्य ज्ञात कीजिए :

प्रश्न (ख)

1. यदि समान्तर श्रेणी में पदों की संख्या अपरिमित है, तो उसे परिमित समान्तर श्रेणी कहते हैं।
2. श्रेणी 21, 18, 15, …………… का 8वाँ पद शून्य है।
3. श्रेणी 5, 9, 13 …………. 185 में अन्तिम पद से पहले पद की और 9वीं पद 152 है।
4. दो अंकों की 30 संख्याएँ उसे भाग्य हैं।
5. समान्तर श्रेणी 3, 15, 27, 39, ….. का 31वाँ पद 21 वे पद से 120 अधिक है।

हल :

1. असत्य,
2. सत्य,
3. असत्य,
4. सत्य,
5. सत्य।

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
एक समान्तर श्रेणी का प्रथम पद p है तथा सार्वअन्तर q है, तो उसका 10वाँ पद है :
(A) q + 9p
(B) p – 9q
(C) p + 9g
(D) 2p + 9q
हल :
a10 = a + (n – 1)d
⇒ a10 = p + (10 – 1) × q
⇒ a10 = p + 9q
अत: सही विकल्प (C) है।

प्रश्न 2.
x का मान जिसके लिए 2x, (x + 10) तथा (3x + 2) एक समान्तर श्रेणी के क्रमिक पद हैं, है:
(A) 6
(B) – 6
(C) 18
(D) – 18
हल :
∵ दिए गए पद स.क्षे. के क्रमिक पद हैं।
∴ (x + 10) – 2x = (3x + 2) – (x + 10)
⇒ – x + 10 – 2x = 3x + 2 – x – 10
⇒ – x + 10 = 2x – 8
⇒ – x – 2x = – 8 – 10
⇒ – 3x = – 18
⇒ x = 6
अत: सही विकल्प (A) है।

प्रश्न 3.
समांतर श्रेणी \(\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}\), ……………… का सार्वअंतर है :
(A) 1
(B) \(\frac{1}{p}\)
(C) – 1
(D) – \(\frac{1}{p}\)
हल :
सार्व अंतर = \(\frac{1-p}{p}-\frac{1}{p}=\frac{1-p-1}{p}\)
= \(\frac{-p}{p}\)
= – 1
अत: सही विकल्प (C) है।

प्रश्न 4.
समान्तर श्रेणी a, 3a, 5a, …………. का n वाँ पद है :
(A) na
(B) (2n – 1) a
(C) (2n + 1)a
(D) 2na
हल :
प्रथम पद, A = a
सार्वअन्तर
D = 3a – a = 2a
∴ n वाँ पद = A + (n – 1) D
= a + (n – 1) × 2a
= a + 2na – 2a
= 2na – a
= (2n – 1)a
अत: सही विकल्प (B) है।

प्रश्न 5.
समान्तर श्रेणी 5, 9, 13, ………., 185 में कितने पद हैं?
(A) 31
(B) 51
(C) 46
(D) 40
हल :
प्रथम पद, a = 5
सार्वअन्तर, d = 9 – 5 = 4
पदों की संख्या, n = ?
∵ an = a + (n – 1)d
⇒ 185 = 5 + (n – 1) × 4
⇒ n – 1 = \(\frac{180}{4}\) = 45
⇒ n = 45 + 1 = 46
अतः सही विकल्प (C) है।

प्रश्न 6.
उस समान्तर श्रेणी, जिसका n वाँ पद an = (3n + 7) है, का सार्वअंतर है :
(A) 3
(B) 7
(C) 10
(D) 6
हल :
n वाँ पद,
an = 3n + 7
प्रथम पद, a1 = 3 × 1 + 7 = 10
द्वितीय पद, a2 = 3 × 2 + 7 = 13
सार्व अंतर, = a2 – a1 = 13 – 10 = 3
अत: सही विकल्प (A) है।

प्रश्न 7.
एक समांतर श्रेणी का प्रथम पद 5 है तथा अंतिम पद 45 है। यदि सभी पदों को योगफल 400 हो, तो पदों की संख्या है :
(A) 20
(B) 8
(C) 10
(D) 16
हल :
दिया है, a = 5, l = 45, Sn = 400
∵ Sn = \(\frac{n}{2}\)(a + l)
⇒ 400 = \(\frac{n}{2}\) (5 + 45)
⇒ n = \(\frac{400 \times 2}{50}\) = 16
अत: सही विकल्प (D) है।

प्रश्न 8.
एक समान्तर श्रेणी – 15 – 11 – 7, …………… 49 का 9 वाँ पद है
(A) 32
(B) 0
(C) 17
(D) 13
हल :
दियाहै,
a = – 15, d = – 11 – (-15) = – 11 + 5 = 4 तथा n = 9
an = a + (n – 1)d
a9 = – 15 + (9 – 1) × 4
= – 15 + 32 = 17
अत: सही विकल्प (C) है।

प्रश्न 9.
निम्नलिखित में से कौन-सी समांतर श्रेढी नहीं है?
(A) – 1.2, 0.8, 2.8, ………….
(B) 3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),….
(C) \(\frac{4}{3}, \frac{7}{3}, \frac{9}{3}, \frac{12}{3}\), ………………..
(D) \(\frac{-1}{5}, \frac{-2}{5}, \frac{-3}{5}\), …………..
हल :
(C) में सार्वअन्तर समान नहीं है।
\(\frac{7}{3}-\frac{4}{3}=\frac{3}{3}\)
तथा \(\frac{9}{3}-\frac{7}{3}=\frac{2}{3}\)
अर्थात् \(\frac{7}{3}-\frac{4}{3} \neq \frac{9}{3}-\frac{7}{3}\)
अंत: सही विकल्प (C) है।

प्रश्न 10.
समान्तर श्रेणी 5, 8, 11, ………. 47 का अंतिम पद से (प्रथम पद की ओर) दूसरा पद है:
(A) 50
(B) 45
(C) 44
(D) 41
हल :
दिया है, a = 5, d = 8 – 5 = 3, l = 47 तथा n = 2
स. क्षे. का अंत से n वाँ पद = l – (n – 1)d
= 47 – (2 – 1) × 3
= 47 – 3 = 44
अत: सही विकल्प (C) है।

प्रश्न 11.
दी गई श्रेढी 2,\(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………….. A.P. में है, तो इसका 3 वाँ पद होगा : (A) – 3
(B) 3
(C) 2
(D) 4
हल :
a = 2, d = \(\frac{5}{2}\) – 2 = \(\frac{5-4}{2}=\frac{1}{2}\)
an = a + (n – 1)d
∴ a3 = 2 + (3 – 1) × \(\frac{1}{2}\)
= 2 + 2 × \(\frac{1}{2}\) = 2 + 1 = 3
अत: सही विकल्प (B) है।

प्रश्न 12.
समान्तर श्रेढी 21, 42, 63, 84, ………….. का कौन-सा पद 210 है ?
(A) 9th
(B) 10th
(C) 11th
(D) 12th
हल :
दी गयी समान्तर श्रेढी है :
21, 42, 63, 84, ………. 210
यहाँ a = 21, d = 42 – 21 = 21 तथा an = 210
∴ an = 210
a + (n – 1)d = 210
⇒ 21 + (n – 1) × 21 = 210
(n – 1) × 21 = 210 – 21 = 189
(n – 1) = \(\frac{189}{21}\) = 9
n = 9 + 1 = 10
अत: विकल्प (B) सही है।

प्रश्न 13.
3 के प्रथम पाँच गुणजों का योगफल है :
(A) 45
(B) 55
(C) 65
(D) 75
हल :
3 के प्रथम पाँच गुणज हैं :
3, 6, 9, 12, 15
यहाँ a = 3, n = 3 तथा l = 15
सूत्र Sn = \(\frac{n}{2}\)(a + l) से
S5 = \(\frac{5}{2}\)(3 + 15)
⇒ S5 = \(\frac{5}{2}\) × 18 = 45
अत: सही विकल्प (A) है।

प्रश्न 14.
m के किस मान के लिए 10, m, – 2 समान्तर श्रेढी में होंगे:
(A) m = 4
(B) m = 3
(C) m = 2
(D) m = 1
हल :
a2 – a1 = a2 – a2
⇒ m – 10 = – 2 – m
⇒ m + m = – 2 + 10
2m = 8
m = \(\frac{8}{2}\)
m = 4
अत: सही विकल्प (A) है।

प्रश्न 15.
यदि एक समान्तर श्रेढी का n वाँ पद (2n + 1) है, तो उसके प्रथम तीन पदों का योगफल है :
(A) 6n + 3
(B) 15
(C) 12
(D) 21
हलं :
दिया गया है:
an = 2n + 1
n = 1, 2, 3 रखने पर
a1 = 2 × 1 + 1 = 3
a2 = 2 × 2 + 1 = 5
a3 = 2 × 3 + 1 = 7
अतः समान्तर श्रेढी है-
3, 5, 7,…
सार्वअन्तर (d) = 5 – 3 = 2
या 7 – 5 = 2
प्रथम तीन पदों का योगफल (Sn)
= \(\frac{n}{2}\) [2a + (n – 1) × d)]
= \(\frac{3}{2}\) [2 × 3 + (3 – 1) × 2]
= \(\frac{3}{2}\) [6 + 4]
= \(\frac{3}{2}\) × 10 = 15
अत: विकल्प (B) सही है।

Jharkhand Board JAC Class 10 Science Important Questions Chapter 14 Sources of Energy Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 14 Sources of Energy

Additional Questions and Answers

Question 1.
Distinguish between :
(1) Solar Cooker and Solar Cell Panel
Answer:

Solar Cooker	Solar Cell Panel
1. It converts solar energy into heat.	1. It converts solar energy into electricity.
2. It is mainly designed for cooking purpose, i.e., for domestic use	2. It is designed for delivering enough electricity for commercial use.
3. It consists of mirror, glass sheet, black wooden box.	3. It consists of large number of solar cells and silicon is used in solar cells.
4. It is useful for cooking only at certain times of the day.	4. The limitation of using solar energy is overcome by using it.

(2) Fossil fuels and Bio fuels
Answer:

Fossil Fuels	Bio Fuels
1. It is a conventional source of energy mainly used in industries and vehicles.	1. It is a conventional source of energy mainly for domestic use.
2. Coal, petroleum and natural gas are fossil fuels.	2. Wood, cow-dung cakes, biogas, etc. are bio fuels.
3. The fossil fuels are non-renewable and exhaustible source of energy.	3. Bio fuels are renewable source of energy in the form of biomass.
4. Fossil fuels are the major fuels used all over the world.	4. Bio fuels are mainly used in rural area.

Question 2.
Give scientific reasons for the following statements:
(1) From environmental point of view, we should reduce the use of fossil fuels.
Answer:
The fossil fuels are non-renewable source of energy and in limited reserves.

• The growing demand for energy at global level are largely meet by fossil fuels.
• Excessive burning of fossil fuels – coal and petroleum increases the level of air pollution. example:
• Carbon dioxide, sulphur dioxide and oxides of nitrogen are released in excess in atmosphere causing such air pollution.
• Carbon dioxide is a main greenhouse gas that causes global warming.
• Sulphur dioxide and oxides of nitrogen lead to acid rain which affects water and soil resources.

So, to reduce the adverse effects on environment, we should reduce the use of fossil fuels.

(2) Though hydropower is renewable source of energy, it creates certain problems.
Answer:
Hydropower is renewable source of energy because water gets refilled in reservoir during raining.

• High-rise dams are constructed to generate hydroelectricity.
• Such dams are constructed in hilly areas causing environmental changes.
• Large eco-systems are submerged under water and destroyed.
• The submerged vegetation decaying anaerobically releases large amount of greenhouse gas, i.e., methane.
• Large areas of agricultural land is also submerged. Many natives are thus displaced.
• Rural-tribal human races may lose their habitat.

So, Hydropower plant construction creates certain problems.

(3) The use of dung as raw material for biogas production is profitable than using it only as fertiliser or fuel.
Answer:
Animal dung is used as raw material c for production of biogas. It provides a safe and i efficient method of waste disposal.

• Besides it, by supplying biogas as a fuel gas, cooking, lighting, etc. we can save natural gas.
• The residual slurry can be used as excellent manure that is rich in nitrogen and phosphorous.

This natural fertiliser maintains soil fertility and agriculture practice becomes more economical.

(4) Inspite of the high cost, solar cells are widely used.
Answer:
A special grade silicon is used for making solar cells and silver is used for interconnection of cells in panel.

• This makes the solar cells expensive.
• But it converts solar energy into electricity and about 0.7 W of electricity is produced when exposed to the Sunlight.
• Solar cells and panels are widely used in artificial satellite, space probes like Mars orbiters, wireless transmission systems, radio or TV relay stations, traffic signals, calculators and in many toys.
• At domestic level, maximum use of solar energy can be achieved by using solar roof top thus saving electricity.
• So, solar cells are widely used.

(5) Generation of nuclear energy is hazardous.
Answer:
Nuclear fission and nuclear fusion are two processes used to generate nuclear energy.

• Radioactive heavy atoms such as uranium, plutonium or thorium are used in a nuclear fission process whereas hydrogen isotopes are used in a nuclear fusion.
• There is a risk of accidental leakage of nuclear radiation.
• There is a possibility of high risk of environmental contamination i.e., radioactive pollution.
• Radioactive waste is hazardous to human health and environment. Radiations which are emitted from the radioactive waste can cause mutation (Change in DNA) in living organisms.
• It creates various disorders and cancer in human beings.

So, generation of nuclear energy is hazardous.

Question 3.
Carefully observe the given diagram / chart and answer the questions based on it.
(1) Observe the following pie-chart showing the major sources of energy.

Questions:
(1) Identify ‘a’ and state where is it used as a fuel to generate electric energy through conversion of heat energy.
Answer:
Coal. It is used as a fuel in thermal power plant.

(2) Label ‘b’ and give the name of processes to generate it.
Answer:
Nuclear energy. The processes to generate nuclear energy are: nuclear fission and nuclear fusion.

(3) Label ‘c’ and state the name of any two place where such power plants are established.
Answer:
Hydropower. Hydropower plants are established at Tehri Dam on the Ganga river and Sardar Sarovar Project on the Narmada river.

(4) Refer to the pie-chart above and arrange the sources of energy in an ascending order.
Answer:
Nuclear < hydro < petroleum and natural gas < coal

Questions:

(1) Label a, b and c in given diagram.
Answer:
a – Digester, b – Slurry, c – Gas outlet

(2) How is biogas obtained from slurry?
Answer:
Slurry consists of bio-wastes / biomass and biogas can be obtained through decomposition by anaerobic microorganisms.

(3) Why is biogas considered as an excellent fuel?
Answer:
Biogas is considered as an excellent fuel because it burns without smoke, leaves no ash residue and has high heating capacity.

(4) Which are the gases present in biogas?
OR
What are components of biogas?
Answer:
Biogas mainly consists gases like methane, carbon dioxide, hydrogen and hydrogen sulphide.

(5) Which are abundant nutrients in the slurry left behind in a biogas plant?
Answer:
Nitrogen and phosphorous are abundant nutrients in the slurry left behind in a biogas plant.

Questions :
(1) Which metal is used in the panel for the interconnection of cells?
Answer:
Silver

(2) Which energy conversion occurs by using solar cells?
Answer:
Solar energy is converted into electricity.

(3) Write any two uses of solar cells for space science.
Answer:
Solar cells are used as a main energy source in an artificial satellites and space probes like Mars orbiters.

(4) If 12 solar cells are interconnected in the panel. How many watt of electricity can be produced from it when exposed to Sun?
Answer:
About 8.4 W

(5) Why is silicon used in a solar cells?
Answer:
Silicon is a natural semiconductor, easily obtained and easily modified to convert solar energy into electricity.

Objective Questions and Answers

Question 1.
Answer the following questions in short:
(1) Name the form of energy obtained from sea.
Answer:
The forms of energy obtained from sea are tidal energy, wave energy and ocean thermal energy.

(2) Which fuel is considered as a cleaner source for vehicles?
Answer:
CNG (Compressed Natural Gas)

(3) State the name of fuels that we are using.
Answer:
We are using wood, coal, petrol, diesel, CNG, kerosene, LPG, etc. as fuels.

(4) What is a source of energy?
Answer:
A source of energy is one which is capable of providing useful energy in sufficient amount.

(5) Which energy conversion occurs in a hydropower plant?
Answer:
In a hydropower plant, potential energy of stored water is converted into electrical energy.

(6) Which is a main constituent of biogas? What is its content?
Answer:
Methane gas is a main constituent of biogas and it is about 75 %.

(7) What are the conventional sources of energy?
Answer:
Fossil fuels, i.e., coal and petroleum are the conventional source of energy, which are used on a large scale.

(8) Name two elements that are used in manufacturing solar cell panels.
Answer:
Silicon, Silver

(9) Name the devices to harness solar energy.
Answer:
Solar cell, solar cooker, solar water heater.

(10) Where are the fossil fuels used directly?
Answer:
In gas stoves, vehicles and thermal power plants, fossil fuels are used directly.

(11) Why is charcoal used as a fuel?
Answer:
Charcoal is used as a fuel because it burns without flames is comparatively smokeless s and has a higher heat generation efficiency.

(12) Why is the term thermal power plant used?
Answer:
The term thermal power plant is used because in such power plants fuel is burnt to produce heat energy which is converted into electrical energy.

(13) State the name of any two greenhouse gases.
Answer:
CO₂ and CH₄

(14) Which fuels leave residue like ash on burning?
Answer:
Wood, charcoal and coal leave residue like ash on burning.

(15) How anaerobic microorganisms help in formation of fuel?
Answer:
Anaerobic microorganisms (Methanogenic bacteria) produce biogas by decomposition of complex compounds that present in cow-dung slurry.

(16) State two advantages of wind energy.
Answer:
Advantages of wind energy :

• It is an environment-friendly energy
• It is efficient source of renewable energy.

(17) What is the age of the sun? What will be its expected lifespan?
Answer:
The sun is 5 billion years old and another 5 billion years will its expected lifespan.

(18) Why is our demand for energy increasing day by day?
Answer:
Our demand for energy is increasing day by day because –

• We use machines to do more and more of our tasks, which needs more energy.
• Industrialisation improves our living standards. The energy requirement therefore constantly rises.

(19) Where is solar cell panels are mounted for domestic use? What is its advantages?
Answer:
The solar cell panels are mounted on specially designed inclined rooftops. More solar rays are incident over it, producing more electric
energy.

(20) Name the two oxides that causes acid rain.
Answer:
Oxides of nitrogen and sulphur cause acid rain.

(21) What is the minimum speed of wind required by a windmill to maintain the necessary speed of turbine in electric generator?
Answer:
More than 15 km/h

(22) Which are the renewable sources of energy?
Answer:
Hydropower, biomass, solar energy, energy from the sea, wind energy, etc. are renewable sources of energy.

Question 2.
Define: OR Explain the terms :
(1) Nuclear fission
Answer:
The process of splitting a heavy nucleus into smaller nuclei along with the release of large amount of energy is called nuclear fission.

(2) Nuclear fusion
Answer:
The process of joining lighter nuclei to make a heavier nucleus along with release of tremendous amount of energy is called nuclear fusion.

(3) Solar constant
Answer:
The solar energy reaching unit area on outer surface of the earth’s atmosphere which is exposed perpendicularly to the rays of sun at the average distance between the sun and earth is called solar constant.

(4) Geothermal energy
Answer:
The heat energy obtained in the form of steam from hot spots region is called geothermal energy.

(5) Biogas
Answer:
A fuel gas that is produced anaerobically from biomass is called biogas.

(6) Conventional sources of energy
Answer:
Those sources of energy which meet our major energy requirement are known as conventional sources of energy.

(7) Non-conventional sources at energy
Answer:
Those sources of energy which meet only a limited energy requirement are known as non-conventional sources of energy.

(8) Wind energy farm
Answer:
A number of windmills erected over a large area for the commercial use of wind energy is known as wind energy farm.

(9) Fossil fuels
Answer:
The fuels such as coal, petroleum and natural gas, that were formed millions of years ago from dead and fossiliged animals or plants in the ground are called fossil fuels.

Question 3.
Fill in the blanks:

1. Biogas obtained from decomposition of biomass by the activity of ……………… microorganisms.
2. Petroleum products are ……………… fuel.
3. Hydropower plants convert the ……………… of falling water into electricity.
4. When wood is burnt in a limited supply of oxygen, ……………… is left behind as a residue.
5. Manure obtained from biogas plant is rich in ……………… and ………………
6. ……………… plant is an efficient method of bio waste disposal.
7. ……………… is the best process to capture solar energy and convert it into biomass.
8. Use of ……………… mirror would be best for in solar cooker.
9. ……………… is abundant in nature but availability of its special grade is limited.
10. The hydrogen bomb is based on ……………… reaction.
11. India is ranked ……………… in harnessing wind-energy for the production of electricity.
12. The total ……………… of the universe remains constant because it can neither be created nor destroyed.

Answer:

1. anaerobic
2. fossil
3. potential energy
4. charcoal
5. nitrogen, phosphorous
6. Biogas
7. Photosynthesis
8. concave
9. Silicon
10. thermonuclear fusion
11. fifth
12. energy

Question 4.
State whether the following sentences are true or false:

1. The air pollution is caused by burning of fossil fuels.
2. The potential energy of flowing water gets, transformed into kinetic energy by collecting the water in dam.
3. In solar cookers, a plain mirror is used to converge the sunlight.
4. Fossil fuels are the major fuels used for generating electricity.
5. Submerged vegetation rots under anaerobic condition producing large amounts of methane.
6. Biogas is derived totally from animal biomass.
7. The level of water in the sea rises and falls due to the gravitational pull of the moon on the spinning earth.
8. Nuclear energy is the best alternative of fossil fuel because of its environment friendly nature.
9. The nuclear bomb is embedded in a deuterium and lithium containing substance.
10. The assembly of the solar cell may be pollution free but the actual operation of it may cause some environmental harm.
11. The energy sources that can be regenerated are called renewable energy sources.
12. Nuclear fusion reactions are the source of energy in the sun and other stars.

Answer:

1. True
2. False
3. False
4. True
5. True
6. False
7. True
8. False
9. True
10. False
11. True
12. True

Question 5.
Graph / diagram based question :
Which of the following graph is correct for Burning of fossil fuel → level of air pollution?

Answer:

Question 6.
Match the following:
(1)

Column I	Column II
1. Silicon	a. nuclear reactor
2. Hot-springs	b. leaves residue like ash
3. Fission of uranium atom	c. used in solar cells
4. Charcoal	d. geothermal energy

Answer:
(1 – c), (2 – d), (3 – a), (4 – b).

(2)

Column I	Column II
1. Fossil fuel	a. Plutonium, thorium
2. Biomass	b. coal, petroleum
3. Nuclear fuel	c. CNG
4. Clean fuel	d. wood, cow-dung cake

Answer:
(1 – b), (2 – d), (3 – a), (4 – c).

(3)

Column I	Column II
1. Geothermal energy plant	a. Traffic signals
2. Solar cell panels	b. New Zealand
3. Wind energy farm	c. Tehri dam
4. Hydropower plant	d. Kanyakumari

Answer:
(1 – b), (2 – a), (3 – d), (4 – c).

Question 7.
Select the correct alternative from those given below each questions:
1. Which of the following is a main component of biogas?
A. Methane
B. Hydrogen
C. Hydrogen sulphide
D. Oxygen
Answer:
A. Methane

2. Which fuel is used in thermal power plant?
A. Biomass
B. Fossil fuel
C. Wood
D. Charcoal
Answer:
B. Fossil fuel

3. Which of the following is the adverse effect caused by burning of fossil fuels?
A. Acid rain
B. Greenhouse effect
C. Both A and B
D. Agricultural land submerged
Answer:
C. Both A and B

4. Find greenhouse gases.
A. Hydrogen and hydrogen sulphide
B. Sulphure dioxide and nitrogen
C. Nitrogen and hydrogen
D. Methane and carbon dioxide
Answer:
D. Methane and carbon dioxide

5. Which of the following fuel does not leave ash-like residue while burning?
A. Wood
B. Charcoal
C. Biogas
D. Coal
Answer:
C. Biogas

6. Which one of the following is a non-renewable source of energy?
A. Bio fuel
B. Fossil fuel
C. Wind power
D. Tidal energy
Answer:
B. Fossil fuel

7. Which of the following exploits energy due to temperature difference at surface water and water at depth?
A. Tidal energy
B. Wave energy
C. Ocean thermal energy
D. All of the given
Answer:
C. Ocean thermal energy

8. Statement A: We need to look for more and more source of energy.
Reason R: There are only limited reserves of fossil fuels.
Which option is correct for statement A and reason R?
A. Both A and R correct, R is explanation of A.
B. Both A and R correct, R is not explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
Answer:
A. Both A and R correct, R is explanation of A.

9. Charcoal is considered as a better fuel compared to wood because …
A. it burns without flames.
B. it is comparatively smokeless.
C. it has a higher heat generation efficiency.
D. all of the given.
Answer:
D. all of the given.

10. From which of following, electricity can be generated without the use of turbine?
A. Solar energy
B. Tidal energy
C. Geothermal energy
D. Wave energy
Answer:
A. Solar energy

11. Which is the fundamental reaction in nuclear weapon for destructive purposes?
A. Fusion chain
B. Fission chain
C. Radiation chain
D. Thermal chain
Answer:
B. Fission chain

12. By which of the following greenhouse effect can be achieved in a solar cooker?
A. Black surface
B. Concave mirror
C. Convex mirror
D. Covered glass plate
Answer:
D. Covered glass plate

13. Which one of following is used as a main source of energy in artificial satellite?
A. Fossil fuel
B. Uranium
C. Solar cells
D. Gravitation pull
Answer:
C. Solar cells

14. Statement X: Ocean thermal energy conversion plant can operate if the temperature difference is 20°C or more between the water at surface and water at depth. Statement Y: Wind energy farm can generate electricity only if the wind speed should be higher than 15 km/h.
Which is correct option for statement X and Y?
A. X is correct, Y is incorrect.
B. Both X and Y are correct.
C. X is incorrect, Y is correct.
D. Both X and Y are incorrect.
Answer:
B. Both X and Y are correct.

15. Which pf the following is the ultimate source of energy on the earth?
A. Solar
B. Wind
C. Ocean
D. Fossil fuel
Answer:
A. Solar

16. Which of the following is an environmental friendly source of energy?
A. Wind energy
B. Fossil fuel
C. Nuclear energy
D. Solar energy
Answer:
A. Wind energy, Solar energy

17. Statement A: Many of the sources ultimately derive energy from the sun.
Reason R: Nuclear fusion reactions are the source of energy in the sun.
Which option is correct for statement A and reason R?
A. Both A and R correct, R is explanation of A.
B. Both A and R correct, R is not explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
Answer:
B. Both A and R correct, R is not explanation of A.

Question 8.
Answer as directed (Miscellaneous):
(1) Write full form of CNG, OTEC; MeV
Answer:
CNG – Compressed Natural Gas
OTEC-Ocean Thermal Energy Conversion
MeV – Mega-electron Volts

(2) State the units of Solar constant, Electricity, Energy.
Answer:
Solar constant – kW / m²
Electricity – MW
Energy – eV or Joules

(3) Find mismatched pair :
I. Fossil fuel – Air pollution
II. Wind energy farm – Large area of land
III. Biogas – Aerobic microorganisms
IV. Geothermal energy – Hot springs
Answer:
III. Biogas – Aerobic microorganisms

(4) Find correct sequence for following events :
I. Winds to blow
II. Kinetic energy of huge waves near the sea-shore
III. Solar radiation
IV. Electricity generated
V. Waves generated
Answer:
III → IV → II → IV

(5) Find mismatched pair :
I. Bio fuel-wood, cow-dung cake
II. Solar cell panel – Silicon, copper
III. Windmill – Large fan, turbine of generator
IV. Nuclear energy – Fission and Fusion reaction
Answer:
II. Solar cell panel – Silicon, copper

(6) Sequentially arrange the events that occur in biogas plant.
p. activity of anaerobic microbes
q. The slurry left behind is used as manure
r. cow-dung and water is mixed and this slurry is fed into the digester
s. Methane gas is generated
t. Breakdown of complex compounds of the cow-dung slurry.
Answer:
r → p → t → s → q

Value Based Questions With Answers

Question 1.
Your father decided to fit CNG kit in his petrol car. Your mother is a working woman. Your father has managed their working time in such a way that he took your mother with him.

Your school is just 3 km away from your home. Your father also insists you to use bicycle instead of fuel using vehicle for going to school.
Questions:

1. Why CNG is prefered as a fuel?
2. Why we need to conserve fossil fuels?
3. What value is reflected in the above family?
4. State any two disadvantages of petrol as a fuel.

Answer:

1. CNG is a cleaner fuel than petrol and it is economical. It is less polluting.
2. The fossil fuels are non-renewable and exhaustible sources of energy. So we need to conserve them.
3. This family shows the values of responsibility to environment and they contribute their efforts to save fuels.
4. Disadvantages of petrol as a fuel: (i) It is expensive, (ii) It causes air pollution.

Question 2.
A group of students visited to a village. They noticed that village people uses dung cakes, agricultural wastes, etc. as a fuel. Students found smoke and ash particles in the environment. They requested the sarpanch to install a biogas plant and insist the people to collect the waste to produce biogas.

Questions:

1. What is the use of slurry left behind?
2. State the two uses of biogas.
3. What are the advantages of biogas plant?

Answer:

1. Slurry is used as nitrogen and phosphorous rich manure.
2. Two uses of biogas : Cooking, lighting
3. Biogas plant is a safe and efficient method for waste-disposal besides supplying energy and manure.

Question 3.
In high rised residential flats, maintainance expense per month per flat is high. Some members suggested to install solar panel and solar water heater. They explained that this would provide some electricity to light the building and may reduce electric bill amount.
Questions:

1. Which energy conversion is achieved by installation of solar panel and solar water heater?
2. What is the disadvantage of appliances which run on solar energy?
3. How much electric power is generated by a solar cell?

Answer:

1. Solar energy is converted to electricity by solar panel and in solar water heater, solar energy is converted to heat.
2. Solar appliances do not work during cloudy and rainy days.
3. 0.7 W

Practical Skill Based Questions With Answers

Question 1.
Two cars are parked In open parking area. One car is of black colour with black film on window glass. Other car is of white colour and without black film on window glass.
You just open the door and sit for 1-2 minute in each car.
Questions:

1. Which car is more hot inside?
2. Do both the cars have warm internal environment? Yes or no, why?
3. Why is black car more hotter than a white s car?
4. What are your suggestions to keep top floor cool in summer?

Answer:

1. Black car
2. Due to absorbing heat the cars are heated, making the inside very warm.
3. A black surface absorbs more heat as compared to a white surface.
4. Suggestions to keep top floor cool in summer:

• • • Paint white colour on the outer I side wall.
    • Mount china mosaic on the s terrace.
    • Prepare terrace garden.

Question 2.

A picture to demonstrate windmill and its function.
Questions :

1. Which energy conversion is shown in the picture?
2. What function does windmill show in the picture?
3. Which is the structure similar to windmill?

Answer:

1. Kinetic energy into mechanical energy.
2. To lift water from a well.
3. The structure of windmill is simillar to electric fan.

Memory Map:

Jharkhand Board JAC Class 10 Science Solutions Chapter 14 Sources of Energy Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 14 Sources of Energy

Jharkhand Board Class 10 Science Sources of Energy Textbook Questions and Answers

Question 1.
A solar water heater cannot be used to get hot water on ………………
A. a sunny day
B. a cloudy day
C. a hot day
D. a windy day
Answer:
a cloudy day

Question 2.
Which of the following is not an example of a biomass energy source?
A. wood
B. gobar-gas
C. nuclear energy
D. coal
Answer:
nuclear energy

Question 3.
Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the Sun’s energy?
A. geothermal energy
B. wind energy
C. nuclear energy
D. biomass
Answer:
nuclear energy

Question 4.
Compare and contrast fossil fuels and the sun as direct source of energy.
Answer:

Fossil Fuel	Sun
It is non-renewable and exhaustible source of energy.	It is renewable and non-exhaustible source of energy.
It causes environmental pollution.	It is non-polluting source.
Fossil fuels have to be extracted for use.	It is easily available for most of the time and at most the places throughout the year in our country.
Sun is indirect source of energy in fossil fuel.	Nuclear fusion reactions are the source of energy in the sun.
It is conventional source of energy.	It is non-conventional source of energy.

Question 5.
Compare and contrast biomass and hydroelectricity as source of energy.
Answer:

Biomass	Hydroelectricity
It causes pollution.	It is pollution free source of energy.
It is less expensive.	It is expensive in terms of construction of dams.
It is in the form of wood, cow-dung cake, agricultural residue etc.	Potential energy of water is used to generate electricity.
A fuel gas (biogas) can be produced by using it.	No gas is produced by using it.

Question 6.
What are the limitations of extracting energy from (a) the wind? (b) waves? (c) tides?
Answer:

Energy Source	Limitations of extracting energy
(a) The wind	The required wind speed should be higher than 15 km/h, the initial cost of establishment of wind farm is high, large area is required, high level of maintenence for wind farm is needed.
(b) Waves	It is costly and difficult to manage, the waves are generated by strong winds blowing across the sea.
(c) Tides	There are limited locations where dams for to harness tidal energy can be built. Efficient commercial exploitation is difficult.

Question 7.
On what basis would you classify energy sources as…
(a) renewable and non-renewable?
(b) exhaustible and inexhaustible?
Are the options given in (a) and (b) the same?
Answer:

• Exhaustible : Fossil fuel, coal will get depleted some day.
• Inexhaustible : Wind, tidal, solar energy, etc. are in the form continuing or repetitive currents of energy.
• Renewable : Biomass, if we manage properly, it can give a constant supply of energy at a particular rate.
• Non-renewable : Fossil fuels as once used up, they are lost forever and cannot be renewed.
  Yes, options given in ( a) and (b ) are same.

Question 8.
What are the qualities of an ideal source of energy?
Answer:
A good source of energy has following properties :

• It would able to do a large amount of work per unit volume or mass.
• It can be easily accessible.
• It can be easy to store and transport.
• It can be economical.

It should be pollution free and should not leave any residue.

Question 9.
What are the advantages and disadvantages of using a solar cooker? Are there places where solar cookers would have limited utility?
Answer:
Using a solar cooker:

Advantages	Disadvantages
It does not cause any pollution.	It cannot be used at night and on cloudy days.
It uses renewable/ inexhaustible source of energy.	It takes comparati-vely more time for cooking
Food nutrients are maintained during cooking, because the food is cooked at a comparatively low temperature.	The position of miror which reflects sunrays need to be monitored and the direction has to be changed again and again.
	It cannot be used for frying and to cook chapattis.

Yes, the place where Sun shine/solar energy is insufficient, the utility of solar cooker would be limited. Also on a rainy and cloudy day solar cooker cannot work.

Question 10.
What are the environmental consequences of the increasing demand for energy? What steps would you suggest to reduce energy consumption?
Answer:
The energy demand is increasing day by day. Exploiting any source of energy may adversely affect the environment more or less.

For example, use of fossil fuels cause air pollution. It may lead to greenhouse effect, acid rain, etc. Use of Hydropower to generate electricity destroys large ecosystem.

Following steps are suggested to reduce energy consumption :

• Minimise the regular use of personal vehicles and use of public transport as far as possible.
• Maximise the use of pollution free source of energy.
• Use of eco-friendly fuels such as biogas, CNG, etc.
• Switch off the light, fan and other electric appliances, whenever not in use.
• Turn off the vehicles at traffic signal while waiting for green signal.
• Use of solar cooker, solar water heater.

Jharkhand Board Class 10 Science Sources of Energy InText Questions and Answers

Question 1.
What is a good source of energy?
Answer:
A good source of energy has following properties :

• It would able to do a large amount of work per unit volume or mass.
• It can be easily accessible.
• It can be easy to store and transport.
• It can be economical.

Question 2.
What is a good fuel?
Answer:
A good fuel is the one.

• Which burns completely without producing smoke or ash.
• Which produces large amount of heat while burning a small quantity of it.
• Which is easily available and economical.

Question 3.
If you could use any source of energy for heating your food, which one would you use and why?
Answer:
If we live in a village, gobar gas will be used as a fuel for heating our food because it has high calorific value and it is easily available and more economic.

If we live in a city, either LPG or microwave or oven can be used as fuel for heating our food, because LPG is pollution free. Microwave or oven is more preferable because nutritive value of food is maintained during heating our food.

Question 4.
What are the disadvantages of fossil fuels?
Answer:
The disadvantages of fossil fuels are :

• It is non-renewable source of energy.
• Air pollution is caused by burning of fossil fuels. Acidic oxides of nitrogen and sulphur are released on burning of fossil fuels which lead to acid rain.
• Gases like carbon dioxide causes green house effect and global warming.

Question 5.
Why are we looking at alternative sources of energy?
Answer:
We are looking at alternative sources of energy because the growing demand for energy was largely met by the fossil fuels. Our technologies were also developed for using fossil fuels i.e., coal and petroleum. The fossil fuels are non-renewable source of energy.

These fuels were formed over millions of years ago and there are only limited reserves. If we are to continue consuming these sources at such alarming rate, we would soon run out energy. To overcome this, alternative sources of energy are explored.

Question 6.
How has the traditional use of wind and water energy been modified for our convenience?
Answer:
The traditional use of wind energy has been modified by constructing wind fans and wind energy farm to generate electricity. The traditional use of water energy has been modified by constructing dams and convert the potential energy of falling water into electricity for our convenience.

Question 7.
What kind of mirror – concave, convex or plain – would be best suited for use in a solar cooker? Why?
Answer:
Use of concave mirror is best suited for solar cooker because it is a converging mirror that converges large amount of sun rays into the solar cooker.

Question 8.
What are the limitations of the energy that can be obtained from the oceans?
Answer:
The limitations of energy obtained from oceans are as follows :

• The locations, where dams can be built for tidal energy are limited.
• The wave energy is obtained only where strong winds are blowing accross the sea.
• Efficient commercial exploitation of ocean thermal energy is difficult.

Question 9.
What is geothermal energy?
Answer:
The heat energy which can be exploited from the steam trapped in rocks near the hot springs is called geothermal energy.

Question 10.
What are the advantages of nuclear energy?
Answer:
The advantages of nuclear energy are :

• It produces much more energy than the other conventional sources. For example, the fission of an atom of uranium produces 10 million times the energy produced by the combustion of an atom of carbon from coal.
• As compared to the space required to harness hydro energy, thermal energy, etc. less space is required to harness nuclear energy.

Question 11.
Can any source of energy be pollution free? Why or why not?
Answer:
Solar energy, wind energy, geothermal energy, etc. are considered as pollution free. But every time a source of energy is exploited or used, it causes envirnomental pollution. So, no source of energy can be pollution free.

Question 12.
Hydrogen has been used as a rocket fuel. Would you consider it a cleaner fuel than CNG? Why or why not?
Answer:
Yes, hydrogen is a cleaner fuel than CNG because hydrogen burnt in presence of oxygen produces water vapours (H₂O_(g)) whereas CNG that contain methane burns to produce carbon dioxide and carbon monoxide.

Question 13.
Name two energy sources that you would consider to be renewable. Give reasons for your choices.
Answer:

• Hydropower, since the water in the reservoir would be refilled each time it rains.
• Wind power, as wind keeps blowing due to unequal heating of the landmass and water bodies by solar radiation.

Question 14.
Give the names of two energy sources that you would consider to be exhaustible. Give reasons for your choices.
Answer:
Fossil fuels, i.e., coal and petroleum are exhaustible because once they are used up they are lost forever.

Activity 14.1 [T. B. Pg. 242]

Questions:

Question 1.
List four forms of energy that you use from morning, when you wake up, till you reach the school.
Answer:

• Muscular energy – To brush, exercise, take bath, to ride bicycle
• Electrical energy – To turn on geyser, to iron uniform, to put on the fan
• Fuel energy/PNG – To boil milk, cooking food; Transport – Going to school by bus / car
• Chemical energy / food – Breakfast, lunch

Question 2.
From where do we get these different forms of energy?
Answer:

Form of energy	Sources
Muscular energy	Energy stored in muscle cells in form of ATP
Electrical energy	From power plant
Fuel energy	From gas station
Chemical energy	From food

Question 3.
Can we call these sources of energy? Why or why not?
Answer:
Yes, we can call all these as sources of energy.

Activity 14.2 [T. B. Pg. 243]

Consider the various options we have when we choose a fuel for cooking our food.

Questions :

Question 1.
What are the criteria you would consider when trying to categorise something as a good fuel?
Answer:
The criteria to categorise good fuel for cooking our food are as follows :

• It should be easily available.
• It should be economical.
• It should not produce smoke or ash.
• It should have high efficiency.

Question 2.
Would your choice be different if you lived …
(a) in a forest?
(b) in a remote mountain village or small island?
(c) in New Delhi?
(d) lived five centuries ago?
Answer:

We live in	Our choice will be
(a) in a forest	wood, dry leaves
(b) in a remote mountain village or small island	wood, dry leaves, cow-dung cake
(c) in New Delhi	LPG, PNG
(d) lived five centuries ago	wood

Question 3.
How are the factors different in each case?
Answer:
In each case, the factors are different on the basis of its availability.

Activity 14.3 [T. B. Pg. 244]

To demonstrate the process of thermoelectric production with a model.

Apparatus and materials: Pressure cooker, pipe, tennis ball, metal sheet, dynamo, bulb.

Procedure:

• Take a table-tennis ball and make three slits into it.
• Put semicirucular fins cut out of a metal sheet into these slits.
• Pivot the tennis ball on an axle through its centre with a straight metal wire fixed to a rigid support.
• Ensure that the tennis ball rotates freely about the axle.
• Now connect a cycle dynamo to this.
• Connect a bulb in series.
• Direct a jet of water or steam produced in a pressure cooker at the fins.
• Note down your observation.
  

Observation : The bulb gets lighted.

Questions :

Question 1.
Which energy conversion do you think is taking place in this model?
Answer:
Heat energy gets converted to electrical energy in this model.

Question 2.
Which structure in the model acts as a turbine?
Answer:
Tennis ball fitted with metal sheets in the model acts as a turbine.

Question 3.
How electricity generated in this simple model?
Answer:
The heat given to pressure cooker produces steam. The steam is used to turn the tennis ball (the turbine). The tennis ball is further linked to a dynamo. Therefore, armature of a dynamo rotates and generates electricity.

Question 4.
What do the simplest turbines have?
Answer:
The simplest turbines have one moving part, a rotor assembly.

Question 5.
Which form of energy is essential in today’s life?
Answer:
Electrical energy is essential in today’s life.

Question 6.
What is your conclusion from this activity?
Answer:
Energy can be converted from one form to another.

Activity 14.4 [T. B. Pg. 248]

Find out from your grandparents or other elders

Questions :
(a) how did they go to school?
(b) how did they get water for their daily needs when they were young?
(c) what means of entertainment did they use?
Compare the above answers with how you do these tasks now.
Is there a difference? If yes, in which case more energy from external sources is consumed?
Answer:

Our grand parents	We are
(a) They went to school by walking.	(a) We use activa, a car or a school bus.
(b) They got water from wells, river, etc.	(b) We get water from bore by using submercible water pump.
(c) Social gathering, group meeting, drama, fairs, festival celebration, etc. were the means of their entertainment.	(c) We are using mobile phones, computers most of times. Besides it television, watching

Yes, today we consume much more energy. This is used for transport and day-to-day living.

Activity 14.5 [T. B. Pg. 249]

To demonstrate, “black surface absorbs more heat as compared to a white surface”.

Apparatus – Materials :
Conical flasks, water, thermometer

Procedure :

• Take two conical flasks and paint one white and the other black.
• Fill water in the both flask.
• Place the conical flasks in direct sunlight for half an hour to one hour.
• Measure the temperature of the water in both conical flask with a thermometer.

Questions :

Question 1.
If you touch the conical flasks, which one is hotter?
Answer:
Conical flask painted black is hotter.

Question 2.
Which property is used in solar cooker and solar water heater?
Answer:
A black surface absorbs more heat as compared to a white or a reflecting surface under identical conditions.

Question 3.
In which ways this finding can be used in our daily life?
Answer:
Wearing black coloured clothes in winter and white colour clothes in summer. White paint on the outer wall as well as white coating on terrace helps to maintain 2 – 3 °C low temperature inside the house during summer.

Activity 14.6 [T. B. Pg. 249]

To study the structure and working of a solar cooker and/or water heater.

Activity is done for understanding how the solar equipments are insulated and how does maximum heat absorption is ensured.

Property: A black surface absorbs more heat.

Structure :

• It consists of an insulated metal box or a wooden box which is painted black from inside.
• The box has a thick glass sheet as a cover over the box.
• A plane or concave mirror is attached to the box that acts as a reflector.
• Mirror focuses the rays of the sun into the box.

Questions :

Question 1.
Why is a glass sheet used?
Answer:
A glass sheet is used to creates greenhouse effect.

Question 2.
How is heat reflected into the box?
Answer:
By using mirror, rays of sun reflect the heat into the box.

Question 3.
State energy conversion in solar cooker/solar water heater.
Answer:
Light energy / solar energy in converted to heat energy.

Question 4.
What temperature is achieved in solar cooker / solar water heater?
Answer:
Typically solar cooker is designed to achieve 65 °C (150 °F) temperature. The temperature of water in the solar water heater is determined by the combination of collector area and the tank capacity. Typically it would be 50-60°C, which is much hotter than the bathing water temperature (around 40 °C).

Question 5.
State the advantages and limitations of using the solar cooker or solar water heater.
Answer:

Advantages	Limitations
1. Use of solar energy, which is available free of cost,	1. These devices are useful only at certain times during the day.
2. We can save fuel gas by using it.	2. These devices are not useful in cloudy days and at night.

Activity 14.7 [T. B. Pg. 252]

Discuss in the class following questions:

Question 1.
What is the ultimate source of energy for biomass, wind and ocean thermal energy?
Answer:
Sun / Solar energy is the ultimate source of energy for biomass, wind and ocean thermal energy.

Question 2.
Is geothermal energy and nuclear energy different in this respect? Why?
Answer:
Yes, geothermal energy is obtained from steam generated in regions of hot spots due to geological changes.
Nuclear energy is obtained by a process called nuclear fission and nuclear fusion of radioactive substances.

Question 3.
Where would you place hydroelectricity and wave energy?
Answer:
Hydroelectricity and wave electricity can be placed under renewable energy sources.

Activity 14.8 [T. B. Pg. 253]

To collect the information about various energy sources and how each one affects the environment.

Debate the merits and demerits of each source and select the best source of energy on this basis.

Sources of energy	Adverse effects on environment
1. Fossil fuel	Air pollution, acid rain, greenhouse effect
2. Thermal power	Air pollution, water pollution
3. Hydropower	Large eco-systems destroy agricultural land get submerged.
4. Biomass	Smoke, ash, and gaseous substances pollute air
5. Wind power	No adverse efffect
6. Solar energy	No adverse efffect
7. Tidal / Wave / Ocean thermal energy	No adverse efffect
8. Geothermal energy	No adverse efffect
9. Nuclear energy	Radioactive pollution, radiation

Based on this, solar energy is best energy soure. wind, ocean energy sources ultimately derive their energy from the sun. Solar energy is inexhaustible.

Activity 14.9 [T. B. Pg. 254]

Debate the following two issues in class: 9uestions:
(a) The estimated coal reserves are said to be enough to last us for another two hundred years. Do you think we need to worry about coal getting depleted in this case? Why or why not?
Answer:
Coal is used in large amount in thermal power stations for to generate electricity. The use of coal generates pollutants as the coal burns. Coal is also non-renewable resource, once depleted, it cannot be restored back. It will last only for 200 years, but formation of coal has taken millions of years and therefore we should use it with great care.

(b It is estimated that the sun will last for another five billion years. Do we have to worry about solar energy getting exhausted? Why or why not?
Answer:
We have not to worry about solar energy getting exhausted because the sun will last for another five billions years. We will have
to develop advanced technology for to trap and store more and more solar energy.

(c) On the basis of the debate, decide which energy sources can be considered –

• Exhaustible
• Inexhaustible
• Renewable
• on-renewable.

Give your reasons for each choice.
Answer:

• Exhaustible : Fossil fuel, coal will get depleted some day.
• Inexhaustible : Wind, tidal, solar energy, etc. are in the form continuing or repetitive currents of energy.
• Renewable : Biomass, if we manage properly, it can give a constant supply of energy at a particular rate.
• Non-renewable : Fossil fuels as once used up, they are lost forever and cannot be renewed.

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 12 वृतों से संबंधित क्षेत्रफल Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 12 वृतों से संबंधित क्षेत्रफल

लयूतरात्मक / निबन्धात्मक प्रश्न

प्रश्न 1.
14 सेमी. व्यास वाले वृत्त की परिधि ज्ञात कीजिए।
हल:
दिया है,
वृत्त का व्यास = 14 सेमी.
हम जानते हैं,
त्रिज्या r = \(\frac{14}{2}\) = 7 सेमी
वृत्त की परिधि = 2πr
= 2 × \(\frac{22}{7}\) × 7
= 44 सेमी.।

प्रश्न 2.
त्रिज्या 21 सेमी वाले वृत्त का एक चाप केन्द्र पर 60° का कोण अन्तरित करता है, तो संगत दीर्घ त्रिज्यखण्ड का क्षेत्रफल ज्ञात कीजिए।
हल:
दिया है :
वृत्त की त्रिज्या (r) = 21 सेमी.
माना कि चाप AB केन्द्र पर 60° का कोण अन्तरित करता है।
संगत दीर्घ त्रिज्यखण्ड का कोण
(θ) = 360° – 60°
= 300°

संगत दीर्घ त्रिज्यखण्ड का क्षेत्रफल
= \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{21^2 \times 300^{\circ}}{360^{\circ}}\)
= \(\frac{22 \times 3 \times 21 \times 5}{6}\)
= 11 × 21 × 5
= 1155 सेमी.²
अतः संगत दीर्घ त्रिज्यखण्ड का क्षेत्रफल = 1155 सेमी.² ।

प्रश्न 3.
आकृति में छायांकित क्षेत्र का क्षेत्रफल ज्ञात कीजिए। यदि AB = 5 सेमी., AC = 12 सेमी. और O वृत्त का केन्द्र है।

हल:
स्पष्टत: ∠BAC अर्द्धवृत्त में बना कोण है।
इसलिए यह समकोण Δ है।
पाइथागोरस प्रमेय से,
BC² = AB² + AC²
BC² = 5² + 12²
= 25 + 144
BC² = 169
∴ BC = 13 सेमी.
वृत्त की त्रिज्या R = \(\frac{13}{2}\) सेमी.
अब छायांकित भाग का क्षे. = अर्द्धवृत्त का क्षे. – ΔABC का क्षे.

प्रश्न 4.
चित्र में छायांकित भाग का क्षेत्रफल ज्ञात कीजिए।

हल:
दिया है बड़े अर्द्धवृत्त की त्रिज्या = 14 सेमी
बड़े अर्द्धवृत्त का क्षेत्रफल = \(\frac{1}{2}\)πr²
= \(\frac{1}{2} \times \frac{22}{7} \times(14)^2\)
= \(\frac{1}{2} \times \frac{22}{7} \times 14 \times 14\)
= 308 वर्ग सेमी
प्रत्येक छोटे अर्द्धवृत्त का क्षेत्रफल = \(\frac{1}{2}\)πr²
= \(\frac{1}{2} \times \frac{22}{7} \times(7)^2\)
= \(\frac{1}{2} \times \frac{22}{7} \times 7 \times 7\)
= 77 वर्ग सेमी
अतः अभीष्ट छायांकित भाग का क्षेत्रफल = (308 + 77 +77) वर्ग सेमी
= 462 वर्ग सेमी
अतः अभीष्ट छायांकित भाग का क्षेत्रफल = 462 सेमी²

प्रश्न 5.
एक 6 सेमी त्रिज्या के वृत्त का व्यास PQRS इस प्रकार है कि PQ, QR और RS बराबर हैं। चित्रानुसार PQ और QS को व्यास मानकर अर्द्धवृत्त खींचे गये हैं। छायांकित भाग का परिमाप और क्षेत्रफल ज्ञात कीजिए।

हल:
दिया है : वृत्त की त्रिज्या = 6 सेमी
∴ वृत्त का व्यास PS = 12 सेमी
PQ = QR = RS = \(\frac{12}{3}\) = 4 सेमी
QS = QR + RS = (4 + 4) = 8 सेमी
अतः अभीष्ट परिमाप = 6 सेमी त्रिज्या वाले अर्द्धवृत्त का चाप + 4 सेमी त्रिज्या वाले अर्द्धवृत्त का चाप + 2 सेमी त्रिज्या वाले अर्द्धवृत्त का चाप
= [π × 6 + π × 4 + π × 2] सेमी
= 12π सेमी
और अभीष्ट क्षेत्रफल = PS व्यास वाले अर्द्धवृत्त का क्षेत्रफल + PQ व्यास वाले अर्द्धवृत्त का क्षेत्रफल – QS व्यास वाले अर्द्धवृत्त का क्षेत्रफल

अतः अभीष्ट छायांकित भाग का क्षेत्रफल = 37.71 सेमी2।

प्रश्न 6.
दी गई आकृति में, एक वृत्त के चतुथांश OAQB के अन्तर्गत एक वर्ग OPQR बना हुआ है। यदि वृत्त की त्रिज्या 6\(\sqrt{2}\) सेमी है, तो छायांकित क्षेत्र का क्षेत्रफल ज्ञात कीजिए। [π = \(\frac{22}{7}\) लीजिए]
हल:
∵ OPQR एक वर्ग है।
माना OP = PQ = QR = OR = x सेमी
दिया है, OR = 6\(\sqrt{2}\) सेमी
समकोण त्रिभुज OPQ में,
OQ² = OP² + PQ² (पाइथागोरस प्रमेय से)

(6\(\sqrt{2}\))² = x² + x²
72 = 2x²
x² = 36
x = 6 सेमी
त्रिज्यखण्ड का कोण (θ) = 90°
चतुर्थांश OPBQ का क्षेत्रफल
= \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{(6 \sqrt{2})^2 \times 90^{\circ}}{360^{\circ}}\)
= \(\frac{396}{7}\) सेमी²
वर्ग OABC का क्षेत्रफल = 6 × 6 = 36 सेमी.²
छायांकित क्षेत्र का क्षेत्रफल = चतुर्थाश OPBQ का क्षेत्रफल – वर्ग OABC का क्षेत्रफल
= \(\frac{396}{7}\) – 36
= 20.5 सेमी.²
अतः छायांकित भाग का क्षेत्रफल = 20.5 सेमी.²

प्रश्न 7.
छायांकित भाग का क्षेत्रफल ज्ञात कीजिए यदि केन्द्र O पर संकेन्द्रीय वृत्तों की त्रिज्या क्रमशः 21 सेमी तथा 42 सेमी तथा ∠AOC = 60° है।

हल:
दिया है,
त्रिज्यखण्ड AOC की त्रिज्या (r₁) = 42 सेमी
त्रिज्यखंड BOD की त्रिज्या (r₂) = 21 सेमी
तथा त्रिज्यखंड कोण (θ) = 60°
क्षेत्र ABDC का क्षेत्रफल = त्रिज्यखण्ड AOC का क्षेत्रफल – त्रिज्यखण्ड BOD का क्षेत्रफल

गोलाकार रिंग का क्षेत्रफल = \(\frac{22}{7}\) × (42)² – \(\frac{22}{7}\) × (21)²
= \(\frac{22}{7}\) × [1764 – 441]
= \(\frac{22}{7}\) × 1323 = 4158 सेमी²
अतः अभीष्ट छायांकित भाग का क्षेत्रफल = 4158 – 693
= 3465 सेमी²
अतः अभीष्ट छायांकित भाग का क्षेत्रफल = 3465 सेमी²

प्रश्न 8.
एक डार्टबोर्ड की प्रथम रिंग (ring I) के अन्तः तथा बाह्य व्यास क्रमश: 32 सेमी तथा 34 सेमी और दूसरी रिंग (ring II) के अन्तः तथा बाह्य व्यास क्रमशः 19 सेमी तथा 21 सेमी हैं। इन दोनों रिगों का कुल क्षेत्रफल ज्ञात कीजिए।

हल:
दिया है,
रिंग के व्यास क्रमशः 34 सेमी तथा 32 सेमी.
अतः रिंग की त्रिज्याएँ होगी R = \(\frac{34}{2}\) = 17 सेमी,
r = \(\frac{32}{2}\) = 16 सेमी.
रिंग II के व्यास क्रमश: 21 सेमी तथा 19 सेमी.
अतः रिंग II की त्रिज्याएँ होगी r₁ = \(\frac{21}{2}\) = 10.5 सेमी r₂ = \(\frac{19}{2}\) = 9.5 सेमी.
प्रथम रिंग का क्षेत्रफल = πR² – πr²
= \(\frac{22}{7}\) × (17)² – \(\frac{22}{7}\)(16)²
= \(\frac{22}{7}\)(17² – 16²)
= \(\frac{22}{7}\) × 33 सेमी²
दूसरी रिंग का क्षेत्रफल = πr₁² – πr₂²
= \(\frac{22}{7}\) × (10.5)² – \(\frac{22}{7}\) × (9.5)²
= \(\frac{22}{7}\) × [(10.5)² – (9.5)²]
= \(\frac{22}{7}\) × 20 सेमी²
अतः दोनों रिंगों का कुल क्षेत्रफल = \(\frac{22}{7}\) × 33 + \(\frac{22}{7}\) × 20
= \(\frac{22}{7}\) × (33 + 20)
= \(\frac{22}{7}\) × 53 = 166.57 सेमी²
अतः दोनों रिंगों का कुल क्षेत्रफल = 166.57 सेमी²।

प्रश्न 9.
एक कागज आयत ABCD आकार का है जिसमें AB = 40 सेमी तथा AD = 28 सेमी है। यदि इसमें से BC व्यास का एक अर्द्ध वृत्ताकार भाग काट लिया जाता है तो शेष भाग का क्षेत्रफल ज्ञात कीजिए।
हल:
आयत ABCD की लम्बाई AB = 40 सेमी और चौड़ाई AD = 28 सेमी
∴ आयत ABCD का क्षेत्रफल = AB × AD
= 40 × 28
= 1120 सेमी²
अर्द्धवृत्त का व्यास AD = 28 सेमी
अर्द्धवृत्त की त्रिज्या = \(\frac{28}{2}\) = 14 सेमी

आयत ABCD से काटे गये अर्द्धवृत्ताकार भाग का क्षेत्रफल
= \(\frac{1}{2}\)πr² = \(\frac{1}{2} \times \frac{22}{7} \times 14 \times 14\)
= 22 × 14 = 308 सेमी²
∴ शेष भाग का क्षेत्रफल
= आयत ABCD का क्षेत्रफल – अर्द्धवृत्त का क्षेत्रफल
= 1120 – 308 वर्ग सेमी
= 812 वर्ग सेमी
अत: शेष भाग का क्षेत्रफल = 812 सेमी²

प्रश्न 10.
14 सेमी. त्रिज्या वाले वृत्त के उस लघु वृत्तखण्ड का क्षेत्रफल ज्ञात कीजिए जिसका केन्द्रीय कोण 60° है। संगत दीर्घ वृत्तखण्ड का क्षेत्रफल भी ज्ञात कीजिए। [π = \(\frac{22}{7}\) लीजिए]
हल:
वृत्त की त्रिज्या = 14 सेमी.
मूल बिन्दु O से जीवा AB द्वारा बना कोण 60° है।
अब वृत्त का क्षेत्रफल = πr² = \(\frac{22}{7}\) × 14²
= 616 सेमी²

त्रिज्यखण्ड AOB का क्षे. = πr² × \(\frac{60}{360}\)
= 616 × \(\frac{1}{6}\)
= 102.67 सेमी.² (लगभग)
ΔOAB में,
AO = OB (वृत्त की त्रिज्या)
∴ ∠OBA = ∠OAB
(त्रिभुज में समान भुजाओं के सम्मुख कोण बराबर होते हैं)
अब ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 2∠OAB = 180° – 60° – 120° [∵ ∠OBA = ∠OAB]
⇒ ∠OAB = 60°
इस प्रकार ΔAOB समबाहु त्रिभुज है।
∴ ΔAOB का है. = \(\frac{\sqrt{3}}{4}\)OA²
= \(\frac{\sqrt{3}}{4}\)(14)²
= 196 × \(\frac{\sqrt{3}}{4}\)
= 84.87 सेमी.² (लगभग)
लघु वृत्त खण्ड का क्षेत्रफल = AOB का क्षे. – ΔAOB का क्षे.
= 102.67 – 84.87
= 17.8 सेमी² (लगभग)
अब दीर्घ वृत्तखण्ड का क्षेत्रफल = वृत्त का क्षे. – लघु खण्ड का क्षे.
= 616 – 17.8
= 598.2 सेमी² (लगभग)।

प्रश्न 11.
56 मीटर भुजा वाले एक वर्गाकार बगीचे ABCD के AB व CD भुजा पर दो वृत्ताकार फूलों की क्यारियाँ बनाई गई हैं। यदि प्रत्येक वृत्ताकार क्यारी का केन्द्र बगीचे के विकणों का प्रतिच्छेद बिन्दु O है, तो बगीचे और क्यारियों के क्षेत्रफलों का योग ज्ञात कीजिए।
हल:
दिया है, वर्ग की भुजा 56 मी.
∵ AC और BD वर्ग के विकर्ण है तथा हम जानते हैं कि वर्ग के विकर्ण एक दूसरे को 90° पर समद्विभाजित करते हैं तथा बराबर होते हैं।

∠AOB = 90°
माना कि AO = OB = x मी. [∵ \(\frac{1}{2}\)AC = \(\frac{1}{2}\)BD]
समकोण त्रिभुज AOB में,
AB² = AO² + OB² (पाइथागोरस प्रमेय से)
⇒ 56² = x² + x²
⇒ 56² = 2x²
⇒ x² = \(\frac{56 \times 56}{2}\)
⇒ x² = 28 × 56
⇒ x = \(\sqrt{28 \times 56}\) मी.
अब त्रिज्यखण्ड OAB की त्रिज्या = \(\sqrt{28 \times 56}\) मी.
तथा त्रिज्यखण्ड कोण (θ) = 90°.
वृत्ताकार क्यारी AB का क्षेत्रफल = त्रिज्यखण्ड OAB का क्षेत्रफल – समकोण ΔAOB का क्षेत्रफल

= 22 × 56 – 28 × 28
= 1232 – 784 = 448 मी.²
इसी प्रकार वृत्ताकार क्यारी CD का क्षेत्रफल = 448 मी.²
वर्गाकार बगीचे ABCD का क्षेत्रफल
= 56 × 56 = 3136 मी.²
अब वर्गाकार बगीचे का क्षेत्रफल + वृत्ताकार क्यारियों का क्षेत्रफल
= 3136 + 448 + 448
= 4032 वर्ग मीटर
अत: वर्गाकार बगीचे का क्षेत्रफल व वृत्ताकार क्यारियों के क्षेत्रफलों का योग = 4032 वर्ग मीटर।

प्रश्न 12.
दी गई आकृति में, दर्शाए गए वृत्त खंड का क्षेत्रफल ज्ञात कीजिए, यदि वृत्त की त्रिज्या 21 सेमी हैं तथा ∠AOB = 120° है। [π = \(\frac{22}{7}\) लीजिए]

हल:
वृत्तखण्ड का क्षेत्रफल = त्रिज्यखण्ड OAYB का क्षेत्रफल – ΔOAB का क्षेत्रफल

अब, त्रिज्यखण्ड OAYB का क्षेत्रफल
= \(\frac{120^{\circ}}{360^{\circ}}\) × π × 21 × 21
= \(\frac{120^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 21 × 21
= 462 सेमी² …..(i)
अब, OM ⊥ AB खींचिए

ΔAMO तथा ΔBMO में,
OM = OM (उभयनिष्ठ) (S)
∠AMO = ∠BMO = 90° (A)
∠MOA = ∠MOB = 60° (A)
अतः AAS सर्वांगसमता द्वारा
∠AMO = ∠BMO
माना OM = x सेमी है।
इसीलिए ΔOMA में,
\(\frac{O M}{O A}\) = cos 60°
\(\frac{x}{21}=\frac{1}{2}\)
x = \(\frac{21}{2}\)
अत: OM = \(\frac{21}{2}\) सेमी

इसलिए वृत्तखंड AYB का क्षेत्रफल = \(\left(462-\frac{441}{4} \sqrt{3}\right)\)
= \(\frac{21}{4}\)(88 – 21\(\sqrt{3}\)) सेमी²

प्रश्न 13.
दी गई आकृति में छायांकित भाग का क्षेत्रफल ज्ञात कीजिए वर्ग ABCD के शीर्षों A, B, C तथा D को केन्द्र मानकर खींची गई चायें भुजाओं AB, BC, CD तथ DA के मध्य बिन्दुओं क्रमश: P, Q, R तथा S पर दो-दो के जोड़ों में काटती हैं तथा वर्ग की भुजा 12 सेमी है। [π = 3.14 लीजिए]

हल:
ज्ञात है ABCD एक वर्ग हैं तथा P, Q, R व S वर्ग को भुजाओं के मध्य बिन्दु हैं।

त्रिज्याखण्ड की त्रिज्या, r = \(\frac{a}{2}\)
= \(\frac{12}{2}\) = 6 सेमी
छायांकित भाग का क्षेत्रफल = वर्ग में क्षेत्रफल – 4 × त्रिज्यखण्ड का क्षेत्रफल
= (a)² – 4 × \(\frac{1}{4}\)πr²
= (12)² – 3.14 × (6)²
= 144 – 11.04 = 30.96 सेमी²

प्रश्न 14.
दी गई आकृति में, प्रत्येक 3 सेमी व्यास के तीन अर्द्धवृत्त, 4.5 सेमी व्यास का एक वृत्त तथा 4.5 सेमी त्रिज्या का एक अर्धवृत्त बनाए गए हैं। छायांकित भाग का क्षेत्रफल ज्ञात कीजिए।

हल:
दिया है, बड़े अर्द्धवृत्त की त्रिज्या = 4.5 सेमी
बड़े अर्द्धवृत्त का क्षेत्रफल = \(\frac{1}{2}\)πR²
= \(\frac{1}{2} \times \frac{22}{7}\) × 4.5 × 4.5
आन्तरिक वृत्त का व्यास = 4.5 सेमी
⇒ r = \(\frac{4.5}{2}\) सेमी
आन्तरिक वृत्त का व्यास = πr²
= \(\frac{22}{7} \times \frac{4.5}{2} \times \frac{4.5}{2}\)
छोटे अर्द्धवृत्त का व्यास = 3 सेमी
⇒ r = \(\frac{3}{2}\) सेमी
छोटे अर्द्धवृत्त का क्षेत्रफल = \(\frac{1}{2}\)πr²
= \(\frac{1}{2} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2}\)
छायांकित भाग का क्षेत्रफल = बड़े अर्द्धवृत्त का क्षेत्रफल + पहले छोटे अर्द्धवृत्त का क्षेत्रफल – अन्तः वृत्त का क्षेत्रफल – दो छोटे अर्द्धवृत्तों पर क्षेत्रफल

= \(\frac{11}{7} \times \frac{90}{4}-\frac{22}{7} \times \frac{29.25}{4}\)
= \(\frac{990-643.5}{28}\)
= 12.37 सेमी² (लगभग)

प्रश्न 14.
दी गई आकृति में, ABCD एक आयत हैं जिसकी विमाएँ 21 सेमी × 14 सेमी हैं। BC की व्यास मान का एक अर्द्ध खींचा गया है। आकृति में छायांकित भाग का क्षेत्रफल तथा परिमाप ज्ञात कीजिए।

हल:
छायांकित भाग का क्षेत्रफल = अद्यतन का क्षेत्रफल – अर्द्धवृत्त का क्षेत्रफल
= l × b – \(\frac{1}{2}\)πr²
= 21 × 14 – \(\frac{1}{2} \times \frac{22}{7}\) × 7 × 7
= 294 – 77
= 217 सेमी²
छायांकित भाग का परिमाप = 2l + b + πr
= 2 × 21 + 14 + \(\frac{22}{7}\) × 7
= 42 + 14 + 22
= 78 सेमी

प्रश्न 15.
दी गई आकृति में, ΔABC एक समकोण त्रिभुज है जिसमें ∠A = 90° है। AB, AC व BC की व्यास मानकर अर्द्धवृत्त खींचे गए हैं। छायांकित भाग का क्षेत्रफल ज्ञात कीजिए।

हल:
समकोण ΔABC में पाइथागोरस प्रमेय से,
BC² = AB² + BC²
= (3)² + (4)²
= 9 + 16 = 25
BC = \(\sqrt{25}\) = 5 सेमी

व्यास BC से निर्मित अर्द्धवृत्त का क्षेत्रफल
= \(\frac{1}{2}\)πr²
= \(\frac{1}{2} \times \pi\left(\frac{5}{2}\right)^2=\frac{25}{8} \pi\) सेमी²
व्यास AB से निर्मित अर्द्धवृत्त का क्षेत्रफल
= \(\frac{1}{2}\)πr²
= \(\frac{1}{2} \pi\left(\frac{3}{2}\right)^2=\frac{9}{8} \pi\) सेमी²
व्यास AC से निर्मित अर्द्धवृत्त का क्षेत्रफल
= \(\frac{1}{2}\)πr²
= \(\frac{1}{2} \pi\left(\frac{4}{2}\right)^2=\frac{16}{8} \pi\) सेमी²
समकोण ΔBAC का क्षेत्रफल = \(\frac{1}{2}\) × AB × AC
= \(\frac{1}{2}\) × 3 × 4 = 6 सेमी²
बिन्दुपूर्ण क्षेत्र का क्षेत्रफल = (\(\frac{25}{8}\)π – 6) सेमी²
छायांकित भाग का क्षेत्रफल
= \(\frac{16}{8} \pi+\frac{9}{8} \pi-\left(\frac{25}{8} \pi-6\right)\)
= \(\frac{16}{8} \pi+\frac{9}{8} \pi-\frac{25}{8} \pi+6\)
= 6 सेमी²

प्रश्न 16.
22 सेमी लम्बी एक तार को एक वृत्त की चाप के रस में इस प्रकार मोड़ा गया कि वह वृत्त के केन्द्र पर 60° का कोण अंतरित करती है। वृत्त की त्रिज्या ज्ञात कीजिए। [π = \(\frac{22}{7}\) लीजिए]
हल:
माना त्रिज्या r सेमी है।
चाप की लम्बाई = 22 सेमी

⇒ \(\frac{\pi r \theta}{180^{\circ}}=22\)
⇒ \(\frac{22 \times r \times 60^{\circ}}{7 \times 180^{\circ}}=22\)
⇒ r = \(\frac{22 \times 180 \times 7}{22 \times 60}=21\) सेमी
अतः वृत्त की त्रिज्या = 21 सेमी

प्रश्न 17.
वृत्त के एक तिज्यखंड को परिधि 16.4 सेमी है। यदि त्रिज्या 5.2 सेमी है, तो त्रिज्यखण्ड का क्षेत्रफल ज्ञात कीजिए।
हल:
दिया है,
r = 5.2 सेमी
त्रिज्यखण्ड की परिधि = 16.4 सेमी

⇒ r + r + चाप AB की लम्बाई = 16.4
⇒ 5.2 + 5.2 + l = 16.4
⇒ l = 16.4 – 10.46 = 6 सेमी
त्रिज्यखंड का क्षेत्रफल = \(\frac{1}{2}\)lr
= \(\frac{1}{2}\) × 5.2 × 6 = 15.6 सेमी²

प्रश्न 18.
निम्न आकृति में, ABCD एक समान्तर चतुर्भुज है। भुजा AB को व्यास तथा बिन्दु O को केन्द्र मानते हुए एक अर्धवृत्त खींचा गया है जो D से होकर गुजरता है यदि AB = 12 सेमी तथा OD ⊥ AB, तो छायांकित भाग का क्षेत्रफल ज्ञात कीजिए। (π = 3.14)

हल:
दिया है, समान्तर चतुर्भुज ABCD की भुजा AB = 12 सेमी
∵ OD ⊥ AB
तथा O, AB का मध्यबिन्दु है।
∴ AO = OB = \(\frac{12}{2}\) = 6 सेमी
∵ OD = OA = 6 सेमी
समान्तर चतुर्भुज ABCD का क्षेत्रफल = AB × OD
= 12 × 6 सेमी
= 72 वर्ग सेमी
चतुर्थांश BOD का क्षेत्रफल = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{3.14 \times(6)^2 \times 90^{\circ}}{360^{\circ}}\)
= 28.26 सेमी
अतः छायांकित भाग का क्षेत्रफल = समान्तर चतुर्भुज ABCD का क्षेत्रफल – चतुर्थाश OBD का क्षेत्रफल
= (72 – 28.26) सेमी²
= 43.64 सेमी²

प्रश्न 19.
एक वृत्त के त्रिज्यखण्ड का परिमाप 31 सेमी है। यदि वृत्त की त्रिज्या 6.5 सेमी है, तो त्रिज्यखण्ड का क्षेत्रफल ज्ञात कीजिए।
हल:
दिया है, वृत्त की त्रिज्या = 6.5 सेमी
वृत्त के त्रिज्यखंड का परिमाप = 31 सेमी

⇒ l + 2r = 31

वस्तुनिष्ठ प्रश्न

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क).
1. नियति बिंदु को वृत्त पर ……………… कहते हैं।
2. वृत्त की परिधि पर स्थित किन्ही दो बिन्दुओं को मिलाने वाले रेखाखण्ड को वृत्त की …………………. कहते हैं।
3. वृत्त के केन्द्र से होकर जाने वाली जीवा, वृत्त का ………………. कहलाती है।
4. वृत्त की परिधि पर किसी सतत् भाग को वृत्त का ………………. कहते हैं।
5. वृत्त के एक-चौथाई भाग को वृत्त का ………………. कहते हैं।
उत्तर:
1. केन्द्र,
2. जीवा,
3. व्यास,
4, चाप,
5. चतुर्थांश।

निम्न में सत्य / असत्य बताइए :

प्रश्न (ख).
1. किसी चाप द्वारा वृत्त के केन्द्र पर अंतरित कोण को चाप का केन्द्रीय कोण कहते हैं।
2. किसी वृत्त का पूरा एक चक्कर चलने में तय की दूरी उसका क्षेत्रफल कहलाती है।
3. किसी वृत्त की त्रिज्या उस वृत्त को दो बराबर भागों में विभाजित करती है।
4. ऐसी रेखा को, जो वृत्त के किन्हीं दो भिन्न बिन्दुओं पर प्रतिच्छेद करती है, वृत्त की छेदक रेखा कहते हैं।
5. लघुचाप से घिरे त्रिज्यखण्ड को लघु त्रिज्यखण्ड कहते हैं।
उत्तर:
1. सत्य,
2. असत्य,
3. असत्य,
4. सत्य,
5. सत्य।

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
10 सेमी त्रिज्या वाले वृत्त की एक जीवा वृत्त के केन्द्र पर समकोण अंतरित करती है, जो जीवा की लम्बाई है :
(A) \(\frac{5}{\sqrt{2}}\)
(B) 5\(\sqrt{2}\)
(C) 10\(\sqrt{2}\)
(D) 10\(\sqrt{3}\)
हल:
दिया है, वृत्त की त्रिज्या OA = OB = 10 सेमी
तथा ∠AOB = 90°

समकोण ΔAOB में,
AB² = OA² + OB²
= (10)² + (10)²
= 100 + 100 = 200
AB = \(\sqrt{100 \times 2}\) = 10\(\sqrt{2}\) सेमी
अतः सही विकल्प (C) है।

प्रश्न 2.
यदि एक वृत्त की परिधि और एक वर्ग का परिमाप बराबर है, तो-
(A) वृत्त का क्षेत्रफल = वर्ग का क्षेत्रफल
(B) वृत्त का क्षेत्रफल > वर्ग का क्षेत्रफल
(C) वृत्त का क्षेत्रफल < वर्ग का क्षेत्रफल
(D) वृत्त और वर्ग के क्षेत्रफलों के बीच के संबंध में निश्चित रूप से कुछ नहीं कहा जा सकता।
हल:
सही विकल्प (B) है।

प्रश्न 3.
वृत्त के चतुर्थाश का परिमाप क्या होगा यदि वृत्त की त्रिज्या r हो :
(A) \(\frac{\pi+2 r}{r}\)
(B) πr + 2r
(C) \(\frac{\pi r+r}{r}\)
(D) \(\frac{\pi r+4 r}{2}\)
हल:
सही विकल्प (D) है।

प्रश्न 4.
एक वृत्त की त्रिज्या 5 सेमी है इस वृत्त के 9 सेमी लम्बाई के चाप द्वारा बने त्रिज्यखण्ड का क्षेत्रफल है:
(A) 45 वर्ग सेमी
(B) 22.5 वर्ग सेमी
(C) 67.5 वर्ग सेमी
(D) 2.25 वर्ग सेमी
हल:
दिया है, वृत्त की त्रिज्या (r) = 5 सेमी
वृत्त के चाप की लम्बाई (l) = 9 सेमी
हम जानते हैं, त्रिज्यखण्ड का क्षेत्रफल
A = \(\frac{1}{2}\) × l × r
= \(\frac{1}{2}\) × 9 × 5 = \(\frac{45}{2}\)
= 22.5 वर्ग सेमी
अत: सही विकल्प (B) है।

प्रश्न 5.
एक वृत्त की परिधि 22 सेमी. है। उसके चतुर्थांश का क्षेत्रफल (वर्ग सेमी. में) है-
(A) \(\frac{77}{2}\)
(B) \(\frac{77}{4}\)
(C) \(\frac{77}{8}\)
(D) \(\frac{77}{16}\)
हल:
दिया है,
वृत्त की परिधि = 22 सेमी.
⇒ 2πr = 22 [जहाँ r वृत्त की त्रिज्या है]
⇒ r = \(\frac{22}{2 \pi}=\frac{22 \times 7}{2 \times 22}=\frac{7}{2}\) सेमी.
वृत्त के चतुर्थाश का क्षेत्रफल = \(\frac{1}{4}\)πr²
= \(\frac{1}{4} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^2\)
= \(\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
= \(\frac{77}{8}\) वर्ग सेमी.
अत: विकल्प (C) सही है।

प्रश्न 6.
चित्र में वृत्त का केन्द्र O है। वृत्त की त्रिज्या 18 सेमी है तथा ∠AOB = 30° है, तो लघु चाप AB की लम्बाई है :

(A) 2π
(B) 3π
(C) 6π
(D) 4π
हल:
दिया है : वृत्त की त्रिज्या (r) = 18 सेमी
∠AOB = θ = 30°
हम जानते हैं कि लघु चाप की लम्बाई = \(\frac{\pi r \theta}{180^{\circ}}\)
= \(\frac{\pi \times 18 \times 30^{\circ}}{180^{\circ}}\)
= \(\frac{\pi \times 18}{6}=3 \pi\)
अत: सही विकल्प (B) है।

प्रश्न 7.
आकृति में, OACB केन्द्र O और त्रिज्या 3.5 सेमी वाले एक वृत्त का चतुर्थाश है। यदि OD = 2 सेमी हो तो छायांकित भाग का क्षेत्रफल ज्ञात, कीजिए:

(A) 6.485 सेमी²
(B) 5.485 सेमी²
(C) 4.485 सेमी²
(D) 3.485 सेमी²
हल:
छायांकित भाग का क्षेत्रफल = \(\frac{1}{4}\)π(R² – r²)
जहाँ R = बाहरी त्रिज्या r = आन्तरिक त्रिज्या है।
= \(\frac{1}{4} \times \frac{22}{7}\) [(3.5)² – (2)²]
= \(\frac{22}{28}\)[12.25 – 4]
= \(\frac{22}{28}\) × 8.25
= \(\frac{181.5}{28}\) सेमी²
= 6.482 सेमी²
अत: सही विकल्प (A) हैं।

प्रश्न 8.
यदि वृत्त के त्रिज्यखण्ड का क्षेत्रफल, वृत्त के क्षेत्रफल का \(\frac{1}{12}\) वाँ भाग हो तो त्रिज्यखण्ड का कोण होगा :
(A) 20°
(B) 30°
(C) 40°
(D) 50°
हल:
वृत्त का क्षेत्रफल = πr²
त्रिज्यखण्ड का क्षेत्रफल = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
प्रश्नानुसार,
त्रिज्यखण्ड का क्षेत्रफल = \(\frac{1}{12}\) (वृत्त का क्षेत्रफल)
⇒ \(\frac{\pi r^2 \theta}{360^{\circ}}=\frac{1}{12} \pi r^2\)
⇒ \(\frac{\theta}{360^{\circ}}=\frac{1}{12}\)
⇒ θ = \(\frac{360^{\circ}}{12}\)
∴ θ = 30°
अत: सही विकल्प (B) है।

प्रश्न 9.
भुजा 6 सेमी. वाले एक वर्ग के अन्तर्गत खींचे जा सकने वाले वृत्त का क्षेत्रफल है-
(A) 36π सेमी.²
(B) 18π सेमी.²
(C) 12π सेमी.²
(D) 9π सेमी.²
हाल:
दिया है :
वर्ग ABCD की भुजा = 6 सेमी.

वर्ग के अन्तर्गत खींचे जा सकने वाले वृत्त की त्रिज्या (r) = \(\frac{6}{2}\) = 3 सेमी.
वृत्त का क्षेत्रफल = πr²
= π × (3)²
= 9π सेमी.²
अत: विकल्प (D) सही है।

प्रश्न 10.
त्रिज्या 8 सेमी वाले एक वृत्त के अन्तर्गत खींचा जा सकने वाले वर्ग का क्षेत्रफल है-
(A) 256 सेमी²
(B) 128 सेमी²
(C) 64\(\sqrt{2}\) सेमी²
(D) 64 सेमी²
हल:
दिया है :
वृत्त की त्रिज्या (r) = 8 सेमी.
∴ वृत्त का व्यास = 2 × 8 = 16. सेमी.

∵ हम जानते हैं कि वृत्त के अन्तर्गत खींचे जा सकने वाले वर्ग के विकर्ण वृत्त के केन्द्र पर समद्विभाजित करते हैं।
अत: AC = 16 सेमी.
माना कि वर्ग की भुजा = x सेमी.
समकोण त्रिभुज ABC मैं
AB² + BC² = AC² (पाइथागोरस प्रमेय से)
⇒ x² + x² = 16²
⇒ 2x² = 256
⇒ x² = \(\frac{256}{2}\) = 128
x = \(\sqrt{128}\) = \(\sqrt{8 \times 8 \times 2}\)
x = 8\(\sqrt{2}\) सेमी.
वर्ग का क्षेत्रफल = 8\(\sqrt{2}\) × 8\(\sqrt{2}\)
= 128 सेमी.²
अतः विकल्प (B) सही है।

प्रश्न 11.
यदि एक वृत्त का परिमाप एक वर्ग के परिमाप के बराबर है, तो उसके क्षेत्रफलों का अनुपात है-
(A) 22 : 7
(B) 14 : 11
(C) 7 : 22
(D) 11 : 14
हल:
माना कि वृत्त की त्रिज्या r तथा वर्ग की भुजा x है।
दिया है :
वृत्त का परिमाप = वर्ग का परिमाप
⇒ 2πr = 4 × x

वृत्त का क्षेत्रफल : वर्ग का क्षेत्रफल = 14 : 11
आत विकल्प (B) सही है।

प्रश्न 12.
यदि π = \(\frac{22}{7}\) लें, तो 35 सेमी, व्यास वाले एक पहिए द्वारा एक चक्कर में तय की गयी दूरी (मीटर में) है-
(A) 2.2
(B) 1.1
(C) 9.625
(D) 96.25
हल:
दिया है :
पहिए का व्यास = 35 सेमी.
∴ पहिए की त्रिज्या (r) = \(\frac{35}{2}\) सेमी.
पहिए द्वारा 1 चक्कर में तय की गयी दूरी = पहिए की परिधि
= 2πr
= 2 × \(\frac{22}{7} \times \frac{35}{2}\)
= 110 सेमी.
= 1.1 मीटर
अतः विकल्प (B) सही है।

प्रश्न 13.
व्यासों 36 सेमी, और 20 सेमी वाले दो वृत्तों की परिधियों के योग के बराबर परिधि वाले एक वृत्त की त्रिज्या है-
(A) 56 सेमी.
(B) 42 सेमी.
(C) 28 सेमी.
(D) 16 सेमी.
हल:
दिया है दो वृत्तों के व्यास 36 सेमी, 20 सेमी. है। अतः इनकी त्रिज्याएँ क्रमशः r₁ = \(\frac{36}{2}\) = 18 सेमी., r₂ = \(\frac{20}{2}\) = 10 सेमी.।
माना वाँछित वृत्त की त्रिज्या R सेमी है।
प्रश्नानुसार
दिये गये वृत्तों की परिधियों का योग = बाँछित वृत्त की परिधि
⇒ 2πr₁ + 2πr₂ = 2πR
⇒ 2π(r₁ + r₂) = 2πR
⇒ (18 + 10) = \(\frac{2 \pi \times R}{2 \pi}\)
⇒ 28 = R
⇒ R = 28 सेमी.
अत: विकल्प (C) सही है।

प्रश्न 14.
AB वृत्त का व्यास है AC = 6 सेमी और BC = 8 सेमी। छायांकित भाग का क्षेत्रफल होगा :

(A) 54.2 सेमी²
(B) 54.3 सेमी²
(C) 54.4 सेमी²
(D) 54.57 सेमी²
हल:
छायांकित भाग का क्षेत्रफल = वृत्त का क्षेत्रफल – ΔABC का क्षेत्रफल
= πr² – \(\frac{1}{2}\) × AC × BC
दिया है : AB वृत्त का व्यास है।
∵ ∠ACB अर्द्धवृत्त में बंना कोण है। ∠ACB = 90°
समकोण ΔACB में,
AB = \(\sqrt{A C^2+B C^2}\)
= \(\sqrt{6^2+8^2}\)
= \(\sqrt{36+64}=\sqrt{100}\)
∴ AB (व्यास) = 10 सेमी
अतः त्रिज्या (r) = \(\frac{10}{2}\) = 5 सेमी
∴ छायांकित भाग का क्षेत्रफल = वृत्त का क्षेत्रफल – समकोण ΔABC का क्षेत्रफल
= πr² – \(\frac{1}{2}\) × AC × BC
= \(\frac{22}{7}\) × 5 × 5 – \(\frac{1}{2}\) × 6 × 8
= \(\frac{550}{7}-\frac{24}{1}\)
= 78.57 – 24
= 54.57 सेमी²
अत: सही विकल्प (D) है।

प्रश्न 15.
त्रिज्याओं 24 सेमी और 7 सेमी. वाले दो वृत्तों के क्षेत्रफलों के योग के बराबर क्षेत्रफल वाले एक वृत्त का व्यास है-
(A) 31 सेमी.
(B) 25 सेमी.
(C) 62 सेमी.
(D) 50 सेमी.
हल:
दिया है,
दो वृत्तों की त्रिज्याएँ r₁ = 24 सेमी., r₂ = 7 सेमी.
माना कि वाँछित वृत्त की त्रिज्या R सेमी है।
प्रश्नानुसार,
दिए गए दो वृत्तों के क्षेत्रफलों का योग = वाँछित वृत्त का क्षेत्रफल
= πr₁² + πr₂² = πR
= π(r₁² + r₂²) = πR2
(24² + 7²) = \(\frac{\pi R^2}{\pi}\)
576 + 49 = R2
R² = 625
R = \(\sqrt{625}\)
= 25 सेमी.
अतः वाँछित वृत्त का व्यास = 2 × 25
= 50 सेमी.
अत: विकल्प (D) सही है।