Bhagya

JAC Board Class 9th Science Important Questions Chapter 8 Motion

Multiple Choice Questions

Question 1.
Which of the following is correct for a car which travels a distance of 100 km in 2 hours?
(a) Its average speed is 50 km/h
(b) The car did not travel at 50 km/h all the time
(c) The car travelled at 50 km/h all the time
(d) All of the above
Answer:
(a) Its average speed is 50 km/h

Question 2.
The distance-time graph of an object shown in the given figure represents that the object is
(a) at rest position
(b) moving with constant speed
(c) moving with constant velocity
(d) moving with constant acceleration
Answer:
(a) at rest position

Question 3.
The SI unit of velocity is
(a) ms^-1
(b) ms^-2
(c) ms^-3
(d) Nn^-1
Answer:
(a) ms^-1

Question 4.
Deceleration of a body is expressed in
(a) m
(b) ms^-1
(c) ms^-2
(d) ms^-3
Answer:
(c) ms^-2

Question 5.
The initial velocity of a body is ‘u’. It is under uniform acceleration ‘a’. Its velocity ‘v’ at any time ‘t’ is given by
(a) v = u + at²
(b) v = u + \(\frac{1}{2}\) at²
(c) v = u + at
(d) v = u
Answer:
(c) v = u + at

Question 6.
A wooden slab starting from rest, slides down a 10m long inclined plane with an acceleration of 5 ms^-2 What would be its speed at the bottom of the inclined plane?
(a) 10 ms^-1
(b) 12ms^-1
(c) 10 cm^-1
(d) 12 cm^-1
Answer:
(a) 10 ms^-1

Question 7.
The velocity of a particle increases from ‘u’ to ‘v’ in time’t’ during which it covers a distance ‘s’. If the particle has a uniform acceleration, which of the following equations does not apply to the motion?
(a) 2s = (v + u) t
(b) v² = u² – 2as
(c) a = v – u/t
(d) s = (u + \(\frac{1}{2}\) at) t
Answer:
(b) v² = u² – 2as

Question 8.
In 12 minutes, a car whose speed is 35 km/h travels a distance of
(a) 7 km
(b) 3.5 km
(c) 2.4 km
(d) none of these
Answer:
(a) 7 km

Question 9.
The area under the distance – time graph gives
(a) uniform speed
(b) non – uniform speed
(c) velocity and speed
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 10.
The odometer of a car measures
(a) speed
(b) velocity
(d) acceleration
(d) distance
Answer:
(d) distance

Question 11.
If a particle moves with a constant speed, the distance time graph is a
(a) straight line
(b) curved line
(c) straight line parallel to time axis
(d) straight line parallel to velocity axis
Answer:
(a) straight line

Question 12.
An object moving with uniform circular motion shows
(a) constant acceleration in speed
(b) constant velocity
(c) constant acceleration in direction
(d) constant change in type of motion
Answer:
(c) constant acceleration in direction

Question 13.
The slope of speed – time graph gives
(a) speed
(b) velocity
(c) acceleration
(d) momentum
Answer:
(c) acceleration

Question 14.
The distance travelled by a freely falling body is proportional to the
(a) mass of body
(b) square of the time of fall
(c) square of the acceleration due to
Answer:
(b) square of the time of fall

Question 15.
The figure shows the displacement time graph of a body moving in a straight line. The velocity of the body during the

(a) 2 ms^-2
(b) zero
(c) 3 ms^-1
(d) 2.5 ms^-1
Answer:
(b) zero

Analysing & Evaluating Questions

Question 16.
In which of the following cases of motion, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on a straight road
(b) If the car is moving in a circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun
Answer:
(a) If the car is moving on a straight road

Question 17.
Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown here. Choose the correct statement.

(a) Car A is faster than car D
(b) Car B is the slowest
(c) Car D is faster than car C
(d) Car C is the slowest
Answer:
(b) Car B is the slowest

Question 18.
Suppose a boy is enjoying a ride on a merry – go – round which is moving with a constant speed of 10 ms^-1. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity
Answer:
(c) in accelerated motion

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.

1. Assertion: Velocity is a scalar quantity.
Reason: Velocity cannot be zero.
Answer:
(D) Both the statements are false.

2. Assertion: A body moving in a circular path is in non – uniform motion.
Reason: The direction of a body moving in a circular path changes at every point.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: A stone thrown vertically upwards has negative acceleration.
Reason: The acceleration of the stone thrown upward is in the direction opposite to the direction of its motion.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

4. Assertion: A freely falling body is in uniform motion.
Reason: A freely falling body covers equal distances in equal intervals of time.
Answer:
(D) Both the statements are false.

5. Assertion: An object under acceleration can have a uniform speed.
Reason: Both speed and acceleration are scalar quantities.
Answer:
(C) The assertion is true but the reason is false.

Very Short Answer Type Questions

Question 1.
Define uniform motion.
Answer:
If an object covers equal distances in equal intervals of time, however small the time intervals may be, the motion of the object is said to be uniform motion.

Question 2.
Define non – uniform motion.
Answer:
If an object covers unequal distances in equal intervals of time, it is said to be in ‘non – uniform motion’.

Question 3.
Define speed.
Answer:
It is the distance travelled by a body per unit time.
Speed. = \(\vec{a}\)

Question 4.
Define average speed.
Answer:
The total distance travelled by an object divided by the total time taken is called its average speed.

Question 5.
Define acceleration.
Answer:
It is defined as the rate of change of velocity with time.
Acceleration = \(\frac{Change in velocity}{Time Taken}\)
or a = \(\frac{v-\mathbf{u}}{t}\)
The SI unit of acceleration is m/s².

Question 6.
Draw a distance – time graph that represents uniform speed.
Answer:

Question 7.
With the help of a distance – time graph show that the object is stationary.
Answer:

Question 8.
Define uniform circular motion.
Answer:
When a body moves in a circular path with uniform speed, its motion is called uniform circular motion.

Question 9.
Define scalar quantities. Give examples.
Answer:
The physical quantities which require only magnitude, and not the direction, for their complete description are called ‘scalars’ or ‘scalar quantities’. Distance, speed, time, area, etc., are all scalar quantities.

Question 10.
Define vector quantities. Give examples.
Answer:
The physical quantities which need both magnitude and direction for their complete description are called ‘vectors’ or vector quantities. Displacement, velocity, force, etc., are all vector quantities.

Question 11.
What is the SI unit of displacement?
Answer:
Metre (m).

Question 12.
A particle moves over three – quarters of a circle of radius. What is the magnitude of its displacement?
Answer:
When the particle covers three quarters of a circle, the magnitude of its displacement is

AB = \(\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}}\) = \(\sqrt{r^{2}+r^{2}}\)
= \(\sqrt{2 r}\) = √2 r

Question 13.
Can the average speed of a moving body ever be zero?
Answer:
Speed is a scalar quantity and is always positive. So the average speed of a moving body can never be zero.

Question 14.
What is the relationship between the distance travelled and the time elapsed for motion with uniform velocity?
Answer:
Distance is directly proportional to the time elapsed. In fact,
Distance travelled = Uniform velocity × Time elapsed

Question 15.
What is the SI unit of acceleration?
Answer:
The SI unit of acceleration is m/s².

Question 16.
What is acceleration of a body moving with uniform velocity?
Answer:
The acceleration of a body moving with uniform velocity is zero.

Question 17.
State a relationship connecting u, v, a and t for an accelerated motion. Give an example of motion in which acceleration is uniform.
Answer:
The relationship between u, v, a and t is v = u + at A body falling freely towards the earth has a uniform acceleration of 9.8 ms^-2.

Question 18.
Express the velocity of a body in uniform circular motion in terms of its time period T.
Answer:
Suppose a body of mass m rotates in a circle of radiusr with velocity v. It completes one revolution in time T. Then,
Velocity = \(\frac{Distance}{Time}\)[/latex] = \(\frac{Circumference}{Time period }\)[/latex]
or v = \(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\)

Analysing & Evaluating Questions

Question 19.
The cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in the figure. Which car is the slowest?
Answer:
Speed = Slope of distance – time graph. The smaller the slope, the smaller is the speed

From the given figure, slope is minimum for car D. So, D is the slowest car.

Question 20.
A rubber ball dropped from a certain height bounce to certain height. This height keeps decreasing in subsequent bounces. What type of motion does the ball exhibit?
Answer:
The ball exhibits non – uniform motion.

Question 21.
A vehicle is moving at a constant velocity of 10 km/h in the north – south direction. It turns and starts moving towards east. What is the velocity of the vehicle towards east before taking the turn?
Answer:
Zero. The component of velocity at right angle to the original direction of motion is zero.

Short Answer Type Questions

Question 1.
Define the term rest and motion.
Answer:

• Rest: If a body does not change its position with respect to its surroundings, the body is said to be at rest, e.g., a table lying in a room is at rest with respect to the walls of the room.
• Motion: A body is said to be in motion if it changes its position with respect to its surroundings, e.g., a car running on the road is in motion with respect to the lamp posts, trees or bus stop on the roadside.

Question 2.
What is meant by a point object? Give some examples.
Answer:
Whenever the size of the object is much smaller than the distance it moves in a given time interval, the size of the object can be neglected. The object can be regarded as a point object in such cases.
Examples:

1. A car covering a distance of 10km can be treated as a point object.
2. Earth can be regarded as a point object for studying its motion around the sun.

Question 3.
Give some examples of straight line motion.
Answer:
Examples:

1. A bus moving on a straight road
2. A train moving on a straight track
3. A runner running along a straight track
4. A ball moving along a straight path
5. An object falling vertically downwards towards the surface of the earth.

Question 4.
How can we specify the position of an object?
Answer:
1. The position of an object can be specified by choosing:

• a fixed point called ‘origin’ or reference point, and
• a fixed line passing through the origin, called reference axis.

2. So the position of an object can be fully described by knowing:

• its distance from origin O and
• the angle 0 which the line joining the origin ‘O’ and the object makes with the reference axis.

In the figure shown below, the position of an object located at point P is 6 cm from the origin and 30° north of east. v (Scale: 1cm = 1m)

Question 5.
A body moves in a circle of radius ‘2R’. What is the distance covered and displacement of the body after 2 complete rounds?
Answer:
Distance covered after 2 complete rounds = 2 × circumference
= 2 × 2π (2R) = 8πR
Displacement = zero, because initial and final positions of the body are the same and displacement is the shortest distance between initial and final positions.

Question 6.
Define the term velocity. What is its SI unit? Is it a scalar or a vector quantity?
Answer:
Velocity is a physical quantity that gives both speed and direction of motion of the body.
Definition: Velocity of a body is defined as the displacement produced per unit time. Velocity is also defined as the speed of a body in a givendirection. If ‘s’ is the distance travelled by a body in a given direction and ‘t’ is the time taken to travel that distance, then the velocity V is given by,
Velocity = \(\frac{Displacement}{Time}\)[/latex]
SI unit of velocity is m/s. Velocity is a vector quantity because it requires both magnitude and direction of a body.

Question 7.
Give the difference between distance and displacement.

Distance	Displacement
(a) It is the total path covered by an object.	(a) It is the shortest path from initial position to the final position.
(b) It does not have any direction.	(b) It has direction.
(c) It cannot be zero.V	(c) It can be positive, negative or zero.

Question 8.
Differentiate between speed and velocity.
Answer:

Speed	Velocity
(a) It is the distance travelled by an object per unit of time.	(a) It is the displacement of the body per unit time.
(b) It is a scalar quantity, direction not required.	(b) It is a vector quantity, direction is required.
(c) It is always positive.	(c) It can be positive, negative or zero.

Question 9.
A cheetah is the fastest land animal and can achieve a peak velocity of 100 km/h upto distances less than 500 m. If a cheetah spots his prey at a distance of 100 m, what is the minimum time it will take to get its prey, if the average velocity attained by it is 90 km/h?
Answer:
Here v = 90Km/h = \(\frac{90 \times 1000 \mathrm{~m}}{3600 \mathrm{~s}}\)
= 25m/s
s = 100m
Minimum time, t = \(\frac{s}{v}\) = \(\frac{100}{25}\) = 4s

Question 10.
The Rajdhani Express travels a distance of 1384 km from Mumbai to Delhi. It starts from Mumbai at 4.00 p.m. and reaches Delhi at 9.00 a.m. the next day. What is its average speed?
Answer:
Total distance travelled = 1384 km
Total time taken = 17 hours
Average speed =\(\frac{Total distance travelled}{Total time taken}\)= \(\frac{1384}{17}\)
= 81.4 km/h

Question 11.
A body starts initially with a velocity ‘u’ and is accelerated at constant rate ‘a’. Find an expression for final velocity after time ‘t’
Answer:
First equation of motion : Let a body start with initial velocity ‘u’ and after time ‘t’, its velocity becomes ‘v’ due to uniform acceleration ‘a’. From the definition of acceleration,
Acceleration = \(\frac{Change in velocity }{Time taken}\)
= \(\frac{Final velocity – Initial velocity}{Time taken}\)
a = \(=\frac{\mathrm{V}-\mathrm{u}}{\mathrm{t}}\)
at = v – u
v = u + at.

Question 12.
Deduce the expression for the distance travelled by a body moving with uniform acceleration in a given time.
Answer:
Second equation of motion:
Suppose a body starts with initial velocity ‘u’ and due to uniform acceleration ‘a’ its final velocity becomes ‘v’ after time ‘t’ Then,
Average velocity = \(\frac{Initial velocity + Final velocity}{2}\)
=\(\frac{u+v}{2}\)
So, the distance covered by the body in time t is
s = Average velocity × Time =\(\frac{u+v}{2}\) × t = \(\frac{\mathrm{u}+(\mathrm{u}+\mathrm{at}) \times \mathrm{t}}{2}\)
= \(\frac{2 u t + a t^{2}}{2}\)
or s = ut + \(\frac{1}{2}\) at 2

Question 13.
Establish the relation v² – u² = 2as, where ‘u’ is the initial velocity, ‘v’ is the final velocity, ‘a’ is the uniform acceleration and ‘s’ is the distance covered by the body.
Answer:
Third equation of motion:
Let a body start with initial velocity ‘u’ and after covering distance ‘s’ under uniform acceleration ‘a’, its velocity becomes ‘v’ in ‘t’ seconds. Then
Average velocity = \(\frac{\mathrm{u}+\mathrm{v}}{2 \mathrm{}}\)
So the distance covered in time t is given by
s = Average velocity x Time
= \(\frac{u+v}{2}\) × t
or v + u = \(\frac{2 s}{t}\) …………..(1)

Using the first equation of motion: v = u + at
or v – u = at … (ii)

Multiplying equations (i) and (ii), we get
(v + u) (v – u) = \(2 \frac{s}{t}\) × at
or v² – u² = 2as

Question 14.
Show that the slope of distance – time graph gives velocity of the body.

Answer:
Given figure shows distance – time graph for a body moving with uniform velocity. Clearly, it covers distances s₁ and s₂ in times t₁ and t₂ respectively.
Slope of line AB = tan θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
= \(\frac{s_{2}-s_{1}}{t_{2}-t_{1}}\) = \(\frac{Displacement}{Time}\) = velocity
Hence, the slope of the distance-time graph gives velocity of the body.

Question 15.
Show that the slope of velocity time graph gives acceleration of the body.
Answer:
Given figure shows the velocity – time graph for a body in uniform acceleration. It is a straight line inclined to the time – axis. Body has velocities ‘u’ and ‘v’ at times ‘t₁’ and ‘t₂’ respectively
Img-1
Slope of line AE = tan θ = \( \frac{E D}{A D}\)
\(\frac{v-u}{t_{2}-t_{1}}\) = \( \frac{Change in velocity}{Time taken}\)
= Acceleration of the body
Hence, the slope of the velocity – time graph gives the acceleration of the body.

Question 16.
Draw velocity – time graph for a body moving with uniform velocity. Hence show that the area under the velocity time graph gives the distance travelled by the body in a given time interval.
Answer:
In the given figure, line PQ is the velocity – time graph of a body moving with a uniform velocity such that OP = v Area of rectangle ABCD = AD × AB
= OP × AB
= v × (t₂ – t₁)
= Velocity × Time
= Distance travelled in time interval (t₂ – t₁)

Hence, the area under the velocity – time graph gives the distance travelled by the body in the given time interval.

Question 17.
The odometer of a car reads 1800 km at the start of a trip and 2400 km at the end of the trip. If the trip took 10h, calculate the average speed of the car in km/h and m/s.
Answer:
Distance covered by the car (s)
= 2400 – 1800 = 600 km.
Trip time = 10h
Average speed = ?

(i) V_av = \(\frac{s}{t}\) = \(\frac{600}{10}\) km = 60 km/h

(ii) In m/s : (60 km/h) = \(\frac{60 \times 1000}{60 \times 60}\)
= 16.7 m/s
The average speed of the car in km/h is 60 km/h and in m/s is 16.7 m/s.

Question 18.
Draw velocity – time graphs to show the following:
(a) Uniform velocity
(b) Uniform acceleration
(c) Non – uniform acceleration
Answer:

Analysing & Evaluating Questions

Question 19.
The table given below shows distance (in km) travelled by bodies A, B and C. Read the data carefully and answer the following questions.

	Distance (in km) covered by different bodies
Time (s)	Body(A)	Body (B)	Body (C)
1st second	20	20	20
2nd second	20	36	60
3rd second	20	24	100
4th second	20	30	140
5 th second	20	48	180

(a) Which of the bodies is moving with
(i) constant speed?
(ii) constant acceleration?
(b) Which of the bodies covers
(i) maximum distance in 3rd second?
(ii) minimum distance in 3rd second?
Answer:
(a) (i) Body A
(ii) Body C

(b) (i) Body C
(ii) Body A

Question 20.
An electron moving with a velocity of 5 × 10⁴ms^-1 enters a uniform electric field and acquires a uniform acceleration of 10⁴ms^-2 in the direction of its initial motion.
(a) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(b) How much distance the electron would cover in this time?
Answer:
(i) Acceleration = \(\frac{ Change in velocity}{Time taken}\)
or Time taken, t = \(\frac{5 \times 10^{4} \mathrm{~m} \mathrm{~s}^{-1}}{10^{4} \mathrm{~m} \mathrm{~s}^{-2}}\)
= 5 s

(ii) s = ut + \(\frac{1}{2}\) at²
= (5 × 10⁴ m s^-1 × 5 s) + \(\frac{1}{2}\) × 10⁴m s^-2 × (5s)²
= 37.5 × 104 m

Long Answer Type Questions

Question 1.
Describe the various types of motion observed in bodies.
Answer:
Various types of motion as observed in bodies are:

1. Translatory motion: When a body moves, as a whole, along a straight or curved path, it is said to be in translatory motion. Translatory motion is again of two types:
   • Rectilinear motion : Here a body moves as a whole along a straight path. For example, a train moving on a straight track has translatory rectilinear motion.
   • Curvilinear motion : In this case a body moves as a whole along a curved path. For example, motion of a bicycle taking a turn along a curved path.
2. Rotatory motion: When a body rotates about a fixed point or axis. it exhibits a rotatory motion. For example, motion of a flywheel about a shaft
3. Vibratory or oscillatory motion: When a body moves to and fro about a mean position, the motion is said to be vibratory or oscillatory motion. For example, the motion of the pendulum of a wall – clock.
4. Complex motion: When the motion of a body may be a combination of more than one type of motion, it is said to be a complex motion. For example, a ball rolling down an inclined plane has both translatory and rotatory motions.

Question 2.
Define average velocity when the velocity of a body changes at a non – uniform rate and a uniform rate.
Answer:
Average velocity : When the velocity of a body changes at a non – uniform rate, its average velocity is defined as the net displacement covered divided by the total time taken.
Average velocity = \(\frac{Net displacement}{Total time taken}\)
When the velocity of a body changes at a uniform rate, the average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time.
Average velocity = \(\frac{Initial velocity + Final velocity}{2}\) :
If u is the initial velocity and v is the final velocity, the average velocity v_ay is given by,
v_ay = \(=\frac{u+v}{2}\)

Question 3.
Explain the difference regarding the nature of acceleration of the three moving bodies as expressed by the following velocity – time graphs:

(a) Uniform acceleration: A body increases velocity by equal amounts in equal intervals of time.
(b) Non – uniform acceleration: A body travels unequal distances in equal intervals of time.
(c) Uniform motion or zero acceleration: A body moves with constant

Analysing & Evaluating Questions

Question 4.
Two stones are thrown vertically upwards simultaneously with their initial velocities u₁ and u₂ respectively. Prove that the heights reached by them would be in the ratio of \(\mathbf{u}_{1}^{2}\) : \(\mathbf{u}_{2}^{2}\) (Assume upward acceleration is – g and downward acceleration to be + g).
Answer:
Stone 1
Initial velocity = u₁
Acceleration = – g
Height = h,
Final velocity, v = 0
Using the equation, v² – u² =2as
0 – \(\vec{a}\) = 2 (- g) × h₁
h₁ = \(\frac{-u_{1}^{2}}{-2 g}\) = \(\frac{u_{1}^{2}}{2 g}\) ……..(1)

Stone 2
Initial velocity = u₁
Acceleration = – g
Height = h2
Final velocity, v = 0
0² \(-u_{2}^{2}\) = 2 (- g) × h₂
h₂ = \(\frac{-u_{2}^{2}}{-2 g}\) =\(\frac{u_{2}^{2}}{2 g}\) ……..(ii)
This gives; h₁ : h₂ = \(\vec{a}\) : \(\vec{a}\)
= \(=\frac{u_{1}^{2}}{2 g}\) : \(\frac{\mathrm{u}_{2}^{2}}{2 \mathrm{~g}}\)
=\(u_{1}^{2}\) : \(\vec{a}\)

Activity 1

1. In your everyday life, you come across a range of motions in which:
   • acceleration is in the direction of motion
   • acceleration is against the direction of motion
   • acceleration is uniform
   • acceleration is non – uniform
2. Observe these motions carefully and identify one example each of the above type of motions.

Observations

• A car moving on a road.
• A ball thrown up.
• Fan blades rotating
• Windmill at time moves fast when the wind speed is more and becomes slow when the wind speed decreases.

Activity 2

• Take piece of thread and tie a small piece of stone at one of its end. Move the stone to describe a circular path with constant speed by holding the thread at the other end.
  
• Now, let the stone go by releasing the thread.
• Observe the direction in which the stone moves after it is released.
• By repeating the activity for a few more times and releasing the stone at different positions of the circular path, carefully observe whether the direction in which the stone moves remains the same or not.

Observations

• When the stone is released, it moves along a straight line, tangential to the circular path.
• By repeating the activity and releasing the stone at different positions of the circular path, the direction in which the stone moves does not remain the same. It changes every time.

Value Based Questions

Question 1.
The speed limits are prescribed for vehicles running on highways. Why is it essential to follow the speed limit rules?
Answer:
Speed limits are prescribed on the highways/expressways because:
1. When a vehicle is made to run at height speeds, its tyres get hot (due to friction). This makes them softer and the air inside the tubes hotter Both. these factors may lead to bursting of tyre/tube.

2. Reflex time varies from person to person. Any running vehicle needs a certain time period to stop after the brakes are applied. At higher speeds, this time period becomes shorter.

3. As a result, the chances of collision increase. Because of these reasons, one should strictly follow the speed limit regulations. Also, it is safer to drive in your lane depending upon the speed of your vehicle.

4. Moral: Drive below the speed limit and in your lane.

Question 2.
Most drivers involved in road accidents are found to be drunk. Give reason.
Answer:
The reflex time of a person increases when drunk, (average reflex time of a normal person is about 1/15s). Therefore, such a driver of vehicle will take more time in applying brakes. As a result, the vehicle may not stop well in time and cause an accident.

JAC Class 9 Science Important Questions

JAC Jharkhand Board Class 9th Sanskrit Solutions शेमुषी भाग 1

• Chapter 1 भारतीवसन्तगीतिः
• Chapter 2 स्वर्णकाकः
• Chapter 3 गोदोहनम्
• Chapter 4 कल्पतरुः
• Chapter 5 सूक्तिमौक्तिकम्
• Chapter 6 भ्रान्तो बालः
• Chapter 7 प्रत्यभिज्ञानम्
• Chapter 8 लौहतुला
• Chapter 9 सिकतासेतुः
• Chapter 10 जटायोः शौर्यम्
• Chapter 11 पर्यावरणम्
• Chapter 12 वाङ्मनः प्राणस्वरूपम्

JAC Class 9 Sanskrit अपठितावबोधनम्

• अपठित अनुच्छेदाः

JAC Class 9 Sanskrit व्याकरणम्

• उच्चारणस्थानानि
• सन्धि-प्रकरणम्
• समास प्रकरणम्
• कारक प्रकरणम्
• उपसर्ग प्रकरणम्
• प्रत्यय प्रकरणम्
• संज्ञा-शब्दरूप-प्रकरणम्
• सर्वनाम शब्दरूप प्रकरणम्
• संख्याज्ञानम्
• धातुरूप-प्रकरणम्

JAC Class 9 Sanskrit रचनात्मक कार्यम्

• पत्र-लेखनम्
• संकेताधारितलघुकथा, चित्रवर्णनम्
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JAC Board Class 9th Science Important Questions Chapter 7 Diversity in Living Organisms

Multiple Choice Questions

Question 1.
Whittaker classified all organisms into
(a) five kingdoms
(b) four kingdoms
(c) three kingdoms
(d) two kingdoms
Answer:
(a) five kingdoms

Question 2.
A housefly belongs to which phylum?
(a) Nematoda
(b) Annelida
(c) Porifera
(d) Arthropoda
Answer:
(d) Arthropoda

Question 3.
The five kingdom classification is based on the complexity of
(a) mode of nutrition
(b) body organisation
(c) cell structure
(d) all of these
Answer:
(d) Arthropoda

Question 4.
Which one of the following belongs to coelenterata?
(a) Sycon
(b) Hydra
(c) Spongilla
(d) Planaria
Answer:
(b) Hydra

Question 5.
In which of the following are the reproductive organs hidden?
(a) Cryptogamae
(b) Phanerogamae
(c) Gymnosperms
(d) Angiosperms
Answer:
(a) Cryptogamae

Question 6.
Which phylum of animals is also called flatworms?
(a) Porifera
(b) Coelenterata
(c) Platyhelminthes
(d) Nematoda
Answer:
(c) Platyhelminthes

Question 7.
The excretory system in annelids consists of tubes called
(a) flame cells
(b) metanephridia
(c) nephridia
(d) protonephridia
Answer:
(c) nephridia

Question 8.
What is the phylum of octopus?
(a) Arthropoda
(b) Mollusca
(c) Annelida
(d) Cnidaria
Answer:
(b) Mollusca

Question 9.
What is the mode of nutrition in bacteria?
(a) Autotrophic
(b) Heterotrophic
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Question 10.
In which organism do flame cells form the excretory system?
(a) Flatworm
(b) Earthworm
(c) Insect
(d) Crab
Answer:
(a) Flatworm

Question 11.
Which sub – group in plant kingdom produces flowers?
(a) Angiosperm
(b) Fungus
(c) Moss
(d) Fern
Answer:
(a) Angiosperm

Question 12.
The mode of nutrition in fungi is
(a) only saprotrophic
(b) saprotrophic or parasitic
(c) only parasitic
(d) none of the above
Answer:
(b) saprotrophic or parasitic

Question 13.
Which among the following is exclusively marine?
(a) Arthropoda
(b) Mollusca
(c) Echinodermata
(d) Coelenterata
Answer:
(c) Echinodermata

Analysing & Evaluating Questions

Question 14.
Meena and Hari observed an animal in their garden. Hari called it an insect, while Meena said it was an earthworm. Which of the following characteristics confirms that it was an insect?
(a) Bilaterally symmetrical body
(b) Body with jointed legs
(c) Cylindrical body
(d) Body with little segmentation
Answer:
(b) Body with jointed legs

Question 15.
After studying the characteristics of Agaric us, Madan noted them as follows:
I. It is fleshy.
II. It has an umbrella-like cap called pileus.
III. Spores are produced by the stipe.
IV. Its body is made of filaments.
Which observation is not correct?
(a) I
(b) II
(c) III
(d) IV
Answer:
(c) III

Question 16.
Preeti studied the following
features in a preserved specimen – bilateral symmetry, true segmentation, clitellum and mouth. It is a/an
(a) tapeworm
(b) cockroach
(c) earthworm
(d) roundworm
Answer:
(c) earthworm

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.

1. Assertion: Bacteria and blue – green algae belong to the same phylum, i.e. Monera.
Reason: Both the bacteria and blue – green algae are unicellular prokaryotic organisms.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

2. Assertion: Most of the amphibians lay their eggs in water.
Reason: Young ones of amphibians have gills at the initial stages of there lives.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: Snake and turtle belong to the same group.
Reason: Both the snake and the turtle are cold – blooded and have four – chambered heart.
Answer:
(C) The assertion is true but the reason is false.

4. Assertion: Dolphins do not belong to pisces.
Reason: Dolphins respire through lungs and possess four – chambered heart.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

5. Assertion: Roundworms lack a digestive tract.
Reason: Roundworms are endoparsites.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
What is evolution?
Answer:
Evolution is the gradual unfolding of organisms, from the pre – existing ones, through changes since the beginning of life.

Question 2.
Name the scientist who described the idea of organic evolution and the book in which he explained it.
Answer:
Charles Darwin first described the idea of evolution in his book ‘The Origin of Species’.

Question 3.
What is lichen?
Answer:
Lichens are a complex life form that is a symbiotic partnership of two separate organisms, i.e., a fungus and an alga The alga can be either a green alga or a blue-green alga (also known as cyanobacteria).

Question 4.
What is a symbiotic relationship or symbiosis?
Answer:
It is a relationship between two organisms in which both of them are benefited, e.g., in the symbiotic association of lichens, fungi get food from blue – green algae and in return, blue-green algae get shelter.

Question 5.
Who proposed the two kingdom classification?
Answer:
Carolus Linnaeus.

Question 6.
What is biodiversity?
Answer:
Different forms of living organisms found in a particular region is known as biodiversity of that region.

Question 7.
Define species.
Answer:
A species includes all organisms that are similar enough to interbreed and perpetuate naturally.

Question 8.
Define mycoplasma.
Answer:
Mycoplasma are the smallest and the simplest organisms. They are prokaryotes having a nucleoi(d) Their body can change shape easily. They are heterotrophs.

Question 9.
Define ‘taxonomy’.
Answer:
The branch of science that classifies living organisms among different categories or groups is called taxonomy. Taxonomy is the science of identifying and naming species and organising them into systems of classification.

Question 10.
Define Taxon.
Answer:
A taxon is a unit of classification of organisms which can be recognised to a definite category at any level of classification, for e.g., fishes, birds, insects, etc.

Question 11.
In how many kingdoms does Carolus Linnaeus divided living beings?
Answer:
Two kingdoms, viz., Plantae (plants) and Animalia (animals).

Question 12.
Name the division of plant kingdom which is also called the amphibians of plant kingdom.
Answer:
Bryophyta, e.g., Lunaria (moss).

Question 13.
What is meant by bilateral symmetry?
Answer:
When the left and right halves of the body have the same design, it is called bilateral symmetry.

Question 14.
What is thallus?
Answer:
Thallophytes have a simple plant body. The plant body is not differentiated into root, stem and leaves and is called thallus.

Question 15.
What is moss?
Answer:
Moss is a radially symmetrical leafy bryophyte having multicellular rhizoids, Funaria, Bryum, Sphagnum.

Question 16.
Why are bryophytes called the amphibians of the plant kingdom?
Answer:
Bryophytes are known as amphibians of the plant kingdom because these plants can live in soil but are dependent on water for sexual reproduction. Usually, they are found in humid and damp areas.

Analysing & Evaluating Questions

Question 17.
The external features of some living organisms are given below as I, II, III and IV. Read these and write the name of their respective animal phylum or class.
I. Streamlined body, covered with scales and divided externally into head, trunk and tail.
II. Skin is dry, covered with epidermal scales and bear two pairs of limbs each with five toes ending in horny claws.
III. Body is externally segmented and covered with a thick exoskeleton made up of chitin. The segments are grouped to form head, thorax and abdomen.
IV. Body is flattened, leaf-like or ribbon – like and bilaterally symmetrical.
Answer:
I – Phylum chordata (superclass pisces)
II – Phylum chordata (class reptilia)
III – Phylum arthropoda
IV – Phylum platyhelminthes

Question 18.
An animal is described as worm-like with unsegmented body and shows some features of  invertebrates and some of chordates. Identify the described animal.
Answer:
The animal is Balanoglossus.

Question 19.
Atrip to Himalayan foothills was organised by the school of Raman and Riya During the trip, they noticed tall trees having needle – like leaves and cones. Name the trees Raman and Riya saw there.
Answer:
They saw Pinus trees.

Short Answer Type Questions

Question 1.
What are the steps in classification of organisms?
Answer:
Organisms are classified into large groups first on the basis of independent characteristics. The characteristic in the next level would be dependent on the previous one and would decide the variety of organisms in the next level. Thus, one can build up a whole hierarchy of mutually related characteristics to be used for classification.

Question 2.
Write the hierarchy of classification proposed by Linnaeus.
Answer:
The hierarchical categories are as follows:

• Kingdom: Plant kingdom and Animal kingdom.
• Phylum (for animals)Division (for plants): A group of closely related classes having certain common characteristics.
• Class: A group of closely related orders having certain common characteristics.
• Order: A group of closely related families having certain common characteristics.
• Family: A group of closely related genera having certain common characteristics.
• Genus: A group of closely related species having certain common characteristics.
• Species: A group of organisms which are similar enough to breed and perpetuate.

Question 3.
What are the characteristics of kingdom Monera?
Answer:

1. These organisms do not have clearly defined nucleus, i.e., nucleus is not enclosed in a nuclear membrane.
2. Cell organelles are not covered with membranes.
3. The organisms do not show multicellular body design, i.e., they are unicellular.
4. Some organisms have cell wall, others do not have cell wall.
5. The mode of nutrition may be autotrophic or heterotrophic.

Question 4.
Why are blue green algae included under Monera and not under Plantae?
Answer:
Monera is a kingdom of prokaryotes while organisms of kingdom Plantae shows a definite nucleus, membrane- bound organelles and multicellular body design. Blue green algae are prokaryotes having nucleoid with naked DNA. The cell organelles are also not enclosed within a membrane. As they also do not possess multicellular body design, these characteristics bring them closer to Monera and exclude them from the kingdom Plantae.

Question 5.
Give four main features of organisms placed under Protista.
Answer:
The four main features of organisms placed under Protista are as follows:

1. This kingdom includes unicellular algae, diatoms and protozoans.
2. The organisms in this kingdom are unicellular, eukaryotic organisms (as well – defined nucleus and other membrane-bound cell organelles are present).
3. Mode of nutrition is either autotrophic (like in algae and diatoms) or heterotrophic (like in protozoans).
4. Some Protists bear hair – like cilia or whip – like flagella for movement. In some protists, like Amoeba, movement takes place by pseudopodia (false feet).

Question 6.
What are the characteristics of fungi? Give examples.
Answer:
Characteristics of fungi are as follows:

1. They are heterotrophic.
2. The organisms are saprophytic, i.e., they use decaying organic material as food or may be parasitic.
3. They have membrane bound nucleus.
4. They have cell walls made of chitin.
5. Some of them have the capacity to become multicellular organisms at certain stages in their lives.
6. Some of the fungal species have symbiotic relationship where they are mutually dependent on the blue – green algae. These are called lichens.
7. Examples: Rhizopus, Yeast, Agaricus (mushrooms), Penicillium.

Question 7.
On what basis are plants divided into two sub-kingdoms?
Answer:
Based on whether the reproductive organs are conspicuous (clearly visible) or not, plants are divided into two subkingdoms:

1. Cryptogamae: Non – flowering or seedless plants which include thallophytes, bryophytes and pteridophytes.
2. Phanerogamae: Flowering plants which include gymnosperms and angiosperms.

Question 8.
How are angiosperms further divided?
Answer:
Angiosperms are divided into two groups on the basis of the number of cotyledons present in the seeds.

1. Monocotyledonous (Monocots): These are the plants with seeds having a single cotyledon, e.g., maize, wheat, rice, etc.
2. Dicotyledonous (Dicots): These are the plants with seeds having two cotyledons, e.g., pea, gram, bean, etc.

Question 9.
List the similarities between plants and animals.
Answer:
In spite of certain external differences, plants and animals show a number of very obvious similarities which are as follows:

1. Plants and animals are made up of microscopic units called cell.
2. Both contain living substance called protoplasm.
3. Certain life processes, i.e., respiration, digestion, reproduction, assimilation, etc., take place in both the groups in an identical manner.
4. Both show response to external stimuli.
5. Both of them show growth.
6. Both reproduce and pass on their hereditary characters to their offsprings by the same mechanism.
7. Both are multicellular and eukaryotic organisms having division of labour in their cells.

Question 10.
Differentiate between algae and fungi.
Answer:

Algae	Fungi
(a) Contain chlorophyll (green in colour).	(a) Do not contain chlorophyll (nongreen).
(b) Autotrophic nutrition.	(b) Heterotrophic nutrition.
(c) Food is stored in the form of starch.	(c) Food is stored in the form of glycogen.
(d) The cell wall is made up of cellulose.	(d) The cell wall is made up of chitin.
(e) Example: Spirogyra.	(e) Example: Rhizopus (bread mould)

Question 11.
What are the conventions followed for writing the scientific names?
Answer:
The conventions followed while writing the scientific names are as follows:

1. The name of the genus begins with a capital letter.
2. The name of the species begins with a small letter.
3. When printed, the scientific name is given in italics.
4. If it is handwritten, the genus name and the species name have to be underlined separately.

Question 12.
Identify the different parts A, B, C and D of fern plant shown in the figure given below.

Answer:
(A) Leaflet
(B) Stipe
(C) Adventitious roots
(D) Rhizome

Question 13.
Define bryophytes. Give some examples.
Answer:

1. Bryophytes:
   • These are also called the amphibians of the plant kingdom.
   • Plant body may be thalloid or leafy.
   • True roots are absent, instead rhizoids develop.
   • No specialised conducting tissue. These grow on damp walls and on the bark of tree.
2. These include two groups:
   • Liverworts
   • Mosses
3. Examples: Marchantia, Riccia, Sphagnum, etc.

Question 14.
Draw a labelled diagram of Spirogyra cell.
Answer:

Question 15.
Define pteridophytes.
Answer:
Pteridophytes are vascular cryptogams and are the first vascular land plants. The main plant body is the saprophyte, which is differentiated into true roots, stems and leaves. Their xylem lacks companion cells. They have specialised tissue for the conduction of water and other substances from one part of the plant body to another. Examples: Mars ilea, ferns and horse tail.

Question 16.
Define coelom.
Answer:
Coelom is a cavity which lies in between the body wall and gut (alimentary canal) and is lined by mesodermal cells. It allows greater body flexibility. In it, the organs of the body can accommodate in a better way.

Question 17.
Differentiate between diploblastic and triploblastic animals.
Answer:

Diploblastic	Triploblastic
The animals whose bodies are developed from two layers of gastrula, i.e., ectoderm and endoderm, are called diploblastic For example, coelenterates (Hydra).	The animals whose bodies develop from three layers of gastrula, i.e., ectoderm, mesoderm and endoderm, are called triploblastic, e.g., tapeworm.

Question 18.
State the features of chordates.
Answer:
The main features of chordates are as follows:

1. They have bilaterally symmetrical body.
2. They have notochord and internal skeleton.
3. They have dorsal nerve cord.
4. They are triploblastic.
5. They are coelomate.
6. Respiration is mainly through lungs in land vertebrates and by gills in aquatic animals.

Question 19.
Discuss the characteristics of sponges.
Answer:
Phylum (Porifera) – Sponges:

1. Animals have pores (called ostia) all over the body.
2. Body is not well differentiated.
3. Non – motile animals, remain attached to solid support.
4. Body is covered with hard skeleton.
5. Reproduction is both asexual (by budding and gemmule formation) and sexual (through fertilisation).
6. Examples: Sycon, Spongilla and Euplectella.

Question 20.
Give specific features of coelenterata.
Answer:
Phylum (Cnidaria) – Coelenterata

1. All coelenterates are found in water.
2. Body is radially symmetrical.
3. These are the first of multicellular animals which possess tissue level of organisation with a distinct division of labour.
4. The body has a single, sac – like central cavity, called coelenteron or the gastrovascular cavity, with only one opening.
5. Some coelenterates live in colonies (Obelia) while others live solitary (Hydra).
6. Examples: Hydra, Aurelia, sea anemone and jelly fish.

Question 21.
Give the general characteristics of Platyhelminthes.
Answer:

1. Their body is dorsoventrally flattened and leaf-like or ribbon like.
2. Body symmetry is bilateral, i.e., left and right halves have the same design.
3. They are mostly hermaphrodite.
4. Body cavity or true coelom is absent.
5. There are three layers of cells from which differentiated tissues can be forme(d) So, animals are triploblastic.
6. They are either free living or parasitic.
7. Examples: Planaria, Fasciola hepatica (liver fluke) and Taenia solium (tape worm) are parasitic.

Question 22.
Give the characteristics of Arthropoda with two examples.
Answer:

1. This is the largest group of animals.
2. They possess jointed legs/ appendages.
3. They have bilaterally symmetrical and segmented body.
4. Body cavity is filled with blood.
5. There is an open circulatory system, i.e., the blood does not flow in definite blood vessels.
6. Examples: Apis (honey bee), Musca (housefly), Anopheles (mosquito), Palaemon (prawn), crabs, etc

Question 23.
Give the characteristics of Echinoder – mata.

1. They are free living, marine animals.
2. They are triploblastic and have a coelomic cavity.
3. They have peculiar water-driven tube system used for moving around.
4. They have a hard calcium carbonate structure that is used as skeleton.
5. Sexes are separate.
6. Examples: Asterias (starfish),

Question 24.
Echinus (sea urchin), Antedon (feather star), etc Give the characteristics of Nematoda.

1. Most of the Nematodes have small cylindrical or round body.
2. Body cavity is not a true coelom. A pseudocoelom is present.
3. Body is bilaterally symmetrical and triploblastic.
4. Sexes are separate.
5. Examples: Ascaris (round worm), Ancylostoma (hook worm) and Wuchereria (filarial worm).

Question 25.
Give the characteristics of phylum Annelida.

1. They have elongated and segmented body.
2. Body bears lateral appendages for locomotion in the form of chitinous setae or parapodia.
3. The body is bilaterally symmetrical and triploblastic.
4. Reproduction by sexual means. Sexes may be either separate or united.
5. They have true coelom (body cavity).
6. Examples: Pheretima (earthworm), Hirudinaria (blood sucking leech) and Nereis.

Question 26.
List the main features of phylum Mollusca.
Answer:

1. They have unsegmented or soft body with little segmentation.
2. The body is divided into three regions—head, dorsal visceral mass and ventral foot.
3. Body is bilaterally symmetrical.
4. The coelomic cavity is reduced.
5. They have open circulatory system and kidney-like organs for excretion.
6. Some molluscs have hard calcareous shell, an outer covering of the body.
7. Respiration is by gills called ctenidia.
8. Examples: Pila (snail), Unio (fresh water mussel) and Octopus.

Analysing & Evaluating Questions

Question 27.
You are given leech, Nereis, Scolopendra, prawn and scorpion and all have segmented body organisation. Will you classify them in one group? If no, give the important characteristics based on which you will separate these organisms into different groups.
Answer:
Leech, Nereis, Scolopendra, prawn and scorpion belong to different groups.

1. Leech and Nereis belong to phylum Annelida They have metameric segmentation, closed circulatory system and unjointed appendages.
2. Scolopendra, prawn and scorpion belong to phylum Arthropoda They have open circulatory system and jointed appendages.

Question 28.
Sakshi’s younger brother frequently suffered from stomach ache and vomiting. Her mother took him to a clinic for treatment. Doctor advised for a stool test before prescribing some medicines. The report diagnosed that the stool is infected with common roundworms. Answer the following questions asked by Sakshi to the doctor:
(a) What are common roundworms?
(b) Are there any other worms that live as parasites in our body and cause diseases?
(c) How do roundworms enter our body? What can be done to prevent such infections?
Answer:

1. Common roundworms are parasites that live inside the human digestive tract and feed on the food present inside the tract. During feeding, they damage the stomach and the intestines. The scientific name of roundworms is Ascaris.
2. Some other common parasitic worms that cause different diseases in our body are Wuchereria (filarial worm), Enterobius (pinworm) and Ancylostoma (hookworm).
3. Roundworms enter our body through contaminated food and water. Hence, one should always maintain hygienic food habits to avoid these parasites.

Long Answer Type Questions

Question 1.
Describe the three basic features for grouping all organisms into five major kingdoms.
Answer:
Basic criteria such as nature of cell, cellularity and mode of nutrition were used by Whittaker to classify the living organisms into five kingdoms,
(a) Nature of cell relates to the pres – ence or absence of membrane bound organelles in it. On the basis of this category, we can clas¬sify the living organisms in two broad categories prokaryotes and eukaryotes.

1. Prokaryotes, such as all bacteria, have poorly developed nuclear material called nucleoid and no membrane – bound organelles.
2. Eukaryotes such as protozoans, fungi, plants and animals have well developed genetic material called nucleus and all membrane – bound organelles.

(b) Cellularity refers to the number of cells present in an organism. Unicellular organisms, such as bacteria, and protozoans, are made up of single cell while multicellular organisms such as many fungi, plants and animals are made up of many cells.

(c) Mode of nutrition categorises all living organisms into autotrophs and heterotrophs. Euglena is a dual organism because of its mode of nutrition. It can act as both an autotroph (makes its own food) as it contains chlorophyll and heterotroph (feeds on other substances). Also, in presence of excess organic matter or darkness, it can act as a saprophytic organism. So, as it shows features of both plants and animals, it is a dual organism.

Question 2.
Give the outline classification of animal kingdom upto the level of phyla

Question 3.
What are the differences between plants and animals?
Answer:

Animals	Animals
(a) Most of the plants around us are stationary and do not move from one place to another.	(A) Most of the animals around us are mobile, i.e., they can move from one place to another.
(b) Plants can synthesise their own food in the presence of sunlight from water and carbon dioxide. In their green parts like leaves,chlorophyll pigment is found which helps in food making by the process of photosynthesis.	(b) Animals cannot synthesise their food because they lack chlorophyll. So, they directly or indirectly depend on plants.
(c) Most of the plants continue growing throughout their life. The growth may be in length (height) or producing lateral branches.	(c) Animals stop growing, especially in length, (height) after attaining a certain maturity.
(d) Plant cells have cell walls as outermost covering.	(d) Animal cells do not have cell walls.

Question 4.
Differentiate between bryophytes and pteridophytes.
Answer:

Bryophytes	Pteridophytes
(a) Plant body is either leafy or thalloid.	(a) Plant body is differentiated into roots, stem and leaves.
(b) The cells in plant body are haploid.	(b) The cells in plant body are diploid.
(c) Vascular tissue like xylem and phloem is absent.	(c) Vascular tissue like xylem and phloem is present.
(d) In life cycle, the gametophyte is dominant.	(d) In life cycle, the sporophytes are dominant.
(e) Sporophytic phase completely depends upon gametophytes.	(e) Sporophytic phase is independent and autotrophic.
(f) The spores are produced in the capsule part of the sporophytes.	(f) The spores are produced in sporangia bom on the leaves called sporophylls.
(g) Example: Moss.	(g) Example: Fem.

Question 5.
Give the classification of plant kingdom.

Question 6.
Write the characteristics of flat worms, round worms and segmented worms. Write their phylum also.
Answer:

Flat worms	Round worms	Segmented worms
1. Phylum – Platyhelminthes.	1. Phylum – Nematoda	1. Phylum – Annelida.
2. Dorsoventrally flat, i.e., flat body.	2. Body is cylindrical.	2. Body is segmented from head to tail.
3. No true body cavity (acoelomates).	3. Pseudocoelom (false body cavity).	3. True body cavity (eucoelomates).
4. Mostly hermaphrodite, i.e., male and female sex organs present in the same individual.	4. Sexes are separate.	4. May be unisexual or bisexual.

Analysing & Evaluating Questions

Question 7.
Observe the plant given below and answer the questions that follow.
1. Identify the plant and name the phylum it belongs to.
2. What is its habitat?
3. List any two of its special features.
4. How does transport of substances take place in this plant?
Answer:

1. It is a Cycas plant. It belongs to the phylum gymmosperms.
2. It is found mainly in colder regions of the earth.
   
3. Plant body is sporophyte. Ovules are not enclosed in ovary, thus produce naked seeds.
4. Vascular tissues are present for the conduction of substances.

Activity 1

Find out the scientific names of the following animals and plants:
(a) Frog
(b) Peacock
(c) Human
(d) Neem
(e) Maize
(f) Honeybee

Observations
It is difficult to remember names of a species in different languages. There was a need for some system to create uniform naming convention.

Animal/Plant	Scientific Name
1. Frog	1. Rana tigrina
2. Peacock	2. Pavo cristatus
3. Human	3. Homo sapiens
4. Honeybee	4. Apis indica
5. Neem	5. Azadirachta indica
6. Maize	6. Zea mays

 

Activity 2

• Soak seeds of green gram, wheat, maize, pea and tamarin(d) Once they become tender, try to split the seeds. Observe if the seeds break into two nearly equal halves.
• Now, observe the roots, leaves and flowers of these plants and record your observations.
  Observations
• All seeds do not break into two nearly equal halves.
• Out of the given seeds, gram, pea and tamarind are dicots while wheat and maize are monocots.
• Dicot plants have tap-roots and monocots have fibrous root system.
• Dicot plant leaves have reticulate venation, whereas monocot leaves have parallel venation.
• In monocot flowers, petals are three or in multiples of three (trimerous). In dicot flowers, petals are five or in multiples of five (pentamerous).

Value Based Questions

Question 1.
Many medicinal plants are getting extinct every year. A group of students who had gone for educational trip clicked photographs of endangered plants. These photographs were used by the school laboratory to study these plants.
1. Name two endangered plants.
2. Name any one medicinal plant and write its medicinal use.
3. What value of students is reflected in the above act?
Answer:

1. Two endangered plants are: Lotus comiculatus and Acacia planifrons.
2. Aloe – vera Juice of Aloe vera is used in case of indigestion, treating skin infections, etc.
3. Students are caring citizens, showing responsible behavior.

Question 2.
Due to global warming, coral is getting diminished in all the oceans/water bodies. People in the Lakshadweep island protect their corals by not allowing people/tourists to take few pieces away.
1. Name the phylum of coral.
2. What is coral made up of?
3. What values of people in Lakshadweep island are reflected?
Answer:

1. Phylum of coral is coelenterata.
2. Coral is made up of calcium carbonate.
3. People in Lakshadweep island reflect the values of being responsible citizens, respecting environment and nature.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 6 Tissues

Multiple Choice Questions

Question 1.
Which are the two types of tissues?
(a) Meristematic and temporary
(b) Meristematic and permanent
(c) Meristems and temporary
(d) None of these
Answer:
(b) Meristematic and permanent

Question 2.
Parenchyma is a type of
(a) complex tissue
(b) simple tissue
(c) xylem
(d) phloem
Answer:
(b) simple tissue

Question 3.
Transpiration and exchange of gases are functions of
(a) stomata
(b) xylem
(c) both (a) and (b)
(d) none of these
Answer:
(a) stomata

Question 4.
Which are the four types of animal tissues?
(a) Epithelial, squamous, muscular, connective
(b) Epithelial, cardiac
(c) Connective, nervous
(d) Cuboidal, columnar
Answer:
(c) Connective, nervous

Question 5.
Which of the myelin sheath?
(a) Blood
(b) Cartilage
(c) Tendon
(d) Neuron
Answer:
(d) Neuron

Question 6.
A group of cells alike in form, function and origin are called
(a) tissue
(b) organ
(c) organelle
(d) none of these
Answer:
(a) tissue

Question 7.
Which of the following tissues is composed of mainly dead cells?
(a) Phloem
(b) Epidermis
(c) Xylem
(d) Endodermis
Answer:
(c) Xylem

Question 8.
Phloem in the plants performs the function of
(a) conduction of food
(b) conduction of water
(c) providing support
(d) photosynthesis
Answer:
(a) conduction of food

Question 9.
Which of the following tissues forms glands?
(a) Epithelial
(b) Connective
(c) Nervous
(d) Muscle
Answer:
(a) Epithelial

Question 10.
The end of a long bone is connected to another bone by
(a) ligament
(b) tendon
(c) cartilage
(d) nuclei
Answer:
(a) ligament

Question 11.
Which of the following cells are living cells?
(a) Fibres
(b) Vessels
(c) Collenchyma
(d) all of these
Answer:
(c) Collenchyma

Question 12.
How many guard cells enclose a stoma?
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(b) Two

Question 13.
Which of the following cells helps sieve tubes to translocate food?
(a) Xylem parenchyma
(b) Phloem parenchyma
(c) Phloem fibre
(d) Companion cell
Answer:
(d) Companion cell

Question 14.
A neuron consists of
(a) cell body
(b) dendrites
(c) axon
(d) all of these
Answer:
(d) all of these

Question 15.
Nerve cell does not contain
(a) nerve endings
(b) tendons
(c) axon
(d) dendrites
Answer:
(b) tendons

Analysing & Evaluating Questions

Question 16.
While doing work and running, you move your organs like hands, legs et(c) Which among the following is correct?
(a) Smooth muscles contract and pull the ligament to move the bones.
(b) Smooth muscles contract and pull the tendons to move the bones.
(c) Skeletal muscles contract and pull the ligament to move the bones.
(d) Skeletal muscles contract and pull the tendon to move the bones.
Answer:
(d) Skeletal muscles contract and pull the tendon to move the bones.

Question 17.
A nail is inserted in the trunk of a tree at a height of 1 metre from the ground level. After 3 years the nail will.
(a) move downwards.
(b) move upwards.
(c) remain at the same position.
(d) move sideways.
Answer:
(c) remain at the same position.

Question 18.
You are asked to observe the permanent slides of parenchyma and sclerenchyma tissues. You will identify the slide containing sclerenchyma tissue by locating.
(a) thickened cell walls
(b) size of the cells
(c) position of nucleus
(d) size of nucleus
Answer:
(a) thickened cell walls

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: Meristematic tissues are found in the growing tips of roots and shoots of a plant.
Reason: Meristematic tissues are composed of cells having the ability of cell division.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

2. Assertion: Outer protective layer of plants is made up of epidermal tissues.
Reason: Epidermal tissue is always composed of dead cells.
Answer:
(C) The assertion is true but the reason is false.

3. Assertion: Xylem shows unidirectional transport of materials.
Reason: Tracheids and vessels of xylem consist of dead cells.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

4. Assertion: Ligament is a fibrous connective tissue.
Reason: Ligament connects muscle with skin.
Answer:
(C) The assertion is true but the reason is false.

5. Assertion: Peristalsis of food pipe is an involuntary action.
Reason: Food pipe is made up of smooth muscles.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
Name the three types of meristematic tissues.
Answer:
Apical, lateral and intercalary tissues.

Question 2.
Where is apical meristem found?
Answer:
It is present at the growing tips of stem and root it increases the length of the stem and roots.

Question 3.
Name the two types of plant tissues.
Answer:
Meristematic tissue and permanent tissue.

Question 4.
Define tracheids.
Answer:
A tracheid is a tubular cell in a vascular plant that carries water and minerals from the roots to the rest of the plant. lt also provides structural support.

Question 5.
Write the names of various types of blood cells.
Answer:
There are three types of blood cells: RBCs, WBCs, and Platelets.

Question 6.
What are guard cells?
Answer:
Each stoma is bounded by a pair of specialised kidney-shaped epidermal cells called guard cells.

Question 7.
Write the main function of meristematic tissue.
Answer:
The main function of meristematic tissue is to form new cells continuously for increasing cell number, length and the girth of the plant.

Question 8.
Define simple permanent tissue.
Answer:
It consists of one type of cells which look like one another.

Question 9.
Give three types of simple permanent plant tissues.
Answer:
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma

Question 10.
What are the main functions of vascular tissues in plants?
Answer:
Vascular tissues transport:

1. Water and dissolved minerals from roots to various parts of the plant (xylem).
2. Prepared food material from leaves to different plant parts (Phloem).

Question 11.
Define vascular bundles.
Answer:
In plants, complex tissues xylem and phloem, together constitute a structure called vascular bundle. Their main function is transportation of water, minerals and food materials within plant body.

Question 12.
Define xylem.
Answer:
Xylem is a complex plant tissue which transports water and dissolved minerals from roots to all other plant parts.

Question 13.
What are muscular tissues?
Answer:
These are specialised tissues which are composed of contractile, fibre – like cells. These tissues are responsible for movement in our body.

Question 14.
What are the main functions of muscular tissue?
Answer:
The main function of muscular tissue is the movement of the body or limbs which is brought about by contraction and relaxation of contractile proteins present in muscle cells.

Question 15.
Which type of muscles is the stomach wall made up of?
Answer:
Smooth or non – striated muscles. They are involuntary muscles.

Question 16.
What are the two main features of connective tissue?
Answer:
Main features:

1. Cells are loosely spaced and are embedded in matrix.
2. Matrix may be jelly – like, fluid, dense or rigid.

Question 17.
Which cell may be longest in the body of an animal?
Answer:
Neuron (nerve cell) which may be up to 1m long (in special cases).

Analysing & Evaluating Questions

Question 18.
If the tip of sugarcane plant is removed from the field, even then it keeps on growing in length. Explain why?
Answer:
The sugaracane plant keeps growing due to the presence of intercalary meristem which is responsible for increase in intemodal length.

Question 19.
If a potted plant is covered with a glass jar, water vapours appear on the inner wall of glass jar. Explain why?
Answer:
Plants lose water through the aerial parts by the process known as transpiration. If the potted pant is covered with a glass jar, the water vapours cannot escape outside and stick to the cooler inner walls of glass jar due to condensation.

Question 20.
Water hyacinth floats on water surface. Explain.
Answer:
In aquatic plants like water hyacinth, a type of parenchyma tissue called aerenchyma is present. It develops air spaces in these plants to provide them buoyancy so that they can float easily on the surface of water.

Short Answer Type Questions

Question 1.
Differentiate between bone and cartilage.
Answer:

Bone	Cartilage
(a) It is hard.	(a) It is soft.
(b) Matrix has an inflexible material.	(b) Matrix has a flexible material.
(c) Matrix always contains calcium salts.	(c) Calcium salt may or may not be present in the matrix.
(d) Bones have rich blood supply.	(d) Cartilages do not have rich blood supply.

Question 2.
What are the functions of areolar tissue?
Answer:
Functions of areolar tissue are as follows:

1.  It fills the space inside the organs, thus acts as a packing tissue between the organs.
2.  It supports many delicate organs in the body.
3.  It plays role in repair of tissues.

Question 3.
Write differences between xylem and phloem.
Answer:

Xylem	Phloem
(a) It contains mainly dead elements.	(a) It consists of mainly living elements.
(b) It conducts water and minerals.	(b) It conducts food.
(c) It provides mechanical strength to the plant.	(c) It does not provide mechanical strength to the plant.

Question 4.
Why does epidermal tissue have no intercellular space?
Answer:
Epidermis is formed of single continuous layered cells. It covers without any intercellular space and protects all parts of the plant. Small pores called stomata are present on the leaf, and help in the exchange of gases and water. Epidermal cells on the aerial parts of the plant secrete waxy, water resistant layers on their outer surface. It checks the loss of water, mechanical injury and invasion by parasitic fungi. Roots commonly bear long hair like parts that increase the total absorptive surface area of water absorption.

Question 5.
Write the functions of different types of cells of xylem.
Answer:
Xylem consists of tracheids, vessels, xylem parenchyma and xylem fibres. Tracheids and vessels allow the transport of water and minerals. Xylem parenchyma stores food and helps in the sideways conduction of water. Xylem fibres are supportive in function.

Question 6.
State the difference between bone and blood.
Answer:

Bone	Blood
(a) It is a hard tissue.	(a) It is a liquid tissue.
(b) It consists of osteocytes.	(b) It consists of plasma, RBC, WBC and blood platelets.
(c) It helps in movement and support of the body.	(c) It helps in the transport of substances.

Question 7.
What are the characteristic features of meristematic cells?
Answer:
Meristematic cells have:

1. Thin cell walls.
2. Abundant or dense cytoplasm and single large nucleus.
3. Spherical, oval, polygonal or rectangular shape.
4. No intercellular spaces between them.
5. Either no vacuole at all or a few vacuoles.

Question 8.
How many types of meristems are present in plants, on the basis of position? Show diagrammatically.
Answer:
On the basis of location of meristem, it is classified into three types:

1. Apical meristem: It is present at the tip of stem, root and their branches.
2. Intercalary meristem: It is found at the leaf base, above the nodes (i.e, at the base of intemodes like in grasses) or below the nodes (i.e., at the uppermost region of intemodes like in mint). Apical meristem
   
3. Lateral meristem: Vascular cambium and cork cambium are examples of lateral meristem. Vascular cambium is found in vascular bundles, while cork cambium is found underneath the bark of trees. Both of these cause increase in girth of plants.

Question 9.
What are permanent tissues?
Answer:
The cells of meristematic tissue lose the ability to divide and get differentiated into specialised cells. These differentiated cells form different types of tissues which are known as permanent tissues. Some examples of permanent tissues are parenchyma, sclerenchyma, etc.

Question 10.
Give the main functions of parenchyma The main functions of parenchyma are as follows:
Answer:

1. Chlorenchyma contains chloroplast which helps in photosynthesis.
2. Parenchyma cell stores food in the form of starch, proteins, oils and fats.
3. It helps in floating of aquatic plants due to presence of aerenchyma tissue.
4. Idioplasmic cell secretes resins, latex, tannin, oils, etc.
5. Parenchyma of xylem and phloem helps in transport of nutrition and water.
6. Parenchyma tissue provides mechanical support.

Question 11.
Define the structure of sclerenchyma Write its major functions.
Answer:
Sclerenchyma (scleras – hard) is the chief mechanical tissue of plants. It is a permanent tissue. The cells are made up of sclerenchymatous tissue. The cells are usually long, narrow, pointed at both ends and uniformly thickened by the deposition of lignin without any space in between the cells. The walls are often very highly thickened so that the lumen or cell cavity is nearly obliterated. They are usually provided with simple pits which may be oblique or straight.

1. Sclerenchyma is made up of dead and lignified cells which provide support to plants.
2. It provides hardness to stony fruits such as nuts, coconut, almonds, etc.

Question 12.
What are the main functions of collenchyma?
Answer:
The main function of collenchyma is to provide mechanical support, tensile strength, elasticity and flexibility to stem, leaf stalks and leaves.

Question 13.
What are the four main functions of epithelial tissues?
Answer:
Epithelial tissue covers the body surface and forms the lining for most internal cavities. The major function of epithelial tissue includes protection, secretion, absorption and filtration. The skin is an organ made up of epithelial tissue which protects the body from dirt, dust, bacteria and other microbes that may be harmful.

Question 14.
Differentiate between voluntary and involuntary muscles.
Answer:
Voluntary muscles are those which are directly under control of our conscious will, e.g., skeletal muscle. They work or move on our command. Involuntary muscles are not directly under command of our will. We cannot stop contraction or restart contraction of the stomach, intestine or heart muscles. They are of two types cardiac and smooth muscles. Smooth muscles are found in stomach, intestine, and iris of eyes, in ureter and in the bronchi of lungs. Another type of involuntary muscles, i.e, cardiac muscles are found in heart.

Question 15.
Name three types of muscle tissues and give functions of each.
Answer:
The three types of muscle tissues are:

1. Striated muscles: These muscles show alternate light and dark bands or striations. They are voluntary muscles and present in skeletal tissues, help in movement of body and bones.
2. Smooth muscles: These are involuntary muscles; control the movement of food in alimentary canal, contraction and relaxation of blood vessels are present in iris, uterus, etc.
3. Cardiac muscles: These muscles are present in heart, help in the rhythmic contraction and relaxation of heart throughout the life.

Question 16.
What are the two main components of blood? Why is blood considered a type of connective tissue?
Answer:
Blood has two main components:

1. Fluid (liquid) matrix called plasma.
2. Suspended red blood cells (RBCs), white blood cells (WBCs) and platelets.
   • Blood is considered as a connective tissue because it has the same origin as the other connective tissues have.
   • It flows to different parts of the body and thus connects different parts of the body with one another to exchange materials and gases.

Analysing & Evaluating Questions

Question 17.
Animals of colder regions and fishes of cold water have thicker layer of subcutaneous fat. Describe why?
Answer:
The animals of colder regions are subject to more heat loss due to very low temperature. The subcutaneous fat under the dermis layer of the skin forms an insulating layer against the extremely cold environment and also acts as a food store.

Question 18.
Due to excessive workout, an athlete was suffering from fatigue. He was suggested to take rest for some time and then again join the practice.
(a) Why did the athlete feel tired after excessive workout?
(b) Why was he asked to take rest and then join the practice again?
Answer:
(a) The skeletal muscles of our limbs are responsible for body movement. These muscles have power and contract very fast. The excessive activity and lack of oxygen supply to the muscles makes them tired soon. This results in the accumulation of lactic acid in the muscles which keeps them in the state of inactivity and tiredness.

(b) While he is on rest, the body regains the oxygen supply to the muscle cells. The lactic acid deposited in the muscles is then transported by blood to the liver, where it is converted to glycogen (a carbohydrate).

Long Answer Type Questions

Question 1.
Give the flow chart of plant tissues.
Answer:

Question 2.
What is connective tissue? Explain its types.
Answer:
A connective tissue consists of different types of cells. The different types of connective tissues are as follows:

1. Blood: It has a fluid matrix called plasma in which RBCs, WBCs and platelets are suspended, Blood transports gases, digested food, hormones and waste materials to different body parts.

2. Bone: This connective tissue forms the framework that supports the body. It also anchors the muscles and supports the main organs of the body

3. Ligament: This connective tissue connects the two bones. This is very elastic having considerable strength.

4. Tendon: It connects muscles to bones. It is a fibrous tissue with great strength but limited flexibility.

5. Cartilage: It has widely spaced cells. It smoothens bone surfaces at joints and is also present in the nose, ear, trachea and larynx.

6. Areolar tissue: It is a connective tissue which consists of matrix, several types of cells, collagen and elastin fibres. It is found between the skin and muscles, around blood vessels and nerves and in the bone marrow. It fills space inside the organs, supports internal organs and helps in repair of the tissues.

7. Adipose tissue: It consists of cells (adipocytes) which are filled with fat globules. These cells remain scattered in a matrix. It is found below the skin and between internal organs. It stores fats and acts as an insulator.

Question 3.
Show the types of animal tissues through flow chart.

Question 3.
Why is epidermis considered a protective tissue?
Epidermis is considered a protective tissue because its prime purpose is to form a physical barrier between the outside and inside of the body. These tissues are usually present in the outermost layer of the plant body such as leaves, stem and roots. It is one cell thick and covered with cutin and protects the underlying tissues present in the plant body. As roots and stems grow older with time, tissues at the periphery become cork cells. Cork cells are dead, have no intercellular spaces and the cell walls are heavily thickened by the deposition of suberin. They prevent loss of water.

Question 5.
Describe the complex tissue of plants.
Answer:
Complex permanent tissue is composed of two or more than two types of cells and contributes to a common function. It is also known as vascular tissue Xylem and phloem are the two complex tissues.

1. ylem: Xylem is composed of four types of cells tracheids, vessels, xylem fibres and xylem parenchyma Most of these cells are dead. Tracheids and vessels help in water transportation, parenchyma cells are mainly supportive in function.
2. Phloem: Phloem is made up of four types of elements sieve tubes, companion cells, phloem fibres and phloem parenchyma It helps in the transportation of food in both the directions, i.e., from leaves to roots and to other parts of the plant.

Question 6.
Observe the diagrams and answer the questions given below.

(a) What kind of cells are there in shoot apex and root apex?
(b) What are the main characteristics of these cells?
Answer:
(a) Meristematic cells.

(b) Cells have big nucleus and dense cytoplasm. These cells lack intercellular spaces and vacuoles.

Question 7.
Describe the characteristics of parenchyma What are its major modifications?
Answer:
Characteristics of parenchyma tissue:

1. Living tissue.
2. Shape spherical, oval, rectangular, polygonal, elongated or irregular in shape.
3. Cell wall a thin wall made up of cellulose, hemicellulose and pectin.
4. Young parenchymatous cells are loosely arranged.
5. Intercellular space is present.
6. Cells store reserve food material.
7. Parenchyma is found in all parts of plant such as cortex, pith, palisade, mesophyll, flower, seed, etc.
8. It is also found in vascular tissues.

Two modifications of parenchyma are chlorenchyma and aerenchyma.

• Chlorenchyma: Sometimes cells of the parenchyma contain chlorophyll and perform photosynthesis. This kind of parenchyma is known as chlorenchyma.
• Aerenchyma: In aquatic plants, parenchyma contains big air spaces in between them. Such a parenchyma tissue is known as aerenchyma

Analysing & Evaluating Questions

Question 8.
Bharti was asked to identify the following blood cells as erythrocytes, thrombocytes and types of leucocytes.
(a) On what basis was she able to identify erythrocytes?
(b) Which cells are shown by the labels B, C and D?
(c) What are the functions of E and F?
(d) Write any characteristic feature of G.

Answer:
(a) Erythrocytes or red blood cells are biconcave, disc – shaped and without nucleus.

(b) B, C and D represent neutrophils, eosinophils and basophils, respectively. These are types of leucocytes called granulocytes.

(c) E – Monocytes engulf bacteria and cellular debris at the injured site on the body. F – Lymphocytes produce antibodies and provide immunity.

(d) G – Blood platelets are colourless, non – nucleated, small or round – shaped cells. They play role in blood coagulation.

Activity 1

• Take two glass jars and fill them with water.
• Now, take two onion bulbs and place one on each jar, as shown in the given figure.
• Observe the growth of roots in both the bulbs for a few days.
• Measure the length of roots on day 1, 2 and 3.

On day 4, cut the root tips of the onion bulb in jar 2 by about 1 cm. After this, observe the growth of roots in both the jars and measure their lengths each day for five more days and record the observations in table, as shown below. Growth of roots in onion bulb:

Length	Day 1	Day 2	Day 3	Day 4	Day 5
Jar 1
Jar 2

 

Observations

• Onion in Jar 1 has longer roots because they keep on growing due to the presence of root tips.
• Roots stop growing in onion in Jar 2 as their root tips are removed.
• The tips stop growing in Jar 2 after we cut them because the apical meristem which causes increase in length of roots is removed.

Activity  2

• Take a plant stem and with the help of your teacher, cut very thin slices or sections of it.
• Now, stain the slices with safranin. Place one neatly cut section on a slide, and put a drop of glycerine.
• Cover with a cover – slip and observe under the microscope. Observe the various types of cells and their arrangement and compare them.

Observations

• All cells are not similar in structure; we see a variety of cells with different shapes and sizes.
• We can see at least 8 – 10 different types of cells in the slide.
• There are varieties of cells so that each group of cells does a specific role in the overall growth of plant.

Activity 3

• Take a freshly plucked leaf of Rhoeo.
• Stretch and break it by applying pressure.
• While breaking it, keep it stretched gently so that some peel or skin projects out from the cut.
• Remove this peel and put it in a petri dish filled with water.
• Add a few drops of safranin.
• Wait for a couple of minutes and then transfer it onto a slide. Gently place a cover slip over it.
• Observe under microscope.

Observations
The epidermis is the outermost covering, usually made of a single layer of cells. The epidermis of leaf bears small pores called stomata.

Value Based Questions

Question 1.
A group of students completed the project of finding the botanical name of all the trees present in the school campus. They prepared metal plates with name carved on it, to fix it on the plant trunks. Shreya was concerned that if the metal plate is fixed into tree many cells of the tree may get damaged. But the group members explained her that the outer layer of trunk does not have living cells and there won’t be any damage to the tree.
1. What types of cells are present on the outer layer of the bark/tree trunk?
2. How does the cork act as a protective tissue?
3. What value of the group is seen in the above case?
Answer:

1. On the outer layer of tree/trunk a thick layer of dead cells is present which acts as a protective tissue.
2. In cork, all cells are dead without intercellular spaces and the walls of the cells have deposition of suberin.
3. The students in a group show team effort, learning and co – operation.

Question 2.
A paralytic patient was unable to walk. The family members of the patient took the utmost care of the patient.
(a) Name two tissues responsible for the movement of a body.
(b) Name the tissues present in brain and spine.
(c) What value of the family members is seen in the above case?
Answer:
(a) The two tissues responsible for movement of the body are muscular tissue and nervous tissue.
(b) The tissues present in brain and spines are nervous tissues.
(c) The family member showed the value of caring, responsibility, dutiful and kindness.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 5 The Fundamental Unit of Life

Multiple Choice Questions

Question 1.
Who coined the term protoplasm for the fluid substance of the cell?
(a) Hugo von Mohl
(b) Robert Brown
(c) W. Flemming
(d) Robert Hooke
Answer:
(a) Hugo von Mohl

Question 2.
Which of the following scientists coined the term ‘cell’?
(a) Leeuwenhoek
(b) J.E. Purkinje
(c) Robert Hooke
(d) Robert Brown
Answer:
(c) Robert Hooke

Question 3.
The powerhouse of a cell is
(a) Chloroplast
(b) Golgi apparatus
(c) Mitochondria
(d) Vacuole
Answer:
(c) Mitochondria

Question 4.
Proteins are formed in
(a) Golgi bodies
(b) nucleus
(c) plastids
(d) ribosomes
Answer:
(d) ribosomes

Question 5.
The solution in which a cell will gain water by osmosis is termed as
(a) isotonic solution
(b) hypertonic solution
(c) hypotonic solution
(d) both (a) and (b)
Answer:
(c) hypotonic solution

Question 6.
The inner membrane of mitochondria is folded because
(a) it has no space inside
(b) it helps in the transportation of materials
(c) it increases the surface area
(d) it stores more food
Answer:
(c) it increases the surface area

Question 7.
The opening and closing of stomata is due to
(a) sunlight
(b) osmosis
(c)plasmolysis
(d) endocytosis
Answer:
(b) osmosis

Question 8.
Which of the following molecules is known as the energy currency of the cell?
(a) DNA
(b) RNA
(c) ATP
(d) Amino acid
Answer:
(c) ATP

Question 9.
The largest cell in the human body is
(a) nerve cell
(b) muscle cell
(c) liver cell
(d) kidney cell
Answer:
(a) nerve cell

Question 10.
The network of endoplasmic reticulum is present in the
(a) nucleus
(b) nucleolus
(c) cytoplasm
(d) chromosomes
Answer:
(c) cytoplasm

Question 11.
Mitochondria are found
(a) in all types of living cells
(b) only in animal cells
(c) only in plant cells
(d) in eukaryotic cells only
Answer:
(d) in eukaryotic cells only

Question 12.
Which organelle is considered as suicide bag of the cell?
(a) Centrosomes
(b) Mesosomes
(c) Lysosomes
(d) Chromosomes
Answer:
(b) Mesosomes

Question 13.
Centriole performs the function of
(a) formation of spindle fibres
(b) nucleolus formation
(c) cell wall formation
(d) cell division initiation
Answer:
(c) cell wall formation

Analysing & Evaluating Questions

Question 14.
A student observed a slide containing onion peel cells under a microscope in low and high magnification. In low magnification, he would have observed
(a) fewer cells in a brighter field of view
(b) more cells in a brighter field of view
(c) more cells in a darker field of view
(d) fewer cells in a darker field of view
Answer:
(a) fewer cells in a brighter field of view

Question 15.
A person with swollen gums rinses his mouth with lukewarm salt water and the swelling of his gums decreases. This is because
(a) the swollen gums absorb the salt water solution
(b) the salt water solution lowers the temperature of the water in the gums
(c) the salt in the solution moves against the concentration gradient
(d) the water in the gums moves out due to the high concentration of salt in the solution
Answer:
(b) the salt water solution lowers the temperature of the water in the gums

Question 16.
In a biology laboratory, Arjun observed two unknown cells namely Cell A and Cell B, by using a microscope. He identified Cell A as a plant cell and Cell B as an animal cell. His teacher told him that he identified the cells correctly. What did Arjun most likely observe to correctly identify the cells?
(a) Cell B had a cell membrane and Cell A did not.
(b) Cell A had a cell wall and Cell B did not.
(c) Cell B had a chloroplast and Cell A did not.
(d) Cell A had a nucleus and Cell B did not.
Answer:
(b) Cell A had a cell wall and Cell B did not.

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: Cell membrane is semi – permeable but cell wall is non – permeable.
Reason: Cell membrane is made up of lipids whereas cell wall is made up of proteins.
Answer:
(D) Both the statements are false.

2. Assertion: Plant cells are more rigid than animal cells.
Reason: Plant cells have cell wall made up of cellulose.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: Mitochondria are semi – autonomous organelles.
Reason: Mitochondria lack their own DNA, RNA and ribosomes.
Answer:
(C) The assertion is true but the reason is false.

4. Assertion: Chromosomes are visible at the time of cell division only.
Reason: Chromosomes in a cell are more condensed at the time of cell division.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

5. Assertion: About 50 – 80 % of a plant cell is coupled by the vacuole.
Reason: Vacuole of a plant cell stores many essential materials.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
What are plastids?
Answer:
Plastids are double membrane – bound organelles found inside plants and some algae, which are primarily responsible for activities related to making and storing food Many plastids are photosynthetic, but some are not.

Question 2.
Name the smallest cell and the longest cell in the human body.
Answer:
The smallest cell is the sperm cell in male. The longest cell is the nerve cell.

Question 3.
What did Robert Hooke observe in the cork of a plant?
Answer:
Robert Hooke observed that cork consists of box – like compartments which form a honeycomb structure.

Question 4.
What is diffusion?
Answer:
The movement of molecules from higher concentration to lower concentration across a semi-permeable membrane is called diffusion.

Question 5.
What is the full form of DNA?
Answer:
The full form of DNA is Deoxyribonucleic acid It is a complex molecule found in all living cells that determines all the characteristics of a living thing.

Question 6.
Name the cell organelle that helps in packaging of substances.
Answer:
Golgi apparatus.

Question 7.
What are cisternae?
membrane – bound vesicles arranged in stacks called cistemae.

Question 8.
Which organelle makes the digestive enzymes of lysosomes?
Answer:
Rough endoplasmic reticulum makes the digestive enzymes of lysosomes.

Question 9.
What is the function of cell wall and plasma membrane?
Answer:
Cell wall gives rigidity, shape and protection to the plant cell. Cell membrane allows only selected materials to move in and out of the cell.

Question 10.
What is endocytosis? Give one example.
Answer:
The process in which a cell, due to flexible cell membrane, engulfs food and other materials from its external environment is known as endocytosis. Amoeba acquires its food through this process.

Question 11.
What is chromatin material?
Answer:
Inside the nucleus, a tangled mass of threadlike structure is present which is called chromatin material. Chromatin material is the combination of DNA and proteins.

Question 12.
What are genes?
Answer:
Genes are segments of DNA present linearly on chromosomes. (They are the functional units of chromosomes.)

Question 13.
Who discovered the first living cell? Give an example of a unicellular organism.
Answer:
Anton von Leeuwenhoek discovered the cells in living organisms. An example of unicellular organism is Amoeba.

Question 14.
Define chromosomes.
Answer:
Chromosomes are rod – shaped structures found in the nucleus of a cell. They contain information for inheritance of features from parents to the next generation in the form of DNA molecules.

Question 15.
What is the function of centrosome?
Answer:
Centrosome helps in cell division in animal cells. The polar caps perform the same function in a plant cell.

Question 16.
What is the function of mitochondria?
Answer:
In mitochondria, digested food is broken down to release energy. This released energy is stored in the form of ATR

Question 17.
Define plasmolysis.
Answer:
When a living plant cell loses water through osmosis, there is shrinkage or contraction of the protoplasm away from the cell wall. This phenomenon is called plasmolysis.

Analysing & Evaluating Questions

Question 18.
Aperson takes concentrated solution of salt and after sometime, he starts vomiting. What is the phenomenon responsible for such situation.
Answer:
The concentrated salt solution is hy – pertonic When consumed, it causes exosmosis of the cells of various digestive organs of the body. This causes uncomfortable stretching of cells resulting into reverse peristaltic movements, ultimately leading to vomiting.

Question 19.
If cells of onion peel and RBC are separately kept in hypotonic solution, what will you observe in each case?
Answer:
When kept in hypotonic solution, the RBCs will swell up and then burst due to endosmosis. In case of onion peel kept in hypotonic solution, a rigid cell wall counteracts the pressure when the cells become turgid The endosmosis stops and onion peel cells do not burst.

Question 20.
Bacteria do not have chloroplast but some bacteria are photoautotrophic in nature and perform photosynthesis. Which part of bacterial cell performs this?
Answer:
In some photoautotrophic (photosyn – thetic) bacteria, photosynthetic pigments are present inside some vesicles which may be attached to the plasma membrane.

Short Answer Type Questions

Question 1.
State two conditions required for osmosis.
Answer:

1. The difference in the concentration of water, i. e., one should have higher concentration than the other,
2. Semi – permeable membrane is also required through which water will flow.

Question 2.
Differentiate between smooth endoplasmic reticulum and rough endoplasmic reticulum.
Answer:

1. Smooth endoplasmic reticulum: It looks smooth. SER helps in the manufacture of fat molecules or lipids.
2. Rough endoplasmic reticulum: It looks rough. Ribosomes are attached to RER which synthesises proteins.

Question 3.
What is the function of nucleus in a cell?
Answer:
Nucleus controls the hereditary characteristics of an organism. It is responsible for protein synthesis, cell division, growth and differentiation. It stores hereditary material in the form of deoxyribonucleic acid (DNA) strands. It also stores proteins and ribonucleic acid (RNA) in the nucleolus. Nucleolus produces ribosomes and is known as protein factory of the cell. It also regulates the integrity of genes and gene expression.

Question 4.
What is the function of plastids?
Answer:
Plastids are present only in plant cells. There are two types of plastids: chromoplast (coloured plastids) and leucoplasts (colourless).

• By trapping solar energy, green plastids manufacture food through photosynthesis.
• Chromoplast provides colour to various flowering parts.
• Leucoplasts help in storage of proteins, starch and fats.

Question 5.
What are ribosomes? Where are they located in the cell? What is their function?
Answer:
Ribosomes are spherical organelles present in the cells which are either freely distributed in the cytoplasm or may be attached to the endoplasmic reticulum. It consists of ribosomal RNA (Ribonucleic acid) and proteins. It helps in the synthesis of proteins.

Question 6.
Define membrane biogenesis.
Answer:
The endoplasmic reticulum helps in the manufacture of proteins and fat molecules or lipids which are important for the cell function. These proteins and lipids help in the building up of the cell membrane. This process is known as membrane biogenesis.

Question 7.
Draw various cells of human body.
Answer:

Question 8.
What is the significance of cell wall in a plant cell?
Answer:
The cell wall performs following functions in the plant cells:
(a) It gives a definite shape to the cell.
(b) It provides rigidity and strength to the cell.
(c) It protects the plasma membrane and inner cell organelles by bounding the cell from outside.
(d) It allows cells of plants, fungi and bacteria to withstand very dilute external media without bursting.
(e) It helps in transport of the materials in and out of the cell.
(f) It maintains the shape of the cell when cytoplasm shrinks away from cell wall during plasmolysis.
(g) It prevents desiccation of cells.

Question 9.
Distinguish between plasma membrane and cell wall.
Answer:

Plasma membrane	Cell wall
(a) It consists of proteins and lipids. It is living.	(a) It is made up of a complex carbohydrate called cellulose. It is dead or non-living.
(b) It is found in both plants and animal cells.	(b) It is found in plant cells only.
(c) It is semi permeable.	(c) It is freely permeable.
(d) It is soft and elastic.	(d) It is hard and rigid.

Question 10.
What is the function of endoplasmic reticulum?
Answer:
(a) ER helps in transporting materials from one part of the cell to the other.
(b) It acts as skeletal framework and provides mechanical strength to cytoplasmic matrix.
(c) It provides larger surface area for synthesis of various metabolic activities.
(d) It contains various enzymes which act as catalyst in synthesis of lipids and proteins.
(e) SER is also involved in the process of detoxification of drugs and poisons.

Question 11.
What is lacking in a virus which makes it dependent on a living cell to multiply?
Answer:
Viruses lack selectively permeable plasma membrane and cell organelles. Thus they lack a basic structural organisation to perform various life processes effectively in their own way. After entering a living cell, a virus utilises its own genetic material and machinery of host cell to multiply.

Question 12.
Write three main points of cell theory as expressed by Scleiden, Schwann and Virchow.
Answer:
Cell Theory: Schleiden and Schwann proposed that:

1. All plants and animals are composed of cells.
2. The cell is the basic unit of life. Later on Virchow added that
3. All cells arise from pre-existing cells.

Question 13.
What are the functions of Golgi apparatus?
Answer:
Functions of Golgi apparatus are as follows:

1. Golgi apparatus is the secretory organelle of the cell. It packages and dispatches the material synthesised in the cell to intracellular (plasma membrane and lysosomes) and extracellular targets.
2. Golgi complex is also involved in the formation of lysosomes.
3. Golgi apparatus is also involved in the synthesis of many substances such as polysaccharides, glycoproteins, etc.

Question 14.
Differentiate between unicellular and multicellular organisms. How is multi-cellularity advantageous over unicellularity?
Answer:
Differences between unicellular and multicellular organisms are as follows:

Unicellular organisms	Multicellular organisms
(a) Their body consists of single cell.	(a) Their body consists of more than one cell.
(b) All functions associated with life take place in a single cell.	(b) Different parts of body perform different functions.
(c) Example: Amoeba and Chlamydomonas.	(c) Banyan tree and horse. Multicellularity provides division of labour. This helps the organisms to perform various life processes in a better way.

Analysing & Evaluating Questions

Question 15.
We eat food composed of all the nutrients like carbohydrates, proteins, fats, vitamins, minerals, and water. After digestion, these are absorbed in the form of glucose, amino acids, fatty acids, glycerol, etc What mechanisms are involved in absorption of digested food and water?
Answer:
The absorption of these digested nutrients and water follows the following mechanisms:
(a) Carbohydrates (glucose), proteins (amino acids) and some ions (minerals) – by active transport.
(b) Fats (fatty acids) and glycerol – by passive transport (diffusion).
(c) Water – by osmosis.

Question 16.
Karan observed onion peel cells in biology laboratory. He could view the cell wall, cytoplasm, and nucleus clearly. Suddenly his friend spilled a few drops of salty water on the slide containing the onion peel. After some time, Karan observed the slide again and found some changes in the onion cells.
1 .What changes would have been observed by karan?
2. Name the process that brought changes in the onion cells.
Answer:

1. Karan observed the process of exosmosis in onion peel cells. Salty water, being a hypertonic solution, causes water to flow out from the cell due to exosmosis. As a result, cytoplasm along with plasma membrane shrinks and separates from the cell wall,
2. Plasmolysis.

Long Answer Type Questions

Question 1.
Draw a neat labelled diagram of animal cell.
Answer:

Question 2.
Draw a neat labelled diagram of plant cell and label its parts.

Question 3.
What will happen if an animal or a plant cell is put into a solution of sugar and water?
Answer:
Osmosis is the net movement of water from an area of higher water potential to an area of lower water potential across a semi – permeable membrane.If a living cell is placed in a solution with a lower water potential (such as concentrated solution of sugar or salt) than the water potential of the cell, water will move out of the cell due to exosmosis. When this occurs, the cytoplasm of the cell shrinks in volume due to water loss. This process is known as plasmolysis.

Question 4.
Describe the role played by the lysosomes.
Answer:
Functions of lysosomes:

1. Sometimes lysosomal enzymes are released outside the cell to break down extracellular material.
2. Lysosomes also destroy any foreign material which enters inside the cell such as bacteria.
3. In damaged cells, ageing cells or dead cells, lysosomes get ruptured and enzymes are release(d) These enzymes digest their own cell.
4. Lysosomes contain about 40 ydrolytic enzymes. When the cell gets damaged, lysosomes burst and their enzymes digest their own cell. So, lysosomes are called ‘suicide bags’ of the cell.
5. Foreign materials entering the cell, such as bacteria or food, as well as old organelles end up in the lysosomes, which break them up into small pieces.

Question 5.
Describe the structure of mitochondria
Answer:
Mitochondria are shaped perfectly to maximise their productivity. They are made up of two membranes. The outer membrane covers the organelle and contains it like a skin. The inner membrane folds over many times and creates layered structures called cristae. The gel like material in the  mitochondria is called matrix. The folding of the inner membrane increases the surface area inside the organelle. Since many of the chemical reactions happen on the inner membrane, the increased surface area creates more space for reactions to occur.

Mitochondria are special because they have their own ribosomes and DNA floating in the matrix. There are also structures called granules which may control the concentration of ions.

Question 6.
Write a note on the structure of a cell.
Answer:
1. Cell is the fundamental unit of all living organisms. It is enclosed by an outer ‘plasma membrane’ (PM) which is selectively permeable. Plant cells have an additional covering called ‘cell wall’, outside to the plasma membrane.

2. Inside the plasma membrane, there is a translucent viscous substance called cytoplasm in which the organelles are embedded The control centre of the cell is the nucleus; it contains all the information necessary for the cell to function and to reproduce. Surrounding the nucleus is the endoplasmic reticulum (ER) in which ribosomes may be embedded Ribosomes are granular structures which are the site of protein synthesis.

3. The powerhouse of the cell is the mitochondria which help in releasing energy by the oxidation of food in cell. There is a membranous secretory organelle in the cell called Golgi body.

4. In plant cells, an additional structure located near the nucleus, called the chloroplast, is also present. They are the site of photosynthesis.

5. Cells also contain lysosomes which are also called suicide bags. They digest and remove the unwanted debris of the cell. Centrioles located near the nucleus help in cell division. Cytoplasm also contains vacuoles filled with cell sap. In plant cells, vacuole is large and centrally placed.

Analysing & Evaluating Questions

The diagram given below shows a beaker containing 30% salt solution and a suspended cell containing 10% salt solution

1. What will happen to the cell in this set – up after 20 minutes?
2. Which part of the cell limits the entry or exit of materials into or from the cell and why?
3. Explain the process that occurred in this set – up.
Answer:

1. Since the solution outside the cell has more salt concentration (30%) than that present inside the cell, the water will move out from the cell and the cell will shrink.
2. Cell membrane limits the entry or exit of materials into or out of the cell due to its selective permeability.
3. Exosmosis occurred in this set – up. It is the process in which water moves out of the cells through the cell membrane, if surrounded by the hypertonic solution.

Activity – 1

• Let us take a small piece of onion peel with the help of a pair of forceps.
• Keep it immediately in a watch glass containing water.
• Take a glass slide, put a drop of water on it at the centre, and transfer a small piece of onion peel on the slide from the watch glass.
• Make sure that the peel is perfectly flat on the slide.
• Now put a drop of iodine solution on this piece followed by a cover slip.
• Take care to avoid the entry of air bubbles while putting the cover slip with the help of a mounting needle.
• Now focus the slide under compound microscope and observe.

Observations

• Small compartment – like structures are observed.
• Each cell shows a cell wall, cell membrane, nucleus and cytoplasm.

Activity-2

• Remove the shell of an egg by dissolving it in dilute hydrochloric acid The shell is
  mostly calcium carbonate. A thin outer skin now encloses the egg. Put the egg in pure water and observe after five minutes.
• Record your observations.
• Place a similar de – shelled egg in a concentrated salt solution and observe after five minutes.

Observations

• The egg swells when kept in pure water because water enters it by osmosis.
• Water passes out of the egg cells into the salt solution because the salt solution is more concentrated Water flows from the region of higher water concentration to the region of lower water concentration.

Activity 3

• Take dried raisins or apricots in normal water and leave them overnight.
• Then place them into a concentrated solution of sugar or salt.
• Record your observations. Observations
• Each raisin/apricot gains water and swells when placed in pure water.
• But, when placed in the concentrated solution, it loses water and shrinks.
• There is no change in water level.

Activity 4

• Mount the peel of a Rhoeo leaf in water on a slide and examine the cells under high power of the microscope. Note the small green granules, called chloroplasts. They contain a green substance called chlorophyll. Put a strong solution of sugar or salt on the mounted leaf on the slide. Wait for a minute and observe under a microscope.
• Now place some Rhoeo leaves in boiling water for a few minutes. This kills the cells. Then mount one leaf peel on a slide and observe it under a microscope. Put a strong solution of sugar or salt on the mounted leaf on the slide. Wait for a minute and observe it again.

Observations

• In this experiment, water from the cells moves out by osmosis thereby shrinking the cells. This process is called plasmolysis.
• On boiling the leaves, the cells are dead and the osmosis does not take place in these leaves.

Activity 5

• Take a glass slide with a drop of water on it. Using an ice cream spoon, gently scrap the inside surface of the cheek.
• With the help of a needle transfer this material on a clean glass slide and spread it evenly. To colour the material, put a drop of methylene blue solution on it. Put a cover slip on it.
• Observe the slide under the microscope.

Observations

• The cells are seen on the slide with no cell wall but prominent nucleus.
• The cells are slightly round/oval in shape. They are cheek cells.

Value Based Questions

Question 1.
A teacher fixed one stained temporary mount of leaf peel cells under one microscope and another slide of cheek cells in the other microscope. He focused both slides under respective microscopes. Then the teacher asked the students to observe the slides under the microscopes and located the following in peel cells and cheek cells:
(a) A densely stained body, called nucleus in both the slides.
(b) A lightly stained substance, called cytoplasm surrounding the nucleus. After observation, the teacher asked the following questions:
1. What do you infer from this activity?
2. What is the key message conveyed about plants and animals?
Answer:

1. Plants are also formed of cells like us or other animals.
2. Plants are also living things. We should not harm them unnecessarily, rather we should protect them.

Question 2.
Our body is formed of millions of cells. Now, you can imagine how minute the cells are(a) We cannot see unicellular bacteria or our cheek cells without the help of a microscope but we can see a type of cell with naked eye. These are shelled eggs, for example, egg of a hen. Eggs of a bird after hatching give rise to young ones (chick).
With this information, answer the following questions:
(a) Egg of which bird is the largest?
(b) Why is it advised to eat one egg daily?
(c) Why should we not eat more eggs daily for a long period?
(d) What is your duty after getting this information?
Answer:
(a) Ostrich (Struthio cameras). Its egg size is about 15 cm x 13 cm, weighing about 3 pounds.
(b) An egg provides sufficient proteins, fats, vitamin A and vitamin D.
(c) Eggs contain lots of cholesterol that may cause heart diseases. We should not eat many eggs in a day for a long period.
(d) Our duty is that we should make our friends, relatives and community aware about useful and harmful effects of eating eggs.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 4 Structure of the Atom

Multiple Choice Questions

Question 1.
Almost the entire mass of an atom is concentrated in the?
(a) proton
(b) electron
(c) nucleus
(d) neutrons
Answer:
(c) nucleus

Question 2.
Positive charge is carried by:
(a) X – rays
(b) cathode rays
(c) Y – rays
(d) anode rays
Answer:
(d) anode rays

Question 3.
Which one is not true for isotopes?
(a) Similar mass number
(b) Similar chemical properties
(c) Similar atomic number
(d) Similar electronic configuration
Answer:
(a) Similar mass number.

Question 4.
Electron was discovered by
(a) Chadwick
(b) Thomson
(c) Goldstein
(d) Bohr
Answer:
(b) Thomson.

Question 5.
The size of an atom is decided by
(a) mass of the atom
(b) number of protons
(c) number of protons and neutrons
(d) number of electrons
Answer:
(d) number of electrons.

Question 6.
Calcium has 20 electrons. These occupy K, L, M and N shells. Which shell or shells are incomplete?
(a) L, M, N shells
(b) M, N shells
(c) N shell
(d) K, M, L, N shells
Answer:
(c) N shell

Question 7.
The K, L and M shells of an atom are full. Its atomic number is:
(a) 18
(b) 20
(c) 10
(d) 12
Answer:
(a) 18.

Question 8.
Cathode rays are deflected towards?
(a) positive electrode
(b) negative electrode
(c) both electrodes
(d) none of the electrode
Answer:
(a) positive electrode.

Question 9.
The absolute charge of an electron is?
(a) -1.6 × 10^-19C
(b) +1.6 × 10^-19C
(c) 0.16 × 10^-19 C
(d) 16 × 10^-19 C
Answer:
(a) -1.6 × 10^-19C.

Question 10.
The proton is heavier than an electron by?
(a) 1850 times
(b) 1840 times
(c) 1000 times
(d) 100 times
Answer:
(b) 1840 times.

Question 11.
Carbon – 12 atom has:
(a) 6 electrons, 6 protons, 6 neutrons
(b) 6 electrons, 12 protons, 6 neutrons
(c) 12 electrons, 6 protons, 6 neutrons
(d) 18 electrons, 6 protons and 6 neutrons
Answer:
(a) 6 electrons, 6 protons, 6 neutrons.

Question 12.
Chadwick got the Nobel prize for the discovery of?
(a) protons
(b) neutrons
(c) electrons
(d) none of these
Answer:
(b) neutrons.

Question 13.
The volume of the nucleus of an atom when compared to the extranuclear part is?
(a) bigger
(b) smaller
(c) same size
(d) unpredictable
Answer:
(b) smaller.

Question 14.
The isobars among the following are
(a) \({ }_{20}^{40} \mathrm{Ca}\), \({ }_{17}^{34} \mathrm{Cl}\)
(b) \({ }_{18}^{40} \mathrm{Ar},{ }_{19}^{40} \mathrm{Ca}\)
(c) \({ }_{8}^{16} \mathrm{O},{ }_{8}^{18} \mathrm{O}\)
(d) \({ }_{7}^{19} \mathrm{X},{ }_{7}^{13} \mathrm{Y}\)
Answer:
(b) \({ }_{18}^{40} \mathrm{Ar},{ }_{19}^{40} \mathrm{Ca}\).

Question 15.
Outermost shell of an atom cannot accommodate more electrons than
(a) 2
(b) 8
(c) 18
(d) 16
Answer:
(b) 8.

Analysing & Evaluating Questions

Question 16.
The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is correct representation of the element?
(a) \({ }_{15}^{31} X\)
(b) \({ }_{31}^{16} X\)
(c) \({ }_{16}^{15} X\)
(d) \({ }_{15}^{17} X\)
Answer:
(a) \({ }_{15}^{31} X\).

Question 17.
The ion of an element has 3 positive charges. Mass number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion?
(a) 13
(b) 10
(c) 14
(d) 16
Answer:
(b) 10.

Question 18.
Which of the following in figures given below do not represent Bohr’s model of an atom correctly?

(a) I and II
(b) II and III
(c) II and IV
(d) I and IV
Answer:
(c) II and IV.

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.

1. Assertion : Nucleus of an atom is positively charged.
Reason : Neutrons lie in the nucleus of an atom.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

2. Assertion : Electrons revolving in the orbit never fall in the nucleus.
Reason : Opposite charges of nucleus and electrons maintain centripetal force.
Answer:
(C) The assertion is true but the reason is false.

3. Assertion : Valency of aluminum is 3.
Reason : Valency is the number of electrons present in the outermost shell.
Answer:
(C) The assertion is true but the reason is false.

4. Assertion : Mass of an atom lies in its nucleus.
Reason : Nucleus of an atom is composed of neutrons and protons.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

5. Assertion : Hydrogen and deuterium have similar chemical properties.
Reason : Hydrogen and deuterium have the same electronic configuration.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
What is an electron? What are its relative mass and charge?
Answer:
An electron is that sub – atomic particle which is negatively charged and has a mass of about 1/1840 times that of an atom of hydrogen.

Question 2.
What are nucleons?
Answer:
The sub – atomic particles (protons and neutrons) present in the nucleus of an atom are known as nucleons.

Question 3.
Why Bohr’s orbits are called stationary states?
Answer:
According to Bohr’s theory, electrons revolve around the nucleus and they have a fixed amount of energy. Thus, they are called stationary states.

Question 4.
What is an orbit?
Answer:
Orbit is the path around the nucleus in which the electron revolves.

Question 5.
What is meant by the electronic configuration of elements?
Answer:
The systematic arrangement of electrons in different orbits or shells of an atom of an element is known as the electronic configuration of that element.

Question 6.
Why did Rutherford select a gold foil in his α – ray scattering experiment?
Answer:
It is because gold has high malleability and it can be hammered into thin sheets.

Question 7.
Will Cl – 35 and Cl – 37 have different valencies?
Answer:
No. It is because these are isotopes of chlorine that have same atomic number but different mass numbers.

Question 8.
Calculate the number of neutrons present in the nucleus of an element X which is represented as X 15.
Answer:
\({ }^{31} \mathrm{X}_{15}\)indicate that no. of protons = 15 and mass number = 31
Mass number = No. of protons + No. of neutrons = 31
Number of neutrons = 31 – number of protons = 31 – 15 = 16

Question 9.
Why do Helium, Neon and Argon have zero valency?
Answer:
Helium, Neon and Argon have 2, 8 and 8 electrons respectively in their outermost shell, so there is no need for them to gain or lose electrons. Hence, they have zero valency.

Question 10.
Name two elements each with same number of protons and neutrons?
Answer:
Carbon (protons = neutrons = 6) and oxygen (protons = neutrons = 8)

Question 11.
Can the value of Z be same for two different atoms? (Z = atomic number)
Answer:
No, two different atoms cannot have same atomic number.

Question 12.
Name the scientist who discovered protons and neutrons in an atom.
Answer:
Protons were discovered by E. Goldstein in 1866 and neutrons were discovered by J. Chadwick in 1932.

Question 13.
What is an octet? Why do the atoms want to complete their octet?
Answer:
When the outermost shell of an atom is completely filled, it is said to be an octet. Atoms want to complete their octet because they want to become stable.

Question 14.
Find the valency of \({ }_{7}^{14} \mathrm{~N}\) and \({ }_{17}^{35} \mathrm{Cl}\)
The atomic number of nitrogen = 7,
number of protons = 7,
number of electrons = 7
Electronic configuration = K L M
= 2, 5,
⇒ Valency = 3
It will either gain three electrons or share 3 electrons to complete its octet.
The atomic number of chlorine = 17,
number of protons = 17,
number of electrons = 17
Electronic configuration = K L M
= 2, 8, 7 = Valency = 1 It will gain 1 electron to complete its octet.

Question 15.
Give the difference between three sub-atomic particles.
Answer:
Three sub – atomic particles are electrons, proton and neutron:

Particle	Electron	Proton	Neutron
Discovered by	J.J Thomson	E. Goldstein	J. Chadwick
Charge	-1	+1	0
Symbol	e	P	n
Mass	1/1840u	1u	1u

Analysing & Evaluating Questions

Question 16.
In a sample of ethyl ethanoate (CH₃COOC₂H₅), the two oxygen atoms have the same number of electrons but different number of neutrons. What can be the possible reason for it?
Answer:
The reason for it is that the two oxygen atoms are isotopes.

Question 17.
In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer.
Answer:
No. The statement is wrong In a neutral atom, the number of protons and electrons are always equal.

Question 18.
In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherford’s model of an atom, about 1.00% of the α – particles were found to deflect at angles > 50°. If one mole of α – particles were bombarded on the gold foil, compute the number of α – particles that would deflect at angles less than 50°.
Answer:
No. of particles deflected by < 50°
=\(\frac{99}{100} × \) × 6.022 x 1023 = 5.96 × 1023.

Short Answer Type Questions

Question 1.
In what way did Thomson propose the atomic model?
Answer:
Thomson proposed the model of an atom to be similar to a Christmas pudding The electrons are studded like currents in a positively charged sphere like Christmas pudding and the mass of the atom is supposed to be uniformly distributed.

Question 2.
What is an electron? State its relative mass and charge.
Answer:
An electron is a negatively charged particle found in the atoms of all the elements. Its relative mass is 1/1840
(a) m.u. and relative charge is negative.

Question 3.
What is a discharge tube?
Answer:
A discharge tube is a glass tube about 70 cm long and having a diameter of 5 cm. Two metal electrodes are sealed at the two ends, one of which is connected to the negative terminal of a battery and the other to the positive terminal. A side tube is fused at the centre of the glass tube which serves to pump out air from it, using a suction pump.

Question 4.
What are anode rays? State three properties of anode rays.
Answer:
Anode rays are stream of positively charged particles shot out from the anode of a discharge tube when a current is passed through a gas.

1. Anode rays travel in straight lines. They cast shadows of the objects placed in their path.
2. Anode rays can produce mechanical effects. This is evident by the fact that they can rotate a light paddle.
3. Anode rays are positively charged as they are deflected towards the negative plate in an electric field.

Question 5.
What happens to the e/m ratio of positive rays and why?
Answer:
The mass and charge of positively charged particles depend upon the gas, which is taken in the discharge tube. Different gases contain particles having different masses and different charges and consequently give different types of positive rays. In other words, the charge – to – mass ratio (e/m) is not constant.

Question 6.
A student weighs 30 kg Suppose his entire body is made up of electrons. How many electrons are there in his body? Compare the total number of electrons in his body with the population of India.
Answer:
Mass of electron = 9.1 × 10^-31kg.
Number of electrons in the body of student = Total mass/Mass of each electron = 30 kg/ 9.1 × 10^-31 kg Therefore, the student is made up of approximately 3 . 29 × 10³¹ electrons. The population of India is about 130 crores.
\(\frac{3.29 \times 10^{31}}{1.3 \times 10^{9}}\)
= 2.53 × 10²²
Therefore, number of electrons in the body of the student is 2.53 × 10²² times the population of India.

Question 7.
When an electron jumps from energy level K to energy level L, why is the energy of the atom increased?
Answer:
When an electron revolves around its nucleus in its orbit, sometimes it gain energy and it jumps from lower energy shell to higher energy shell. The energy of electrons increases hence, the energy of atom also increases.

Question 8.
State the similarities and dissimilarities between protons and neutrons.
Answer:
1. Similarities:

• Both protons and neutrons are present in the nucleus of an atom.
• The mass of a neutron is approximately equal to the mass of a proton, i.e., 1.67 × 10^-27 kg.

2. Dissimilarities:

• Protons possess a unit positive charge (1.6 × 10^-19coulomb).
• Neutron carries no charge, i.e., it is an electrically neutral particle.

Question 9.
Argon atom has 18 electrons. How many energy shells or orbits are incomplete?
Answer:
Distribution of 18 electrons in argon atom is K – shell = 2 electrons, L – shell = 8 electrons and M – shell = 8 electrons. K – shell and L – shell have maximum capacity of 2 and 8 electrons respectively. M – shell has a capacity of 18 electrons. Since, it is the outermost orbit and there is a rule that ‘the maximum number of electrons that can be accommodated in the outermost orbit is 8. In the present case, M – shell is also complete. Thus, in argon atom, no shell is incomplete. Argon atom has three comletely filled energy shells.

Question 10.
An element has 16 protons. How many electrons will be present in K, L and M shells of its atom? What will be its electrovalency?
Answer:
No. of electrons = no. of protons = 16
Electronic configuration: K = 2, L= 8, M = 6.
Valency or electrovalency is 8 – 6 = 2.

Question 11.
Define the term isotope. State the similar properties of isotopes.
Answer:
Atoms of the elements having the same atomic number but different atomic masses are called isotopes. This means isotopes of an element have the same number of protons but different number of neutrons in their nuclei. For example,

1. Carbon has two isotopes:

• \({ }_{6}^{12} \mathrm{C}\) – Atomic number 6 and mass number 12 and thus has 6 neutrons.
• \({ }_{6}^{14} \mathrm{C}\) – Atomic number 6 and mass number 14 and thus has 8 neutrons.

Properties of isotopes are:

• Isotopes of an element have same atomic number.
• Isotopes of an element have similar chemical properties.
• Isotopes of an element have similar number of electrons.
• Isotopes of an element have similar electronic configurations.

Question 12.
Why is carbon – 14 isotope radioactive whereas carbon – 12 is not?
Answer:
C – 12 atom has 6 protons and 6 neutrons, so it is not radioactive. C – 14 atom has 6 protons but 8 neutrons and thus has more neutrons than protons in its nucleus, therefore, its nucleus becomes unstable and shows radioactivity.

Analysing & Evaluating Questions

Question 13.
An atom ‘M’ of an element reacts with oxygen to form M203. Calculate the valency of the element ’M’.
Answer:
Two atoms of element ‘M’ combine with 3 atoms of oxygen.
∴ Number of oxygen atoms combining with one atom of element ‘M’ =\(=\frac{3}{2}\)
Therefore, the valency of element ‘M’ = \(=\frac{3}{2}\) × 2 = 3.

Question 14.
When high electrical potential is applied across a gas at very low pressure (≈10^-5 s atm), the walls of the discharge tube opposite to the cathode start glowing with a faint greenish light. What is the cause behind this observation?
Answer:
Fluorescence of the glass walls due to bombardment by the rays emitted from the cathode is responsible for faint greenish light.

Long Answer Type Questions

Question 1.
What are cathode rays? How are they formed?
Answer:
Cathode rays are a stream of negatively charged particles. These particles, called electrons, are shot from the metal cathode of a discharge tube when an electric current is passed through a gas at a very low pressure. A discharge tube is a long glass tube having two metal electrodes. When the pressure of air in the discharge tube is reduced to 0.001 mm of mercury and a high voltage is applied to the electrodes, the emission of light by air stops. But it is noticed that the wall of the discharge tube at the end opposite to the cathode begins to glow with greenish light. Since these rays originate at the cathode, they are known as cathode rays.

Question 2.
Who discovered the nucleus within an atom? How?
Answer:
Ernest Rutherford discovered the nucleus within an atom in his alpha – ray scattering experiment. The arrangement of the alpha – particle scattering experiment is as follows:

Rutherford produced a narrow beam of particles from a radioactive source (e.g., radium or polonium), which was allowed to strike an extremely thin gold foil. Rutherford proposed that if the spherical model proposed earlier, which made a uniform distribution of positive and negative particles was correct, then the alpha particle striking the gold atoms would be uniformly deflected However the observations were:

1. Most of the alpha particles passed straight through the gold foil without suffering any deflection from their original path
2. A few of them were deflected through small angles, while a very few deflected to a large extent.
3. A very small percentage (1 in 12,000) was deflected through 180° (turned back).
   This showed that positive mass is concentrated only at the centre of the atom. This was identified as nucleus.

Question 3.
Describe Rutherford’s model of an atom.
Answer:
Rutherford’s model of an atom:

1. The atom of an element consists of a small positively charged nucleus which is situated at the centre of the atom and which carries almost the entire mass of the atom.
2. The electrons are distributed in the empty space of the atom and are revolving around the nucleus at high speed.
3. The number of electrons in an orbit is equal to the number of positive charges (protons) in the nucleus. Hence, the atom is electrically neutral.
4. The volume of the nucleus is negligibly small as compared to the volume of the atom.
5. Most of the space in the atom is empty.
6. The arrangement is just like the solar system.

Question 4.
What are the important properties of cathode rays?
Answer:
The important properties of cathode rays:

1. Cathode rays travel in straight lines. That is why, cathode rays cast shadow of any solid object placed in their path.
2. Cathode rays set a paddle wheel into motion when it is placed in the path of these rays on the bladder of the paddle wheel. Hence, they are made up of material particles.
3. Cathode rays consist of negatively charged particles. When cathode rays are subjected to an electrical field, these get deflected towards the positively charged plate (anode).
4. Cathode rays heat the object only on which they fall. The cathode ray particles possess kinetic energy. When these particles strike an object, a part of the kinetic energy is transferred to the object. This causes a rise in the temperature of the object.
5. Cathode rays cause green fluorescence on glass surface, i.e., the glass surface on which the cathode rays strike shows a coloured shine.
6. Cathode rays can penetrate through thin metallic sheets.
7. Cathode rays ionise the gases through which they travel.
8. When cathode rays fall on certain metals, such as copper, X – rays are produced The X – rays are not deflected by electrical or magnetic fields. X – rays pass through opaque materials such as black paper, but stopped by solid objects such as bones.
9. Cathode rays travel with speed nearly equal to that of light.

Question 5.
State the postulates of Bohr’s model of an atom. Draw a diagram of Bohr’s model of an atom.
Answer:
The postulates of Bohr’s model of atom are as follows:
(a) In an atom, the electrons revolve around the nucleus in certain definite circular paths called orbits or shells. These are represented by the letters K, L, M, N… or the numbers n = 1,2, 3. 4…..

(b) The maximum number of electrons present in a shell is given by the formula 2n², where ‘n’ is the orbit number or energy level index, 1,2,3……. Hence, the maximum number of electrons in different shells is as follows, First orbit (K shell) will be = 2 × 12 = 2, second orbit (or L shell) will be = 2 × 22 = 8, third orbit (M shell) will be = 2 × 32 = 18 and so on.

(c) The maximum number of electrons that can be accommodated in the outermost orbit is 8.

(d) Electrons are not accommodated in a given shell, unless the inner shells are filled That is, the shells are filled in a step – wise manner.

(e) While revolving in discrete orbits, the electrons do not radiate energy.

Question 6.
Atom A has a mass number of 238 and atomic number of 92 and atom B has mass number 235 and atomic number 92.
1. How many protons do atoms A and B have?
2. How many neutrons do atoms A and B have?
3. Are atoms A and B isotopes of the same element? How?
Answer:

1. Both atoms A and B have the same number of protons = atomic number = 92
2. Number of neutrons in atom A = Mass number – atomic number = 238 – 92 = 146 Number of neutrons in atom B = 235 – 92 = 143
3. Atoms A and B are isotopes of the same element as these have similar chemical properties. This is because both have same number of electrons.

Analysing & Evaluating Questions

Question 7.
The radiation emitted by the cathode in a discharge tube at very low pressure (≈ 10 – 5 atm) is found to get deflected towards positive field, can set a paddle wheel into motion, cast shadow of any solid object placed in its path and travel with the speed of light.
(a) Identify the radiation.
(b) Mention its properties which are responsible for the observed phenomena.
Answer:

1. The radiation is cathode rays.
2. The properties of the emitted radiation which are responsible for the observed phenomena are:
   • get deflected towards positive field indicates that it consists of negatively charged particles.
   • can set a paddle into motion indicates that the cathode rays consist of material particles.
   • casts shadow of any solid when placed in its path indicates that cathode rays travel in straight line.
   • travel with the speed of light which indicates that particles in it are associated with electrical and magnetic fields, that is, the rays are electromagnetic radiation.

Activity – 1

1. Rub a comb on your dry hair.
2. Does the comb then attract the small pieces of paper?
   Now bring the comb close to some small pieces of paper.
   Observation
3. On rubbing, some electrons will move from hair to comb Thus the comb has negative charge.
4. The negatively charged comb will attract the paper pieces.

Activity – 2

1. Rub a glass rod with a silk cloth and bring the rod near an inflated balloon.
2. Observe what happens.

Observation
On rubbing, some electrons will move from glass rod to the silk cloth. Thus the glass rod becomes positively charged This charged glass rod now attracts the inflated balloon.

Value Based Questions

Question 1.
The distribution of electrons into different orbits of an atom was suggested by Bohr and Bury. The maximum number of electrons in a shell is given by 2n where n is the orbit number or energy level, i.e., K, L, M, N Further, shells are filled in a stepwise manner and that the outermost shell cannot accommodate more than 8 electrons. Shyam was asked to give the distribution of 20 electrons contained in an atom. He suggested the distribution as 2, 8, 10 as per formulation (2n²). The teacher – guided him and explained him the correct distribution.
1. What should be the correct distribution? Which shells are completely filled?
2. What moral values were given by the teacher to Shyam?
Answer:

1. The correct distribution is 2, 8, 8, 2. The shells (K, L) are completely filled.
2. The teacher told Shyam that a principle should be applied in totality to solve a problem.

Question 2.
00Isotops are the variants of a particular element. While all isotopes of a given element show the same number of protons, each isotope differs from the other in the number of neutrons. Due to particular proton – neutron ratio in the nucleus, some isotopes are stable and some are unstable. The latter are called radioactive elements. In the natural element, the amount of radioactivity is quite small and remains ineffective. As radio – elements find use in chemical reactions, medicines and as fuel and even face – wash, these are enriched for specific purpose.
1. Should the enrichment/isolation of radio – elements be banned or continued?
2. As a common man, show your concern to the production and use of isotopes?
Answer:

1. The isolation/enrichment of isotopes should be continued as these are very useful.
2. The common concern is that their exposure is dangerous to life and vegetation. These should be handled with care and stored properly.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 3 Atoms and Molecules

Multiple Choice Questions

Question 1.
The value 6.022 × 10²³ is also called:
(a) Dalton number
(b) Avogadro number
(c) Atomic number
(d) Mass number
Answer:
(b) Avogadros number

Question 2.
The number of moles in 52g of He is:
(a) 4
(b) 13
(c) 52
(d) 1
Answer:
(b) 13

Question 3.
1 amu means:
(a) 1/2th mass of C – 12atoms
(b) Mass of C – 12atom
(c) Mass of O – 16atom
(d) Mass of hydrogen molecules
Answer:
(a) 1/2th mass of C – 12atoms.

Question 4.
Formula for magnesium chloride is:
(a) MgCl
(b) Mg₂Cl
(C) MgCl₂
(d) MgCl₄
Answer:
(C) MgCl

Question 5.
Formula for aluminium oxide is:
(a) AlO
(b) AlO₄
(c) AlO₄
(d) Al₂O₃
Answer:
(d) Al₂O₃

Question 6.
The percentage of hydrogen in H₂O molecule is:
(a) 5.55%
(b) 11.11%
(c) 4445%
(d) 88.89%
Answer:
(b) 11.11%

Question 7.
Which of the following is not correctly matched?
(a) Calcium – Ca
(b) Silver – Ag
(e) Gold – Au
(d) Copper – Co
Answer:
(d) Copper – Co

Question 8.
Which of the following will have maximum mass?
(a) 0.1 mole NH₃
(b) 10²² atoms of carbon
(c) 10²² molecules of CO₂
(d) 1g of Fe
Answer:
(a) 0.1 mole NH₃

Question 9.
Which of the following satements is not true about an atom?
(a) Atoms are not able to exist independently.
(b) Atoms are the basic units from which molecules and ions are formed.
(C) Atoms are always neutral in nature.
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.
Answer:
(a) Atoms are not able to exist independently.

Question 10.
Which of the following has maximum number of atoms?
(a) 18g of H₂O
(b) 18g of O₂
(c) 18g of CO₂
(d) 18g of CH₄
Answer:
(d) 18g of CH₄

Question 11.
All noble gas molecules are
(a) monoatomic
(b) diatomic
(c) triatomic
(d) both (a) and (b)
Answer:
(a) monoatomic

Question 12.
The Valency of nitrogen in NH₃ is?
(a) 1
(b) 3
(C) 4
(d) 5
Answer:
(b) 3

Question 13.
How many moles are present in 28g nitrogen atoms?
(a) 1 mole
(B) 2.3 moles
(c) 0.5 mole
(d) 2 moles
Answer:
(d) 2 moles

Question 14.
Molecules of which of the following elements are made up of only one atom of that element?
(a) Sodium
(b) Iron
(c) Helium
(d) Chlorine
Answer:
(c) Helium

Question 15.
Which of the following elements gives polyatomic molecules?
(a) Ne
(b) P
(e) Ni
(d) Si
Answer:
(b) P

Analysing & Evaluating Questions

Question 16.
Chemical formula of a metal sulphate is MSO₄. What will be the formula of its chloride?
(a) MCl₂
(b) MCl
(e) M₂Cl
(d) M₂Cl₃
Answer:
(a) MCl₂

Question 17.
Valenc of an clement X is 3. Write the chemical formula of its oxide.
(a) X₂O₃
(b) X₂O
(c) XO₂
(d) X₃O₂
Answer:
(a) X₂O₃

Question 18.
A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
(a) 182.4 : 1
(b) 104 : 2
(c) 155 : 1
(d) 18 : 7
Answer:
(a) 182.4 : 1

Assertion Reason Questions

Directions:
In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.

1. Assertion : Non – metals usually form anions.
Reason : Non – metals have the tendency to lose electrons.
Answer:
(C) The assertion is true but the reason is false.

2. Assertion : When a substance undergoes a chemical change, there is no change in total mass.
Reason : In a chemical substance, elements are always present in definite proportion by mass.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

3. Assertion : Atoms cannot exist independently.
Reason : Atoms are the building blocks of elements.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

4. Assertion : Carbon – 12 isotopes is taken as the standard element for measuring atomic mass.
Reason : Carbon lacks the tendency to form a large number of compounds.
Answer:
(C) The assertion is true but the reason is false.

5. Assertion : Noble gases are inert in nature.
Reason : Noble gases have zero valency.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

Very Short Answer Type Questions 

Question 1.
What is the law of constant proportions?
Answer:
The law of constant proportions states that “a pure chemical compound always consists of the same elements that are combined together in a fixed proportion by mass.”

Question 2.
Define atom.
Answer:
The smallest particle of matter, which can take part in a chemical reaction is called atom.

Question 3.
What is molar mass? What are its units?
Answer:
The mass of one mole of a substance, i.e., Avogadro’s number of particles (6.022 x 10²³ particles) is called its mass. Its unit is gram per mole (g/mol) or kilogram per mole (kg/mol).

Question 4.
Define atomicity.
Answer:
Atomicity is defined as the number of atoms present in a molecule. For example, atomicity of NH₃ is 4 because there is one atom of N and three atoms of H.

Question 5.
Calculate the formula unit mass of Na₂CO₃. (Atomic mass of Na = 23u, C = 12u, O = 16u).
Answer:
The formula unit mass of a substance is the sum of the atomic masses of all the atoms in the formula unit of an ionic compound.
Therefore, formula unit mass of Na₂CO₃ = (2 × 23) + (1 × 12) + (3 × 16) = 46 + 12 + 48 = 106u.

Question 6.
Write the definition of a cation and an anion.
Answer:

1. Cation: It is the positively charged ion, e.g., Na⁺ which is attracted towards cathode in an electric field.
2. Anion: It is the negatively charged ion, e.g., Cl^– which is attracted towards anode in an electric field.

Question 7.
Define law of conservation of mass.
Answer:
It states that “mass can neither be created nor destroyed in a chemical reaction”.

Question 8.
‘Atoms of most elements are not able to exist independently. Name two atoms which exist as independent atoms.
Answer:
Noble gases such as argon (Ar) and helium (He) exist independently as atoms.

Question 9.
Write the atomicity of the following:
(a) Sulphur
(b) Phosphorus
Answer:
(a) Polyatomic (Octa – atomic)
(b) Tetra – atomic

Question 10.
State the number of hydrogen atoms in 1g of hydrogen.
Answer:
One gram of hydrogen = One mole = 6.022 × 10²³ atoms.

Analysing & Evaluating Questions

Question 11.
An element ‘Z’ forms the following compounds when it reacts with hydrogen, chlorine and oxygen.
ZH₃, ZCl₃ and Z₂O₃
(a) What is the valency of element ‘Z’?
(b) Is element ‘Z’ a metal or a non-metal?
Answer:
(a) The valency of ‘Z’ is 3.
(b) Element ‘Z’ is a metal because it is electropositive and is reacting with non – metals.

Question 12.
You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?
Answer:
This can be done by heating a little sample of each on a nickel spatula.

1. If it gets charred, the sample is sugar.
2. If it remains unaffected, it is salt.

Question 13.
An atom of element A is 1.92 times heavier that an atom of carbon – 12. What is the atomic mass of the element A?
Answer:
Atomic mass of A = 12.0g / mol × 1.92 = 23.04g/mol

Short Answer Type Questions

Question 1.
Write two drawbacks of Dalton’s atomic theory.
Answer:
Drawbacks of Dalton’s atomic theory:

1. According to modem theory, atom is not the ultimate indivisible particle of matter. Today, we know that atoms are divisible, i.e., they are themselves made up of particles (protons, electrons, neutrons, etc.),
2. In case of isotopes of an element, the assumption that atoms of the same element have same mass does not hold good.

Question 2.
What is meant by the term ‘molecule’? Give example.
Answer:
A molecule is the smallest particle of an element or a compound capable of independent existence under ordinary conditions. It shows all the properties of the substance, e.g., molecule of oxygen is O_2, Ozone is O_3, phosphorus is P_4, sulphur is S_8, etc.

Question 3.
Define one mole. Illustrate its relationship with Avogadro’s constant.
Answer:
One mole of any species (atoms, molecules, ions or particles) is that quantity in number, having a mass equal to its atomic or molecular mass in grams. The number of particles (atoms, molecules or ions) present in 1 mole of any substance is fixed with a value of 6.022 × 10²³. This number is called Avogadro’s number.

Question 4.
What is formula unit mass? How is it different from molecular mass?
Answer:
The formula unit mass of a substance is a sum of the atomic masses of all the atoms in one formula unit of a compound. The constituent particles of formula unit mass are ions and the constituent particles of molecular mass are atoms.

Question 5.
Find the number of atoms in each of the following:
(a) 0.5moles of C atoms
(b) 2moles of N atoms
Answer:
(a) 0.5mole of C atoms : Number of atoms in 1 mole of C atoms = 6.022 × 10²³ atoms
Number of atoms in 0.5 mole of C atoms = 6.022 × 10²³ × 0.5 atoms = 3.011 × 10²³ atoms.

(b) 2moles of N atoms : Number of atoms in 1 mole of N atoms = 6.022 × 10²³ atoms Number of atoms in 2 moles of N atoms
= 6.022 × 10²³ × 2
= 1.2044 × 10²³ atoms.

Question 6.
What do you mean by symbols of elements?
Answer:
In order to write the chemical reactions conveniently, each element is represented by a ‘letter’ or a group of two letters called symbol. Thus, hydrogen is represented as ‘H’ and calcium is represented as ‘Ca’.

Question 7.
With the help of an example, state the significance of a symbol of an element.
Answer:
The symbol of oxygen is O. As an example, let us give the significance of the symbol O.

1. O represents oxygen element.
2. O represents one atom of oxygen element.
3. O represents 16 atomic mass unit.

Question 8.
How do atoms exist?
Answer:
Individual atoms of most elements do not exist independently. These either aggregate to form elements or form molecules or ions which also aggregate in large numbers to form the matter that is available to us. Noble gases like helium, neon, etc., are an exception.

Question 9.
How does an atom differ from a molecule?
Answer:
An atom is the smallest particle of an element which may or may not have independent existence. On the other hand, molecule is the smallest particle of the element or compound which is capable of independent existence. For example, helium (H) is an atom and can exist as such, whereas hydrogen atom (H) cannot exist as such but exists as a molecule, i.e., Hexist as such, whereas hydrogen atom (H) cannot exist as such but exists as a molecule, i.e., H₂. A molecule may be made up of similar atoms (homoatomic molecule such as O₂) or dissimilar atoms (heteroatomic molecule such as HCl).

Question 10.
How many hydrogen and oxygen atoms are obtained when one molecule of water decomposes?
Answer:
It takes two molecules of the diatomic hydrogen gas to combine with one molecule of the diatomic oxygen gas to produce two molecules of water. In other words the ratio of hydrogen to oxygen is 2 : 1 the ratio of hydrogen to water is 1 : 1 and the ratio of oxygen to water is 1 : 2.

Question 11.
What are ionic compounds?
Answer:
Compounds containing charged species are known as ionic compounds. Such compounds are formed from combination of metals with non – metals. For example, sodium chloride is an ionic compound. Its constituent particles are positively charged sodium ions (Na⁺) and negatively charged chloride ions  (CF^–)

Question 12.
Name the ions in the following compounds. Sodium fluoride, potassium bromide, calcium oxide, silver sulphide, magnesium oxide
Answer:
(a) Sodium fluoride : Sodium cation and fluoride anion
(b) Potassium bromide : Potassium cation and bromide anion
(c) Calcium oxide : Calcium cation and oxide anion
(d) Silver sulphide : Silver cation and sulphide anion
(e) Magnesium oxide : Magnesium cation and oxygen anion

Question 13.
What is valency? What is its use?
Answer:
The valency of an element is the number of electrons an atom of the element uses to combine with atoms of other elements – it is the combining power of an atom of the element. In an atom, the valence electrons are the electrons that can be used in combining with other atoms. These are the electrons in the orbitals of the outermost shell (also called valency shell). It is not in all cases that the valency of an atom equals the total number of its valence electrons. For example, oxygen has 6 valence electrons, but its valency is 2. Some elements may have more than one combining power (or valency), while others have just one.

Question 14.
Do all elements have a charged valency? Explain.
Answer:
All elements do not form ions so their valencies do not have a charge. For example, carbon and silicon have a valency of 4 and nitrogen has a valency of 3. Further, molecules containing only non – metals are formed without a charged valency. For example, in carbon tetrachloride (CCl₄), carbon has a valency of 4 and chlorine has a valency of 1.

Question 15.
Write the symbol and valency of the following polyatomic ions. Ammonium, hydroxide, nitrate carbonate, sulphate, sulphite bicarbonate and phosphate. A molecule may be made up of similar atoms (homoatomic molecule such as O₂) or dissimilar atoms (heteroatomic molecule such as HCl).
Answer:

Polyatomic ion	Symbol	Valency
Ammonium	(NH4)+	+1
Hydroxide	(OH)–	-1
Nitrate	(NO3)–	-1
Carbonate	(CO3)2-	-2
Sulphate	(SO4)2-	-2
Sulphite	(SO3)2-	-2
Bicarbonate	(HCO3)–	-1
Phosphate	(PO4)3-	-3

Analysing & Evaluating Questions

Question 16.
In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C₆H₁₂ O₆. How many grams of water would be required to produce 18g of glucose? Compute the volume of water so consumed assuming the density of water to 1g cm 3.
Answer:
The photosynthesis can be described by the reaction.

(6 × 12) + (12 × 1) + (6 × 16)g = (72 + 12 + 96)g = 180g
180g of glucose requires 108g water. 18g of glucose will require \(\frac{108 \mathrm{~g}}{180 \mathrm{~g}} \) x 8g = 10.8g
Then volume of water used
=\(\frac{\text { Mass }}{\text { Density }}=\frac{10.8}{1 \mathrm{~g} \mathrm{~cm}^{-3}}\)
= 10.8 cm3.

Question 17.
A compound was found to contain: Carbon = 55%, Oxygen = 36% and Hydrogen = 9%. What is the simplest whole number atomic ratio in the compound? The given data are processed as follows: Let us assume that the mass of the given compound is 100 gram.
Answer:
Carbon = 55% = \(\frac{55}{100}\) = 55 gram
Oxygen = 36% = \(\frac{36}{100}\) = 36 gram
Hydrogen = 9% \(\frac{9}{100}\) = 9 gram
\(\frac{55}{12}\) = 4.6 12
\(\frac{36}{16}\) 36/16 = 2.25 16
\(\frac{9}{1}\) 9/1 = 9.

Simplest atomic ratio
\(\frac{4.6}{2.25}\) = 2
\(\frac{2. 25}{2.25}\) = 1
\(\frac{9}{2.25}\) = 4
Thus, the simplest atomic ratio for the given compound is C : O : H = 2 : 1 : 4.

Long Answer Type Questions

Question 1.
What are the rules for writing the symbol of an element?
Answer:
IUPAC, i.e., International Union of Pure and Applied Chemistry approves names of elements. Symbols are the first one or two letters of the element’s name. The first letter of a symbol is always written as a capital letter (upper case) and second letter as a small letter (lower case).
For example:
Hydrogen → H
Helium → He
Some symbols are taken from the names of elements in Latin, German or Greek. For example,

• Symbol of iron is Fe, its Latin name is Ferrum.
• Symbol of sodium is Na, its Latin name is Natrium.

Question 2.
Describe the postulates of Dalton’s atomic theory.
Answer:

1. Every element is composed of extremely small particles called atoms.
2. Atoms of a given element are identical, both in mass and properties. Different chemical elements have different kinds of atoms. In particular, their atoms have different masses.
3. Atoms can neither be created nor be destroyed or transformed into atoms of other elements.
4. Compounds are formed when atoms of different elements combine with each other in the ratio of small whole numbers.
5. The relative number and kinds of atoms in a given compound are constant.

Question 3.
Explain relative atomic mass and relative molecular mass.
Answer:
The atomic mass of an element is the relative mass of its atom as compared with the mass of a particular atom of carbon – 12 which is taken as 12 units. Thus, the atomic mass of an element indicates the number of times one atom of element is heavier than \(\frac{1}{2}\)  of a carbon – 12 atom or (C – 12). For example, the atomic mass of oxygen is 16 which indicates that an atom of oxygen is 16 times heavier than \(\frac{1}{2}\) th mass of a carbon – 12 atom.
One atomic mass unit = \(\frac{1}{2}\) the mass of C – 12 atom.
Relative molecular mass is defined as the number of times one molecule of a substance or given element is heavier than \(\frac{1}{2}\) th of the mass of one atom of carbon – 12. For example, the relative molecular mass of water is 18. This means that the average mass of one molecule of water is 18 times the mass of one – twelfth of a carbon – 12 atom.

Question 4.
What is the utility of the mole concept?
Answer:
The utility of mole concept:

1. From the number of moles of a substance, we can calculate the number of elementary particles
   because the number of moles of a substance is directly proportional to the number of elementary particles.
2. One mole of a gas occupies 22. 4 litres at S. T. P (273 K and 1 atm).
3. One mole of any gas particles in the same conditions of temperature and pressure occupies the same volume.
4. One mole is equal to molecular mass in grams which is equal to 6.022 x 10²³ atoms, molecules, ions, etc. Thus, we can calculate absolute masses of atoms and molecules.

Question 5.
(a) Calculate the mass of 0.72g molecules of CO₂.
(b) Calculate the number of moles of iron in iron sheet containing 1022 atoms of iron.
Answer:
(a) Mass of one mole or 1 g molecules of CO₂ = 12 + 32 = 44g
Mass of 0.72 g molecules of CO₂ = 44 × 0. 72 = 31.68g.

(b) Number of moles containing 6.022 × 10²³ atoms = 1 mole
Number of moles containing 10²² atoms of iron = \(\frac{1}{6.022 \times 10^{23}}\) × 10²² = 0.0166 mole
Moles of iron in iron sheet = 0. 0166 mole.

Question 6.
Define the term gram atom. How is it related to the mole and Avogadro’s number?
Answer:
The atomic mass of an element expressed in grams is called gram atomic mass or gram atom. One gram atom of any element contains 6.022 × 10²³ (Avogadro’s number) atoms of the element. It is equal to one mole of atoms.
One gram atomic mass = 6.022 × 10²³ atoms (Avogadro’s number) = 1 mole. The gram atomic mass of an element is equal to the mass of its = 6.022 × 10²³ atoms.

For example, one gram atom of hydrogen atom weighs one gram and contains 6.022 × 10²³ hydrogen atoms. One mole of hydrogen atom also weighs one gram. In other words, one atomic unit hydrogen means only one atom of hydrogen whereas one gram hydrogen has one mole atoms or 6.022 × 10²³ atoms of hydrogen. Similarly, 16 u oxygen means only one atom of oxygen whereas 16 g oxygen has 6.22 × 10²³ atoms of oxygen.

Question 7.
Calculate following:
(a) The mass of one atom of oxygen
(b) The mass of one molecule of oxygen
(c) The mass of one mole of oxygen gas
(d) The mass of one ion of oxygen
(e) The number of atoms in 1 mole of oxygen molecules
Answer:
(a) Mass of one atom of oxygen:
1 mole oxygen atom = 16g = 6.022 × 10²³ atoms.
Mass of one atom of oxygen = 16/6.022 × 10²³ =2.656 × 10^-23g

(b) Mass of one molecule of oxygen:
1 molecule of oxygen = O² = 2 × 16 = 32u

(c) Mass of one mole of oxygen gas:
1 mole of oxygen gas is O² = 32g

(d) Mass of one ion of oxygen:
1 mole of oxygen = 6. 022 × 10²³ atoms = 16g
Mass of one ion of oxygen =16/6.22 × 10²³ = 2.65 × 10^-23g

(e) Number of atoms in one mole of oxygen molecules:
1 mole of oxygen molecules, i. e.,
O₂ = 6. 022 × 10²³ molecules,
1 molecule of O₂ = 2 atoms,
Number of atoms in 1 mole of oxygen molecule
= 6. 022 × 10²³ × 2 atoms = 1.2044 × 10²⁴ atoms.

Analysing & Evaluating Questions

Question 8.
Take 25 mL of 5% solution of lead nitrate in a beaker (100 mL). Take about 10 mL of dilute HCl in another beaker (100 mL). Weigh both the beakers together with their contents. Transfer dilute HCl to the lead nitrate solution and mix the solution. Allow the reaction to take place. Weigh both the beakers (one with the reaction mixture and the other empty) together.
1. What happens when HCl is added into the solution of lead nitrate?
2. Write the balanced chemical equation for the reaction that takes place.
3. What happens to the mass of the setup after the reaction?
4. Which law of chemical combination is illustrated by this activity?
Answer:
Weigh both the beakers (one with the reaction mixture and the other empty) together:

1. When HCl is added to the solution of lead nitrate, white precipitate of lead chloride is formed.
2. The white precipitate settles down.
   
3. There is no change in the mass of the setup after the reaction.
4. The activity supports / illustrates the law of conservation of mass.

Activity
Take one of the following pairs of chemicals X and Y:

X	Y
(a) Copper sulphate	Sodium carbonate
(b) Barium chloride	Sodium sulphate
(c) Lead nitrate	Sodium chloride

Prepare separately a 5% solution of any one pair of substances listed under X and Y each in water:

• Take a little amount of solution of Y in a conical flask and some solution of X in an ignition tube.
  Hang the ignition tube in the flask carefully such that the solutions do not get mixed. Put a cork on the month of the flask as shown in the figure.
• Weigh the flask with its contents carefully.
• Now tilt and swirl the flask, so that the solutions X and Y get mixed.
• Weigh again.
• What happens in the flask?
• Do you think that a chemical reaction has taken place?
• Why should we put a cork on the mouth of the flask?
• Does the mass of the flask and its contents change?

Observations

• Some precipitate is formed in all cases.
• A chemical reaction takes place in all the cases.
• Cork is put on the mouth of the flask so that no material escapes out or splits out on swirling.
• The mass of the flask and its contents are the same before the reaction and after the reaction.

Value Based Questions

Question 1.
The idea of divisibility of matter was considered long back in India around 500 B, C. An Indian philosopher Maharishi Kanad, postulated that if we go on dividing matter, we shall end up getting smaller and smaller particles. Ultimately, a time will come when we shall come across the smallest particles, beyond which further division will not be possible. Another Indian philosopher Pakudha Katayama, eleborated his doctrine and said that these particles normally exist in a combined form and give us various forms of matter which find diverse use in our daily life.
1. Later on what name was given to these indivisible particles by Greek philosopher, Democritus.
2. What conclusion can you draw from the above paragraph in the social context?
Answer:

1. Atoms.
2. United we stand, divided we fall. Being together we can hope to achieve great heights.

Question 2.
Lavoisier, along with other scientists, observed that many compounds were formed from two or more elements. All these compounds had the same elements in the same proportions, irrespective of the source of compounds. For example, in a compound of water taken from sea, well or rain, hydrogen and oxygen are present in the ratio 1 : 8 by mass. Similarly, in ammonia, nitrogen and hydrogen are always present in the ratio 14 : 3 by mass, whatever be its source.
1. What name was given to this behaviour of chemical compounds? How is it formulated?
2. Does such properties of a compound related to human behaviour?
Answer:

1. Above behaviour led to the law of definite proportions. This law states that in a chemical substance, the elements are always present in definite proportions by mass,
2. Human behaviour is quite complicated. To a good extent, it also depends on the environment from where a person comes.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 2 Is Matter Around Us Pure

Multiple Choice Questions

Question 1.
Which of the following shows Tyndall effect?
(a) Starch solution
(b) Salt solution
(c) Sugar solution
(d) Copper sulphate solution
Answer:
(a) Starch solution

Question 2.
To obtain toned and double toned milk from full cream milk we can perform
(a) distillation
(b) filtration
(c) sedimentation
(d) centrifugation
Answer:
(d) centrifugation

Question 3.
Fog and clouds are examples of
(a) suspension
(b) emulsion
(c) aerosol
(d) colloid
Answer:
(c) aerosol

Question 4.
A mixture of ammonium chloride and sand can be separated by
(a) decantation
(b) sublimation
(c) centrifugation
(d) evaporation
Answer:
(b) sublimation

Question 5.
…………. may be termed as a pure substance.
(a) Soft drink
(b) Aerosol
(c) Sodium chloride
(d) Soil
Answer:

Question 6.
…………. is a chemical change.
(a) Evaporation
(b) Heating of copper and sulphur
(c) Freezing of water
(d) Mixing of H2 and 02
Answer:
(b) Heating of copper and sulphur

Question 7.
A colloidal solution is a
(a) heterogeneous and transparent mixture
(b) homogeneous mixture
(c) heterogeneous mixture in which particles can be seen with naked eyes
(d) heterogeneous mixture in which particles cannot be seen with naked eyes
Answer:
(d) heterogeneous mixture in which particles cannot be seen with naked eyes

Question 8.
Shaving cream is a colloidal solution of
(a) gas in liquid
(b) liquid in liquid
(c) solid in liquid
(d) gas in solid
Answer:
(a) gas in liquid

Question 9.
………… represents a physical change.
(a) Electric current passed through water
(b) Electric current is passed through a bulb and it glows
(c) Burning of a candle
(d) Making of curd from milk
Answer:
(b) Electric current is passed through a bulb and it glows

Question 10.
10 gram each of iron filings and sulphur podwer is taken in a china dish. Which of the following activities gives a compound?
(a) Separate iron filings with the help of a magnet
(b) Add carbon disulphide, stir well and filter
(c) Mix and crush the mixture
(d) Heat strongly till red hot while mixing
Answer:
(d) Heat strongly till red hot while mixing

Question 11.
Choose a colloid out of the following.
(a) Air
(b) NaCl in water
(c) Emulsion
(d) Alloy
Answer:
(c) Emulsion

Question 12.
Chalk powder dissolved in water is an example of
(a) suspension
(b) true solution
(c) colloid
(d) saturated solution
Answer:
(a) suspension

Analysing & Evaluating Questions

Question 13.
Two substances, A and B were made to react to form a third substance, A2B according to the following reaction: 2A + B → A2B Which of the following statements concerning this reaction are incorrect?
I. The product A2B shows the properties of substances A and B.
II. The product will always have a fixed composition.
III. The product so formed cannot be classified as a compound.
IV. The product so formed is an element.
(a) I, II and III
(b) II, III and IV
(c) I, III and IV
(d) I, II, and IV
Answer:
(c) I, III and IV

Question 14.
Two chemical species X and Y combine together to form a product P which contains both X and Y: X + Y → P X and Y cannot be broken down into simpler substances by simple chemical reactions. Which of the following statements concerning the species X, Y and P are correct?
I. P is a compound.
II. X and Y are compounds.
III. X and Y are elements.
IV. P has a fixed composition.
(a) I, II and III
(b) I, II and IV
(c) II, III and IV
(d) I, III, and IV
Answer:
(d) I, III, and IV

Question 15.
Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g of NaCl + 100 g of water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl+99.90 g of water
(d) 0.10 g of NaC 1 +99.90 g of water
Answer:
(c) 0.01 g of NaCl+99.90 g of water

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.

1. Assertion: Colloids are heterogeneous mixtures.
Reason: Particles of colloids can scatter a beam of light.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

2. Assertion: A solution contains solute dissolved in a solvent.
Reason: Solutions always exist in liquid state.
Answer:
(C) The assertion is true but the reason is false.

3. Assertion: Ethanol and water can be separated from their mixture by distillation process.
Reason: There is much difference between the boiling points of ethanol and water.
Answer:
(D) Both the statements are false.

4. Assertion: On heating, a saturated solution becomes unsaturated.
Reason: Solubility of a substance decreases with increasing temperature.
Answer:
(C) The assertion is true but the reason is false.

5. Assertion: Human blood is an impure substance.
Reason: The composition of human blood is variable.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
Which of the following materials are pure substances? Butter, ghee, ink, water, banana, glass, sugar, blood, marble, wood Answer:Pure substances are sugar and water.

Question 2.
Define solution. If 10 mL of sulphuric acid is dissolved in 90 mL of H₂O, calculate the concentration of sulphuric acid in the solution.
Answer:
A solution is a homogeneous mixture of two or more substances. Concentration of sulphuric acid in solution = \(\frac{10 \times 100}{100}\)
= 10% volume by volume.

Question 3.
Can we separate alcohol dissolved in water by using a separating funnel? Yes or No. Why?
Answer:
No, water and alcohol, when mixed, form a single layer being miscible and thus separating funnel cannot be used to separate alcohol dissolved in water.

Question 4.
What is the important criterion used in chromatography for separating the components of a mixture?
Answer:
Chromatography can be used for separating the components of a mixture if the components of the mixture exhibit different solubilities in the same solvent or a mixture of solvents.

Question 5.
Differentiate between the solvent and the solute.
Answer:
The component of the solution that dissolves the other component into it is called the solvent. The component of the solution that is dissolved in the solvent is called solute.

Question 6.
What is ‘tincture of iodine’?
Answer:
A solution of iodine in alcohol is known as tincture of iodine, ft has iodine (solid) as the solute and alcohol (liquid) as the solvent.

Question 7.
What is concentration of a solution?
Answer:
The amount of solute dissolved in the given amount of solution is called the concentration of that solution.

Question 8.
Why is water called a universal solvent?
Answer:
Water is called the “universal solvent” because it dissolves more substances than any other liquid.

Question 9.
Define Tyndall effect.
Answer:
Tyndall effect is the scattering of light as a light beam passes through a colloid.

Question 10.
Differentiate between emulsion and aerosol.
Answer:
When both the dispersed phase and the dispersing medium are liquid, the colloid is called an emulsion. For example, milk and face cream. When solid or liquid is dispersed in a gas, it is called aerosol. Example, smoke and fog.

Question 11.
How can you separate two liquids that have difference of boiling points less than 25K?
Answer:
To separate a mixture of two or more miscible liquids for which the difference in boiling points is less than 25K, fractional distillation is used.

Question 12.
What is crystallisation?
Answer:
Crystallisation is a process that separates a pure solid in the form of its crystals from a solution. The crystallisation method is used to purify solids. For example, the salt we get from sea water can have many impurities in it. To remove these impurities, the process of crystallisation is used.

Question 13.
Write any two applications of crystallisation.
Answer:

1. Purification of salt that we get from sea – water.
2. Separation of crystals of alum (phitkari) from impure samples.

Question 14.
What is an alloy?
Answer:
The homogeneous mixture of two or more metals or a metal and a non-metal is called an alloy. For example, steel is an alloy of iron and carbon.

Analysing & Evaluating Question

Question 15.
While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice.
Answer:
Acetone can be obtained back by distillation and can be collected as distillate. This is because the boiling point of acetone (56°C) is much lower than that of water (100°C).

Question 16.
Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it be a pure substance or a mixture? Give reasons for the same.
Answer:
It will be a pure substance because the chemical composition of sugar (sucrose) is always the same, whatever be its source.

Question 17.
Mixtures of three substances in water are filtered and the observations are shown in the figure given alongside. Which of the mixtures is:
(a) true solution?
(b) suspension?
(c) colloidal solution?

Answer:
(a) Mixture in C is a true solution.
(b) Mixture in B is a suspension.
(c) Mixture in A is a colloidal solution.

Short Answer Type Questions 

Question 1.
Differentiate between mixtures and compounds.
Answer:

Mixtures	Compounds
(a) Elements or compounds just mix together to form a mixture and no new substance is formed.	(a) Elements react to form new substances, i.e. compound
(b) A mixture has a variable composition.	(b) The composition of a compound is always fixed.
(c) A mixture shows the properties of its constituent substances.	(c) The properties of a compound are totally different from those of its constituents.
(d) The constituents can be separated fairly easily by physical methods.	(d) The constituents can be separated only by chemical or electrochemical reactions.

Question 2.
Distinguish between a physical change and a chemical change.
Answer:

Physical Change	Chemical Change
(a) No new substance is formed.	(a) New substance is formed.
(b) It is a reversible change.	(b) It is an irreversible change.
(c) The properties of constituents are retained.	(c) The properties of constituents are not retained.

Question 3.
How will you separate a mixture of oil and water?
Answer:
To separate a mixture of oil and water, we need a separating funnel as both are immiscible liquids. Pour the mixtures in separating funnel and let the funnel stand undisturbed for a few minutes, so that separate layer of oil and water are formed. Open the stopcock of the separating funnel and pour out the lower layer of water carefully.

Question 4.
How will you establish that air is a mixture and not a pure substance?
Answer:
Air is a homogeneous mixture of several gases. The amounts of gases present in air at different places vary. No definite formula can be assigned to air. The different gases present in air are separable by a physical process, i.e., fractional distillation of liquid air. Hence, air is a mixture and not a pure substance.

Question 5.
Define the following terms:
(a) True solution
(b) Solute
(c) Solvent
(d) Solubility
Answer:
(a) True solution: A true solution is defined as a homogeneous mixture of two or more substances. It means that every portion of the solution has same properties, e.g., sugar dissolved in water, iodine in ethyl alcohol (tincture of iodine).

(b) Solute: The substance present in smaller proportion in a solution is called solute.

(c) Solvent: The substance present in larger proportion in a solution is called solvent. For example, if a homogeneous mixture or a solution is formed by dissolving 2 g salt in 100 g of water, then salt is the solute and water is the solvent.

(d) Solubility: The maximum amount of solid that can be dissolved in a given amount of the solvent at a particular temperature is termed a its solubility at that temperature.

Question 6.
State the difference between aqueous and non – aqueous solutions.
Answer:
True solutions obtained in water are aqueous solutions, e.g, vinegar. True solutions obtained without water, but in organic liquids, like alcohol, acetone, etc., are non-aqueous solutions, e.g., amino acids dissolved in acetone.

Question 7.
State the properties of a solution.
Answer:
Properties of a solution are as follows:
(a) A solution is a homogeneous mixture.
(b) Particles of a solution are smaller than 1 nm. They cannot be seen with naked eyes.
(c) Particles do not scatter a beam of light.
(d) Solute particles cannot be separated from the mixture by the process of filtration and thus, solution is stable.

Question 8.
Why is crystallisation better than evaporation?
Answer:
Crystallisation is better than evaporation because during evaporation:
(a) some solids, like sugar, are decomposed on heating to dryness.
(b) some impurities may remain dissolved in the solution even after filtration, which on evaporation, contaminates the solid.

Question 9.
What is a suspension? Give examples.
Answer:
Suspension is a heterogeneous mixture of two or more substances. In suspension, particles are suspended throughout the bulk and can be seen with naked eyes. In suspension, particles of solute do not dissolve and are rather suspended. Particles of suspension are large enough to scatter the rays of light and hence, the path of light is visible through a suspension.
Example of suspension – mixture of chalk powder and water, muddy water, mixture of flour and water, mixture of dust particles and air, etc.

Question 10.
What are colloidal solutions? Give examples.
Answer:
Colloidal solution is a heterogeneous mixture of two or more substances. Colloidal solutions appear homogeneous because of relatively small size of particles in comparison with suspension. Particles of colloidal solution are called colloid. Colloids are dispersed throughout the solvent. Particles of colloidal solution are not visible to naked eye but scatter the ray of light, i.e., show Tyndall effect. The size of the molecules of the solute, also called dispersed phase, is between 1 nm and 1000 nm and the molecules remain suspended in the dispersing medium. Colloids cannot be separated by filtration but can be separated using centrifugation. Milk, ink, blood, solution of soap or detergent, etc., are some common examples of colloidal solution.

Question 11.
State the characteristics of colloids.
Answer:
Colloids exhibit the following characteristics:
(a) Brownian motion: Colloidal particles move randomly in zigzag paths like gas particles. This is called Brownian motion. This type of motion is caused due to the collisions between the particles of the dispersion medium and the dispersed phase.

(b) Tyndall effect: When a strong beam of light is passed through a colloidal solution taken in a beaker placed in a dark room, the path of light through the colloidal solution becomes visible. This is called Tyndall effect. The particles become illuminated because they scatter the light falling on them in all directions.

(c) Charge: Colloidal particles carry charge. When electric current is passed through a colloidal solution, the particles move either towards the positive (+) or the negative (-) electrode. This phenomenon is known as electrophoresis. Using the direction of the movement of colloidal particles, we can know the nature of the charge on them.

Question 12.
State the principle of the process of centrifugation.
Answer:
Sometimes, the solid particles in a liquid are very small and pass through a filter paper. For such particles the filtration techniques cannot be used for separation. Such mixtures are separated by centrifugation. The denser particles are forced to the bottom and the lighter particles stay at the top when spun rapidly. The mixture is taken in a closed bottle and rotated at a high speed. The heavy particles settle at the bottom while lighter particles remain behind. For example, to separate cream from milk, milk is churned for 2 – 3 minutes. Cream being lighter than milk, floats at the top of the mixture.

Question 13.
What is chromatography? State its applications.
Answer:
Chromatography is the technique whereby different soluble components of a mixture are separated due to their differential movement over a stationary phase under the influence of a mobile phase, i.e., solvent. If water is the solvent, then the component that is more soluble in water will move faster. In this way, the different components get separated. The technique of chromatography is used to separate:
(a) components of dyes
(b) pigments from natural colours
(c) amino acids
(d) sugar from urine samples
(e) drugs in blood

Question 14.
What is decantation? Explain.
Answer:
Decantation is the process of separating insoluble solids from liquids. A suspension of solid particles in a liquid is allowed to stand for a few minutes. Insoluble solid particles settle down at the bottom due to their weight. This forms the sediment. The clear liquid is then transferred into another container, without disturbing the settled particles. In other words, clear liquid is decanted and separated from the solid.

Question 15.
Draw a labelled diagram of water purification system in water works.
Answer:
A flow – diagram of a typical water purification system is given below:

Analysing & Evaluating Questions

Question 16.
The table given below shows the number of grams of five different solids dissolving in 100 g of the solvents: water, alcohol and
chloroform (all at 20°C).

Solvent	Salt	Sugar	Iodine	Chalk	Urea
Water	36.0	204.0	0.6	0.0	100.0
Alcohol	0.0	0.0	20.0	0.0	16.0
Chloro	0.0	0.0	3.0	0.0	0.0

(a) Which solid dissolves best in water at 20°C?
(b) Which solid has maximum solubility in alcohol?
(c) Which solid is insoluble in all three solvents?
Answer:
(a) Sugar dissolves best in water at 20°C.
(b) Iodine dissolves maximum in alcohol.
(c) Chalk is insoluble in all the three solvents.

Question 17.
The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively, to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?
Answer:
‘C’ has made the desired solution. Mass by volume percentage
\(⇒\frac{\text { Mass of solute }}{\text { Volume of solution }} \times 100\)
\(⇒\frac{50}{100} \times 100=50 \%\)

Long Answer Type Questions

Question 1.
Describe the process of filtration.
Answer:
Filtration is a process by which insoluble solids can be removed from a liquid using a filter paper. A filter paper is a special type of paper which has pores that are tiny enough to let only liquids pass through it. If you pass a solution through filter paper, the undissolved solid particles will be left behind on the paper whereas the liquid will pass through the filter paper. The liquid that passes through is called the filtrate and the undissolved solid particles left behind are called residue.For example, a mixture of chalk powder and water can be separated by this method.

Experiment Set up the apparatus as shown in the figure. The solution is passed through filter paper fitted in the funnel. During filtration, insoluble solid substance is retained in the filter paper as residue while the liquid, free from any suspended matter, passes through the filter paper and is collected as filtrate. This filtrate may be heated to dryness to obtain soluble component.

Question 2.
Describe the method of nitration with the help of a diagram to separate a mixture of two immiscible liquids- kerosene and water.
Answer:
When two liquids do not mix, they form two separate layers and are known as immiscible liquids. These two liquids can be separated using a separating funnel. Experiment: A separating funnel is a special type of glass funnel, which has a stop – cock in its stem to regulate the flow of liquid. It will separate the immiscible liquids into two distinct layers depending on their densities. The heavier liquid forms the lower layer while the lighter one forms the upper layer. Open the stopcock of separting funnel and let the liquid forming lower layer run into the a beaker. Close the stopcock as the liquid forming upper layer reaches the stopcock. You will be left behind with just the liquid forming upper layer in the funnel. Collect this liquid into another beaker. For example, a mixture of kerosene and water is separated by using separating funnel method. This method is also used to extract iron from its ore.

Separation of immiscible liquids using separating funnel

Question 3.
Explain sublimation process with a labelled diagram.
Answer:
Sublimation is the property of a substance in which it is converted directly from solid to gaseous state. Such substances are known as sublime. Some examples of solids which sublime are ammonium chloride, camphor, naphthalene and anthracene. Following is an activity to separate a mixture of ammonium chloride and salt.

Experiment: Take a mixture of ammonium chloride and salt in a china dish and cover it with an inverted conical transparent funnel. At the other end of the funnel, put a cotton plug so that vapours could not come out. Now place the china dish on a burner. As the ammonium chloride is sublime, after heating, it will be converted directly into vapours and this vapours will again condense at the upper colder part of the funnel to form solid ammonium chloride. In this way, the mixture of ammonium chloride and salt can be separated by the

funnel to form solid ammonium chloride. In this way, the mixture of ammonium chloride and salt can be separated by the sublimation method.

Question 4.
Define the term ‘element’. Give an illustration in support of your definition.
Answer:
An element is a substance which contains atoms of the same kind. It cannot be converted into anything visibly simpler than itself. For example, the element mercury is composed of only one kind of atoms with atomic number 80. Gold, copper, hydrogen are other examples of elements. After the discovery of isotopes, the definition of an element is modified. It is more correct to say that an element is a substance made up of atoms, all having the same atomic number.

Illustration:
1. Take about 5.0 g of mercury (II) oxide in a hard glass test tube fitted with cork and delivery tube.

2. Heat the tube over a Bunsen humer flamc first slowly and then strongly. Afier sometime you would observe a gas coming out of che capillary tube.
3. If you bring a glowing maichstick near the mouth of the tube, the slick starts burning with a flame. That is due to esolsed oxygen which is a supporter of combustion,
4. Continue heating the glass lube. After sometime, no more gas is evolved and a shining liquid is left at the bottom of the glasscube. That is mcrcury.
Mercury (II) oxide → Mercury + Oxygen

Conclusion: Rcd mercuric oxide on heating breaks up into two simpler substances, mercury and oxygen which are not further broken. Thus, mercury and oxygen arc elements.

Question 5.
What is a fractionating column? Esplain Its advantage.
Answer:
In case, the difference in the boiling points of the hquids is less than 25 K, WC USC the fractional distillation method. The apparatus is almost the same as used in distillation. hie only diffcrencc is that a fractionating column is tiLted in the distillation flask and the condenser is attached to che fractionating column. A simple fractionating column is made up of a tube packcd with glass beads. The beads provide the surface for the vapours to cool and condense again and again. The fractionating columns obstruct the smooth upward flow of vapours. For example, a mixture of n-hexane and n – heptane can be separated through the process of fractional distillation.

Experiment:
Put the mixture into a distillation flask. Heat the mixture. The vapours of n-hexane has a lower boiling point, so they pass through and get condensed, n – heptane, which has a higher boiling point, condenses and flows back into the distillation flask.

Analysing & Evaluating Questions

Question 6.
Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’ Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?
Answer:
Part ‘A’ of the mixture on heating gives iron (II) sulphide.
Fe+S FeS Part ‘B’ of the mixture contains only sulphur and iron filings.
On reactions with dii. HCL the gases given by the two samples are:
Part ‘A’: H₂S
(i) Smell of rotten eggs
FeS + 2HCl(aq) → FeCI₂(aq) + H₂S(g)

(ii) Turns lead acetate paper black
H₂S+Pb(CH₃COO)₂ → PbS + 2CH₃COOH
Part ‘B’: H₂
(i) Burns with a blue flame and popping sound.

Activity 1

1. Divide the class into four groups A, B, C and D.
2. Distribute the following samples to each group:
   • Few crystals of copper sulphate to group A.
   • One spatula full of copper sulphate to group B.
   • Chalk powder to group C.
   • Few drops of milk or ink to group D.
3. Each group should add the given sample in water and stir properly using a glass rod. Are the particles in the mixture visible?
4. Direct a beam of light from a torch through the beaker containing the mixture and observe from the front. Is the path of the beam of light visible?
5. Leave the mixture undisturbed for a few minutes and set up the filtration apparatus in the meantime.
6. Is the mixture stable or do the particles begin to settle after sometime?
7. Filter the mixture. Is there any residue on the filter paper?
8. Discuss the results and form an opinion.
   
9. Particles of mixture are visible only in case of group C.
10. The path of beam of light was visible in case of groups C and D.
11. The particles settle down after sometime in case of group C.
12. Residue will be left in case of group C. Groups A and B have got a solution. Group C has got a suspension. Group D has got a colloidal solution.

Activity 2

• Fill a beaker half with water.
• Put a watch glass on the mouth of the beaker containing water as shown in the figure.
• Put a few drops of ink on the watch glass.
• Now start heating the beaker. We do not want to heat the ink directly. You will see that evaporation is taking place from the watch glass.
• Continue heating as the evaporation goes on and stop heating when you do not see any further change on the watch glass.
• Observe carefully and record your observations.
• What do you think has got evaporated from the watch glass?
• Is there a residue on the watch glass?
• What is your interpretation? Is ink a single substance (pure) or is it a mixture?

Observations

• Water and volatile substances get evaporated from the watch glass.
• Residue is obtained on the watch glass.
• Ink is a mixture of a dye in water.

Activity 3

• Take a thin strip of filter paper.
• Draw a line on it using a pencil, approximately 3 cm above the lower edge.
• Put a small drop of ink (water – soluble, that is from a sketch pen or fountain pen) at the centre of the line. Let it dry.
• Lower the filter paper into jar/glass/beaker/ test tube containing water so that the drop of ink on the paper is just above the water level and leave it undisturbed.
• Watch carefully, as the water rises up on the filter paper. Record your observations.
• What do you observe on the filter paper as the water rises on it?
• Do you obtain different colours on the filter paper strip?
• What according to you, can be the reason for the rise of the coloured spot on the paper strip?

Observations

• With water, ink also rises up, though with smaller speed.
• Different colours will be observed on the filter paper.
• Ink is a mixture of different coloured substances. Each substance moves with a different characteristic speed with respect to the speed of water.

Activity 4

• Take some (approximately 5 g) impure sample of copper sulphate in a china dish.
• Dissolve it in minimum amount of water.
• Filter the impurities out.
• Evaporate water from the copper sulphate so as to get a saturated solution.
• Cover the solution with a filter paper and leave it undisturbed at room temperature to cool down slowly, for a day.
• You will obtain the crystals of copper sulphate in the china dish.
• This process is called crystallisation.
• What do you observe in the china dish?
• Do the crystals look alike? How will you separate the crystals from the liquid in the china dish?

Observations

• The crystals of copper sulphate are obtained in the china dish.
• The crystals look alike to some extent.
• Crystals can be separated from the liquid by filtration. When the liquid passes through the funnel, crystals are retained on the filter paper.

Value Based Questions

Question 1.
Purification of drinking water is a big problem. In your locality, vendors of various companies are promoting RO units for purificaiton of water for domestic use irrespective of whether it is required or not. Answer the following questions:
(a) What is the meaning of Ro?
(b) Would you recommend the installation of Ro unit in all the areas? Give reason.
(c) What values are associated with your approach?
Answer:
(a) RO means reverse osmosis.

(b) Installation of RO units for purifying the drinking water in all the areas is not recommended. An RO plant generally removes about 95% of TDS (total dissolved salts) in water that is treated by RO. The desirable level of TDS as per Indian standard is 500 ppm. Before installing RO unit, it is important to check the TDS level of your input water.

(c) Awareness of drinking TDS in potable water supply to the people.

Question 2.
After cleaning the refrigerator thoroughly well, Rohit closed it and kept it switched off for two days. After that, on opening it, he got a foul smell. His neighbourer advised him to keep a piece of charcoal (C) in the refrigerator.
(a) As a student of chemistry, explain why this suggestion is effective?
(b) Name another substance which can be used in place of charcoal.
(c) What values can be drawn from this advice?
Answer:
(a) The charcoal piece absorbs the foul – smelling gases developed in refrigerator.
(b) Coal.
(c) Help your friends and neighbours/ use of scientific knowledge.

JAC Class 9 Science Important Questions

JAC Jharkhand Board Class 9th Social Science Notes in Hindi & English Medium

JAC Board Class 9th Social Science Notes in English Medium

Jharkhand Board Class 9th History Notes

• Chapter 1 The French Revolution Notes
• Chapter 2 Socialism in Europe and the Russian Revolution Notes
• Chapter 3 Nazism and the Rise of Hitler Notes
• Chapter 4 Forest Society and Colonialism Notes
• Chapter 5 Pastoralists in the Modern World Notes

Jharkhand Board Class 9th Geography Notes

• Chapter 1 India-Size and Location Notes
• Chapter 2 Physical Features of India Notes
• Chapter 3 Drainage Notes
• Chapter 4 Climate Notes
• Chapter 5 Natural Vegetation and Wildlife Notes
• Chapter 6 Population Notes

Jharkhand Board Class 9th Civics Notes

• Chapter 1 What is Democracy? Why Democracy? Notes
• Chapter 2 Constitutional Design Notes
• Chapter 3 Electoral Politics Notes
• Chapter 4 Working of Institutions Notes
• Chapter 5 Democratic Rights Notes

Jharkhand Board Class 9th Economics Notes

• Chapter 1 The Story of Village Palampur Notes
• Chapter 2 People as Resource Notes
• Chapter 3 Poverty as a Challenge Notes
• Chapter 4 Food Security in India Notes

JAC Board Class 9th Social Science Solutions in Hindi Medium

JAC Board Class 9th History Notes in Hindi

• Chapter 1 फ्रांसीसी क्रांति Notes
• Chapter 2 यूरोप में समाजवाद एवं रूसी क्रांति Notes
• Chapter 3 नात्सीवाद और हिटलर का उदय Notes
• Chapter 4 वन्य समाज एवं उपनिवेशवाद Notes
• Chapter 5 आधुनिक विश्व में चरवाहे Notes

JAC Board Class 9th Geography Notes in Hindi

• Chapter 1 भारत-आकार वे स्थिति Notes
• Chapter 2 भारत का भौतिक स्वरूप Notes
• Chapter 3 अपवाह Notes
• Chapter 4 जलवायु Notes
• Chapter 5 प्राकृतिक वनस्पति एवं वन्य जीवन Notes
• Chapter 6 जनसंख्या Notes

JAC Board Class 9th Civics Notes in Hindi

• Chapter 1 लोकतंत्र क्या? लोकतंत्र क्यों? Notes
• Chapter 2 संविधान निर्माण Notes
• Chapter 3 चुनावी राजनीति Notes
• Chapter 4 संस्थाओं का कामकाज Notes
• Chapter 5 लोकतांत्रिक अधिकार Notes

JAC Board Class 9th Economics Notes in Hindi

• Chapter 1 पालमपुर गाँव की कहानी Notes
• Chapter 2 संसाधन के रूप में लोग Notes
• Chapter 3 निर्धनता : एक चुनौती Notes
• Chapter 4 भारत में खाद्य सुरक्षा Notes

JAC Jharkhand Board Class 9th Social Science Important Questions in Hindi & English Medium

JAC Board Class 9th Social Science Important Questions in English Medium

Jharkhand Board Class 9th History Important Questions

• Chapter 1 The French Revolution Important Questions
• Chapter 2 Socialism in Europe and the Russian Revolution Important Questions
• Chapter 3 Nazism and the Rise of Hitler Important Questions
• Chapter 4 Forest Society and Colonialism Important Questions
• Chapter 5 Pastoralists in the Modern World Important Questions

Jharkhand Board Class 9th Geography Important Questions

• Chapter 1 India-Size and Location Important Questions
• Chapter 2 Physical Features of India Important Questions
• Chapter 3 Drainage Important Questions
• Chapter 4 Climate Important Questions
• Chapter 5 Natural Vegetation and Wildlife Important Questions
• Chapter 6 Population Important Questions

Jharkhand Board Class 9th Civics Important Questions

• Chapter 1 What is Democracy? Why Democracy? Important Questions
• Chapter 2 Constitutional Design Important Questions
• Chapter 3 Electoral Politics Important Questions
• Chapter 4 Working of Institutions Important Questions
• Chapter 5 Democratic Rights Important Questions

Jharkhand Board Class 9th Economics Important Questions

• Chapter 1 The Story of Village Palampur Important Questions
• Chapter 2 People as Resource Important Questions
• Chapter 3 Poverty as a Challenge Important Questions
• Chapter 4 Food Security in India Important Questions

JAC Board Class 9th Social Science Important Questions in Hindi Medium

JAC Board Class 9th History Important Questions in Hindi

• Chapter 1 फ्रांसीसी क्रांति Important Questions
• Chapter 2 यूरोप में समाजवाद एवं रूसी क्रांति Important Questions
• Chapter 3 नात्सीवाद और हिटलर का उदय Important Questions
• Chapter 4 वन्य समाज एवं उपनिवेशवाद Important Questions
• Chapter 5 आधुनिक विश्व में चरवाहे Important Questions

JAC Board Class 9th Geography Important Questions in Hindi

• Chapter 1 भारत-आकार वे स्थिति Important Questions
• Chapter 2 भारत का भौतिक स्वरूप Important Questions
• Chapter 3 अपवाह Important Questions
• Chapter 4 जलवायु Important Questions
• Chapter 5 प्राकृतिक वनस्पति एवं वन्य जीवन Important Questions
• Chapter 6 जनसंख्या Important Questions

JAC Board Class 9th Civics Important Questions in Hindi

• Chapter 1 लोकतंत्र क्या? लोकतंत्र क्यों? Important Questions
• Chapter 2 संविधान निर्माण Important Questions
• Chapter 3 चुनावी राजनीति Important Questions
• Chapter 4 संस्थाओं का कामकाज Important Questions
• Chapter 5 लोकतांत्रिक अधिकार Important Questions

JAC Board Class 9th Economics Important Questions in Hindi

• Chapter 1 पालमपुर गाँव की कहानी Important Questions
• Chapter 2 संसाधन के रूप में लोग Important Questions
• Chapter 3 निर्धनता : एक चुनौती Important Questions
• Chapter 4 भारत में खाद्य सुरक्षा Important Questions

JAC Jharkhand Board Class 9th Social Science Solutions in Hindi & English Medium

JAC Board Class 9th Social Science Solutions in English Medium

Jharkhand Board Class 9th Social Science History: India and the Contemporary World – I

• Chapter 1 The French Revolution
• Chapter 2 Socialism in Europe and the Russian Revolution
• Chapter 3 Nazism and the Rise of Hitler
• Chapter 4 Forest Society and Colonialism
• Chapter 5 Pastoralists in the Modern World

Jharkhand Board Class 9th Social Science Geography: Contemporary India – I

• Chapter 1 India-Size and Location
• Chapter 2 Physical Features of India
• Chapter 3 Drainage
• Chapter 4 Climate
• Chapter 5 Natural Vegetation and Wildlife
• Chapter 6 Population

Jharkhand Board Class 9th Social Science Civics: Democratic Politics – I

• Chapter 1 What is Democracy? Why Democracy?
• Chapter 2 Constitutional Design
• Chapter 3 Electoral Politics
• Chapter 4 Working of Institutions
• Chapter 5 Democratic Rights

Jharkhand Board Class 9th Social Science Economics

• Chapter 1 The Story of Village Palampur
• Chapter 2 People as Resource
• Chapter 3 Poverty as a Challenge
• Chapter 4 Food Security in India

JAC Board Class 9th Social Science Solutions in Hindi Medium

JAC Board Class 9th Social Science History: India and The Contemporary World – I (इतिहास : भारत और समकालीन विश्व-I)

• Chapter 1 फ्रांसीसी क्रांति
• Chapter 2 यूरोप में समाजवाद एवं रूसी क्रांति
• Chapter 3 नात्सीवाद और हिटलर का उदय
• Chapter 4 वन्य समाज एवं उपनिवेशवाद
• Chapter 5 आधुनिक विश्व में चरवाहे

JAC Board Class 9th Social Science Geography: Contemporary India – I (भूगोल : समकालीन भारत-I)

• Chapter 1 भारत-आकार वे स्थिति
• Chapter 2 भारत का भौतिक स्वरूप
• Chapter 3 अपवाह
• Chapter 4 जलवायु
• Chapter 5 प्राकृतिक वनस्पति एवं वन्य जीवन
• Chapter 6 जनसंख्या

JAC Board Class 9th Social Science Civics: Democratic Politics – I (राजनीति विज्ञान : लोकतांत्रिक राजनीति-I)

• Chapter 1 लोकतंत्र क्या? लोकतंत्र क्यों?
• Chapter 2 संविधान निर्माण
• Chapter 3 चुनावी राजनीति
• Chapter 4 संस्थाओं का कामकाज
• Chapter 5 लोकतांत्रिक अधिकार

JAC Board Class 9th Social Science Economics (अर्थशास्त्र)

• Chapter 1 पालमपुर गाँव की कहानी
• Chapter 2 संसाधन के रूप में लोग
• Chapter 3 निर्धनता : एक चुनौती
• Chapter 4 भारत में खाद्य सुरक्षा