Month: March 2023

Jharkhand Board JAC Class 10 Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.3

Question 1.
Prove that \(\sqrt{5}\) is irrational.
Solution:
Let us assume that \(\sqrt{5}\) is rational.
So, we can find co-prime integers a and b such that \(\sqrt{5}\) = \(\frac{a}{b}\)
Then, squaring on both the sides, we get
5 = \(\frac{a^2}{b^2}\)
∴ a² = 5b²
This suggests that 5 is a factor of a².
Since 5 is a prime number, by theorem 1.3, 5 is also a factor of a.
Let a = 5c, where c is an integer.
∴ a² = 25c²
∴ 25c² = 5b²
∴ b² = 5c²
This suggests that 5 is a factor of b². Since 5 is a prime number, by theorem 1.3, 5 is also a factor of b.
Thus, a and b have a common factor 5.
But, this contradicts the fact that a and b are co-prime.
Hence, our assumption that \(\sqrt{5}\) is rational is incorrect.
So, we conclude that \(\sqrt{5}\) is irrational.

Question 2.
Prove that 3 + 2\(\sqrt{5}\) is irrational.
Solution:
Suppose 3 + 2\(\sqrt{5}\) is rational.
Then, 3 + 2\(\sqrt{5}\) = \(\frac{a}{b}\); where a and b are co-prime integers.
Now, 3 + 2\(\sqrt{5}\) = \(\frac{a}{b}\)
∴ 2\(\sqrt{5}\) = \(\frac{a}{b}\) – 3
∴ 2\(\sqrt{5}\) = \(\frac{a-3 b}{b}\)
∴ \(\sqrt{5}\) = \(\frac{a-3 b}{2 b}\)
As a and b are integers, \(\frac{a-3 b}{2 b}\) is a rational number and so is \(\sqrt{5}\).
This contradicts the fact that \(\sqrt{5}\) is irrational.
Hence, our assumption is incorrect.
So, we conclude that 3 + 2\(\sqrt{5}\) is irrational.

Question 3.
Prove that the following are irrationals:
1. \(\frac{1}{\sqrt{2}}\)
2. 7\(\sqrt{5}\)
3. 6 + \(\sqrt{2}\)
Solution:
1. \(\frac{1}{\sqrt{2}}\)
Suppose \(\frac{1}{\sqrt{2}}\) is rational.
Then, \(\frac{1}{\sqrt{2}}=\frac{a}{b}\) where a and b are co-prime integers.
Squaring on both the sides, we get
\(\frac{1}{2}=\frac{a^2}{b^2}\)
∴ b² = 2a²
Hence, 2 is a factor of b².
Since 2 is a prime number, by theorem 1.3, 2 is also a factor of b.
Let b = 2c, where c is an integer.
∴ b² = 4c²
∴ 4c² = 2a²
∴ a² = 2c²
Hence, 2 is a factor of a².

Since 2 is a prime number, by theorem 1.3, 2 is also a factor of a.
Thus, a and b have a common factor 2. which contradicts our assumption that a and b are co-prime integers.
Hence, our assumption that \(\frac{1}{\sqrt{2}}\) is rational is incorrect.
So, we conclude that \(\frac{1}{\sqrt{2}}\) is irrational.

2. 7\(\sqrt{5}\)
Suppose 7\(\sqrt{5}\) is rational.
Then, 7\(\sqrt{5}\) = \(\frac{a}{b}\) where a and b are co-prime integers.
This gives, \(\sqrt{5}\) = \(\frac{a}{7 b}\)
As a and b are integers, \(\frac{a}{7 b}\) is a rational number and so is \(\sqrt{5}\).
This contradicts the fact that \(\sqrt{5}\) is irrational.
Hence, our assumption is incorrect.
So, we conclude that 7\(\sqrt{5}\) is irrational.

3. 6 + \(\sqrt{2}\)
Suppose 6 + \(\sqrt{2}\) is rational.
Then, 6 + \(\sqrt{2}\) = \(\frac{a}{b}\) where a and b are co-prime integers.
This gives, \(\sqrt{2}\) = \(\frac{a}{b}\) – 6
As a and b are integers, \(\frac{a}{b}\) – 6 is a rational number and so is \(\sqrt{2}\).
This contradicts the fact that \(\sqrt{2}\) is irrational.
Hence, our assumption is incorrect. So, we conclude that 6 + \(\sqrt{2}\) is irrational.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 1 Real Numbers Ex 1.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.2

Question 1.
Express each number as a product of its prime factors:
1. 140
2. 156
3. 3825
4. 5005
5. 7429
Solution:
1. Using factor tree method, we have

Thus, 140 = 2 × 2 × 5 × 7
= 2² × 5 × 7

2. Using factor tree method, we have

Thus, 156 = 2 × 2 × 3 × 13
= 2² × 3 × 13

3. Using factor tree method, we have

Thus, 3825 = 3 × 3 × 5 × 5 × 17
= 3² × 5² × 17

4. Using factor tree method, we have

Thus, 5005 = 5 × 7 × 11 × 13

5. Using factor tree method, we have

Thus, 7429 = 17 × 19 × 23

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers:
1. 26 and 91
2. 510 and 92
3. 336 and 54
Solution:
1. Using factor tree method, we have

∴ 26 = 2 × 13 and 91 = 7 × 13
Then, LCM (26, 91) = 2 × 7 × 13 = 182 and HCF (26, 91) = 13
Now, LCM × HCF = 182 × 13 = 2366 and 26 × 91 = 2366.
Hence, LCM × HCF = product of the two numbers.

2. Using factor tree method, we have

∴ 510 = 2 × 3 × 5 × 17 and
92 = 2 × 2 × 23 = 2² × 23
Then,
LCM (510, 92) = 2² × 3 × 5 × 17 × 23
= 23,460
and HCF (510, 92) = 2
Now, LCM × HCF = 23,460 × 2 = 46,920
and 510 × 92 = 46,920
Hence, LCM × HCF = product of the two numbers.

3. Using factor tree method, we have

∴ 336 = 2 × 2 × 2 × 2 × 3 × 7
= 2⁴ × 3 × 7 and
54 = 2 × 3 × 3 × 3 = 2 × 3³
Then, LCM (336, 54) = 2⁴ × 3³ × 7 = 3024
and HCF (336, 54) = 2 × 3 = 6
Now, LCM × HCF = 3024 × 6 = 18,144 and
336 × 54 = 18,144.
Hence, LCM × HCF = product of the two numbers.

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method:
1. 12, 15 and 21
2. 17, 23 and 29
3. 8, 9 and 25
Solution:
1. Using factor tree method, we have

∴ 12 = 2 × 2 × 3 = 2² × 3, 15 = 3 × 5 and 21 = 3 × 7
Then,
LCM (12, 15, 21) = 2² × 3 × 5 × 7 = 420 and HCF (12, 15, 21) = 3.

2. 17 = 17 × 1, 23 = 23 × 1 and 29 = 29 × 1 as each of the given numbers is a prime.
Then, LCM (17, 23, 29) = 17 × 23 × 29
= 11,339
and HCF (17, 23, 29) = 1.

3. Using factor tree method, we have

∴ 8 = 2 × 2 × 2 = 2³, 9 = 3 × 3 = 3²
and 25 = 5 × 5 = 5²
Then, LCM (8, 9, 25) = 2³ × 3² × 5²
= 1800
and HCF (8, 9, 25) = 1.

Question 4.
Given that HCF (306, 657) = 9 find LCM (306, 657).
Solution:
We know, LCM (a, b) =

Taking a = 306 and b = 657, we get
LCM (306, 657) = \(\frac{306 \times 657}{\mathrm{HCF}(306,657)}\)
= \(\frac{306 \times 657}{9}\)
= 34 × 657
= 22,338
Thus, LCM (306, 657) = 22,338.

Question 5.
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
If a number ends with digit 0, it would be divisible by 5 as well as 2. Hence, any number ending with digit 0, must have 2 and 5 both in its prime factorisation.
Now, 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ for any natural number n. Thus, 6n has only two prime factors 2 and 3. So, the prime factorisation of 6ⁿ does not include 5 and hence 6ⁿ cannot end with digit 0 for any natural number n.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
7 × 11 × 13 + 13 = 13 (7 × 11 + 1)
= 13 (77 + 1)
= 13 (78)
= 13 × 2 × 3 × 13
(∵ 78 = 2 × 3 × 13)
= 2 × 3 × 13²
Thus, 7 × 11 × 13 + 13 can be expressed as a product of primes. Hence, 7 × 11 × 13 + 13 is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 (1008 + 1)
= 5 × 1009
Thus, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 can be expressed as a product of primes. Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Here, the LCM of the timings (in minutes) taken by Sonia and Ravi will answer the question satisfying all the conditions as mentioned in the question.
Now, 12 = 2 × 2 × 3 = 2² × 3 and
18 = 2 × 3 × 3 = 2 × 3².
Then, LCM (12, 18) = 2² × 3² = 36
Hence, after 36 minutes, Sonia and Ravi meet again at the starting point if they both start at the same point and at the same time and go in the same direction.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1

Question 1.
Find the distance between the following pairs of points :
1. (2, 3), (4, 1)
2. (- 5, 7), (1, 3)
3. (a, b), (- a, – b)
Answer:
1. Let A (2, 3) and B (4, 1) be the given points. Then,
AB = \(\sqrt{(2-4)^2+(3-1)^2}\)
= \(\sqrt{4+4}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
Thus, the distance between the given points is 2\(\sqrt{2}\).

2. Let A (- 5, 7) and B (- 1, 3) be the given points. Then,
AB = \(\sqrt{(-5+1)^2+(7-3)^2}\)
= \(\sqrt{16+16}\) = \(\sqrt{32}\) = 4\(\sqrt{2}\)
Thus, the distance between the given points is 4\(\sqrt{2}\).

3. Let P (a, b) and Q(- a, – b) be the given points. Then,
PQ = \(\sqrt{(a+a)^2+(b+b)^2}\)
= \(\sqrt{4 a^2+4 b^2}\)
= 2\(\sqrt{a^2+b^2}\)
Thus, the distance between the given points is 2\(\sqrt{a^2+b^2}\).

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in section 7.2.
Answer:
Let A (0, 0) and B (36, 15) be the given points.
Then,
AB = \(\sqrt{(0-36)^2+(0-15)^2}\)
= \(\sqrt{1296+225}\) = \(\sqrt{1521}\) = 39
Thus, the distance between the given points is 39.
Yes, now we can find the distance between the two towns A and B discussed in section 7.2 in the textbook.

Town B is located 36 km east and 15 km north of the town B. So, if we take town A to be situated at the origin with coordinates (0, 0) then the coordinates of town B become (36, 15). Now, as calculated above, the distance between A(0, 0) and B (36, 15) is 39. Hence, the distance between town A and town B is 39 km.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Answer:
Let A(1, 5), B(2, 3) and C (- 2, – 11) be the given points. Then,

Now, 14.56 + 2.24 = 16.80 ≠ 16.28
Thus, AB + BC ≠ AC
Moreover, BC + AC ≠ AB and AC + AB ≠ BC are obvious.
Hence, the given points are not collinear.

Question 4.
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Answer:
Let the given points A (5, – 2), B (6, 4) and C(7, – 2) be the vertices of ΔABC. Then,

Here, in ΔABC, AB = BC ≠ AC.
Hence, ABC is an isosceles triangle.
Thus, the given points (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in the given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Answer:
Here A(3, 4), B(6, 7), C (9, 4) and D(6, 1) give the position of points where those four friends are seated. Then,

Thus, in quadrilateral ABCD,
AB = BC = CD = DA and AC = BD.
In other words, all the sides of quadrilateral ABCD are equal and its diagonals are also equal.
Hence, ABCD is a square.
So. Champa is correct stating that ABCD is a square.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
1. (- 1, – 2), (1, 0), (1, 2), (- 3, 0)
2. (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
3. (4, 5), (7, 6), (4, 3), (1, 2)
Answer:
1. If possible, let A (- 1, – 2), B(1, 0), C(- 1, 2) and D(- 3, 0) be the vertices of quadrilateral ABCD. Then,

Thus, in quadrilateral ABCD,
AB = BC = CD = DA and AC = BD.
Hence, ABCD is a square.

2. If possible, let A (- 3, 5), B(3, 1), C(0, 3) and D(- 1, – 4) be the vertices of quadrilateral ABCD. Then,

Hence, A, B and C are collinear points in which C lies between A and B.
Hence, A, B, C and D do not form a quadrilateral.

3. If possible, let A (4, 5), B(7, 6), C(4, 3) and D (1, 2) be the vertices of quadrilateral ABCD. Then,


Thus, in quadrilateral ABCD, AB = CD, BC = DA, i.e., both the pairs of opposite sides are equal, but AC ≠ BD. i.e.. diagonals are not equal.
Hence, ABCD is a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2,-5) and (-2, 9).
Answer:
Here, A (2, – 5) and B(- 2, 9) are two given points.
Coordinates of any point of the x-axis are of the form (x, 0).
Let P(x, 0) be the required point on the x-axis which is equidistant from A and B.
So, PA = PB
∴ PA² = PB²
∴ (x – 2)² + (0 + 5)² = (x + 2)² +(0 – 9)²
∴ x² – 4x + 4 + 25 = x² + 4x + 4 + 81
∴ -8x=56
∴ x = – 7
Thus, the required point on the x-axis which is equidistant from (2, – 5) and (- 2, 9) is (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P (2, – 3) and Q(10, y) is 10 units.
Answer:
The distance between P(2, – 3) and 9(10, y) is given to be 10.
∴ PQ = 10
∴ PQ² = 100
∴ (2 – 10)² + (- 3 – y)² = 100
∴ 64 + 9 + 6y + y² = 100
∴ y² + 6y – 27 = 0
∴ (y + 9) (y – 3) = 0
∴ y + 9 = 0 or y – 3 = 0
∴ y = – 9 or y = 3
Thus, the required values of y are – 9 and 3.

Question 9.
If Q(0, 1) is equidistant from P (5, – 3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Answer:
Q(0, 1) is equidistant from P(5, -3) and R (x, 6).
∴ PQ = RQ
∴ PQ² = RQ²
∴ (5 – 0)² + (- 3 – 1)² = (x – 0)² + (6 – 1)²
∴ 25 + 16 = x² + 25
∴ x² = 16
∴ x = ± 4
∴ x = 4 or x = – 4

Thus, x = ± 4, QR = \(\sqrt{41}\) and PR = \(\sqrt{82}\) or 9\(\sqrt{2}\).

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Answer:
Here, point P(x, y) is given to be equidistant from points A (3, 6) and B (- 3, 4).
∴ PA = PB
∴ PA² = PB²
∴ (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
∴ x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 + y² – 8y + 16
∴ – 12x – 4y + 20 = 0
∴ 3x + y – 5 = 0 (Dividing by – 4)
Thus, 3x + y – 5 = 0 is the required relation between x and y.

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 7 निर्देशांक ज्यामिति Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 7 निर्देशांक ज्यामिति

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
बिन्दु (- 5, 4) की x – अक्ष से दूरी लिखिये ।
हल :
बिन्दु (- 5, 4) की x-अक्ष से दूरी = 4 इकाई ।

प्रश्न 2.
बिन्दु (3, – 2) की -अक्ष से दूरी लिखिए।
हल :
बिन्दु (3, – 2) की y-अक्ष से दूरी 3 इकाई है।

प्रश्न 3.
k के मान ज्ञात कीजिए जिनसे (1, – 1), (- 4, 2k) तथा (- k, – 5) शीर्षों वाले त्रिभुज का क्षेत्रफल 24 वर्ग इकाई हो ।
हल :
माना A(1, – 1), B (- 4, 2k) तथा (- k, – 5) दिये गये ΔABC के शीर्ष हैं।
ΔABC का क्षेत्रफल
= \(\frac{1}{2}\)[1(2k + 5) + (- 4) (- 5 + 1) + (-k) (- 1 – 2k)]
= \(\frac{1}{2}\)[2k + 5 + 16 + k + 2k²]
= \(\frac{1}{2}\)[2k² + 3k + 21] वर्ग इकाई
दिया है
ΔABC का क्षेत्रफल = 24 वर्ग इकाई
∴ \(\frac{1}{2}\)[2k² + 3k + 21] = 24
2k² + 3k + 21 = 48
2k² + 3k – 27 = 0
2k² + 9k bk – 27 = 0
k(2k + 9) – 3(2k + 9) (k – 3) (2k + 9) =0
तथा k = 3
अतः k के मान 3 और हैं।

प्रश्न 4.
x- अक्ष पर वह बिन्दु ज्ञात कीजिए जो बिन्दुओं 4 (6, 5) और B (- 4, 5) से समदूरस्थ है।
हल :
x- अक्ष पर किसी बिन्दु P के निर्देशांक (x, 0) हैं।
∴ बिन्दु P(x, 0) की बिन्दुओं A(6, 5) और B(- 4, 5) से दूरी समान होगी।
PA = PB
⇒ PA² = PB²
⇒ (x – 6)² + (0 – 5)² = (x + 4)² + (0 – 5)²
⇒ x² – 12x + 36 + 25 = x² + 8x + 16 + 25
⇒ – 12x + 36 = 8x + 16
⇒ – 12x – 8x = 16 – 36
⇒ – 20x = – 20
⇒ x = \(\frac{-20}{-20}\) = 1
अतः अभीष्ट बिन्दु के निर्देशांक = (1, 0).

प्रश्न 5.
यदि M(4, 5), रेखाखण्ड AB का मध्य बिन्दु है तथा 4 का निर्देशांक (3, 4) है, तो बिन्दु B के निर्देशांक ज्ञात कीजिए।
हल :
माना बिन्दु B के निर्देशांक (x, y) हैं।
दिया है (3, 4) तथा (x, y) के मध्य बिन्दु के निर्देशांक (4, 5) है।
4 = \(\frac{3+x}{2}\) ⇒ 3 + x = 8
x = 8 – 3 = 5
तथा 5 = \(\frac{y+4}{2}\) ⇒ 10 = y + 4
y = 10 – 4 = 6
अत: B के निर्देशांक B(5, 6) है।

प्रश्न 6.
यदि बिन्दु 4(x, y), B (- 5, 7) तथा C (- 4, 5) सरेखीय हों तो x तथा में सम्बन्ध ज्ञात कीजिए।
हल :
माना कि दिया गया बिन्दु (x, y), B(- 5, 7) और C(- 4, 5) है।
यहाँ x₁ = x, x₂ = – 5, x₃ = – 4
y₁ = y, y₂ = 7, y₃ = 5
तीन बिन्दु सरेखी होने के लिए ΔABC का क्षेत्रफल 0 होता है।
∴ \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac{1}{2}\)[x(7 – 5) + (-5)(5 – y) + (-4) (y – 7)] = 0
⇒ \(\frac{1}{2}\)[2x – 5(5 – y) – 4(y – 7)] = 0
⇒ 2x – 25 + 5y – 4y + 28 = 0
⇒ 2x + y + 3 = 0
अत: 2x + y + 3 = 0 अभीष्ट सम्बन्ध हैं।

प्रश्न 7.
सिद्ध कीजिए कि बिन्दु A(2, – 1), B(3, 4), C (- 2, 3) तथा D(- 3, – 2) एक समचतुर्भुज ABCD के शीर्ष बिन्दु हैं। क्या ABCD एक वर्ग है?
हल :

= \(\sqrt{(-4)^2+(4)^2}\)
= \(\sqrt{16+16}\) = 4\(\sqrt{2}\)
∴ AB = BC = CD = AD
तथा विकर्ण AC ≠ विकर्ण BD
अत: ABCD वर्ग नहीं एक समचतुर्भुज है।
यही सिद्ध करना था ।

प्रश्न 8.
यदि बिन्दु A(1, – 2), B(2, 3), C(a, 2) और D (- 4, – 3) एक समान्तर चतुर्भुज बनाते हैं। a का मान तथा AB को आधार मानते हुए चतुर्भुज की ऊँचाई ज्ञात कीजिए।
हल :
बिन्दु 4 (1 , 2), B(2, 3), C(a, 2) तथा D(- 4, – 3) दिए गए समान्तर चतुर्भुज के शीर्ष हैं।
∵ समान्तर चतुर्भुज के विकर्ण परस्पर समद्विभाजित करते हैं।

∴ विकर्ण AC के मध्य बिन्दु O के निर्देशांक विकर्ण BD के मध्य बिन्दु O के निर्देशांक

अत: बिन्दु C के निर्देशांक = (- 3, 2)
आधार AB पर लम्ब DM खींचा। त्रिभुज ABD का क्षेत्रफल
= \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[1(3 – 3) + 2(3 + 2) + (- 4) (- 2 – 3)]
= \(\frac{1}{2}\)[0 + 10 + 20]
= \(\frac{1}{2}\) × 30
= 15 वर्ग इकाई
आधार (AB) की लम्बाई

प्रश्न 9.
बिन्दु A (4, 7), B (P, 3) तथा C (7, 3) एक समकोण त्रिभुज के शीर्ष हैं जिसमें B पर समकोण है। P का मान ज्ञात करो ।
हल :
समकोण ΔABC में,
AB² + BC² = AC² (पाइथागोरस प्रमेय से)

(3 – 7)² + (P – 4)² + (3 – 3)² + (7 – P)² = (3 – 7)² + (7 – 4)²
⇒ (P – 4)² + (7 – P)² = 9
⇒ P² + 16 – 8P + 49 + p² – 14P = 9
⇒ 2P² – 22P + 56 = 0
⇒ P² – 11P + 28 = 0
⇒ P² – 4P – 7P + 28 = 0
= P(P – 4) – 7(P – 4) = 0
= (P – 4) (P – 7) = 0
P = 4, 7
अत: P का मान 4, 7 है।

प्रश्न 10.
यदि बिन्दु 4 (6, 1), B(8, 2), C(9, 4) और D(x, y) क्रम में एक समान्तर चतुर्भुज के शीर्ष हैं, तो बिन्दु D(x, y) ज्ञात कीजिए।
हल :
बिन्दु 4 (6, 1), B(8, 2), C(9, 4) तथा D(x, y) दिए गए समान्तर चतुर्भुज के शीर्ष हैं।
∵ समान्तर चतुर्भुज के विकर्ण परस्पर समद्विभाजित करते हैं।

∴ विकर्ण AC के मध्य विन्दु के निर्देशांक = विकर्ण BD के मध्य बिन्दु के निर्देशांक

⇒ 8 + x = 15 और 2 + y = 5
⇒ x = 15 – 8 और y = 5 – 2
⇒ x = 7 और y = 3
अतः बिन्दु D के निर्देशांक = (7, 3).

प्रश्न 11.
बिन्दुओं P(- 3, 4) और Q(4, 5) को जोड़ने वाले रेखाखण्ड को समत्रिभाजित करने वाले बिन्दुओं के निर्देशांक ज्ञात कीजिए।
हल :
माना कि A(x₁, y₁) और B(x₂, y₂) अभीष्ट बिन्दु हैं जो बिन्दुओं P(- 3, 4) और Q( 4, 5) को जोड़ने वाले रेखा खण्ड को समत्रिभाजित करते हैं।

माना कि
PA = AB = QB = x
AQ = x + x = 2x
PB = x + x = 2x
\(\frac{PA}{AQ}=\frac{x}{2x}\) = \(\frac{1}{2}\) = 1 : 2
\(\frac{PB}{BQ}=\frac{2x}{x}\) = \(\frac{2}{1}\) = 2 : 1
अर्थात् बिन्दु A, PQ को 1 : 2 के अनुपात में तथा बिन्दु B, PQ को 2 : 1 के अनुपात में विभाजित करती है।
बिन्दु A के लिए:

अतः A और B के निर्देशांक क्रमश (\(\frac{5}{3}\), \(\frac{14}{3}\)) तथा (\(\frac{-2}{3}\), \(\frac{13}{3}\)) है।
यही सिद्ध करना था ।

प्रश्न 12.
यदि K(5, 4) रेखाखण्ड PQ का मध्यबिन्दु है तथा Q के निर्देशांक (2, 3) है तो P के निर्देशांक ज्ञात कीजिए ।
हल :
माना बिन्दु P के निर्देशांक (x, y) है।

तब बिन्दु P(x, y) तथा Q(2, 3) के मध्यबिन्दु K के निर्देशांक (5, 4).
5 = \(\frac{x+2}{2}\)
⇒ 10 = x + 2
⇒ x = 10 – 2 = 8
और 4 = \(\frac{y+3}{2}\)
⇒ 8 = y + 3
⇒ y = 8 – 3 = 5
P के निर्देशांक = (8, 5).

प्रश्न 13.
उस त्रिभुज का क्षेत्रफल ज्ञात कीजिए जिसके शीर्ष (- 3, – 2), (5, 2) और (5, 4) हैं। यह भी सिद्ध कीजिए कि यह समकोण त्रिभुज हैं।
हल :
माना ABC एक त्रिभुज है जिसके शीर्ष A(- 3, – 2), B (5, – 2) तथा C(5, 4) हैं।
ΔABC का क्षेत्रफल
= \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[- 3(- 2 – 4) + 5(4 + 2) + 5(- 2 + 2)]


∵ AB² + BC² = (8)² + (6)² = 100
= (10)² = CA²
⇒ AB² + BC² = CA²
अत: ΔABC एक समकोण त्रिभुज है।

प्रश्न 14.
त्रिभुज ABC, जिसमें A(1, – 4) तथा A से जाने वाली भुजाओं के मध्य बिन्दु (2, – 1) तथा (0, – 1) है, का क्षेत्रफल ज्ञात कीजिए।
हल
माना ABC एक त्रिभुज है, जहाँ B (x, y) तथा C(z, 1) है।
दिया है, P, AB का मध्य-बिन्दु है।
(2, – 1) = (\(\frac{1+x}{2}\), \(\frac{-4+y}{2}\))

∴ C के निर्देशांक = (- 1, 2)
∴ ΔABC के शीषों के निर्देशांक A(1, – 4), B(3, 2) तथा C(1, – 2) हैं।
ΔABC का क्षेत्रफल
= \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[1(2 – 2) + 3(2 + 4) – 1(- 4 – 2)]
= \(\frac{1}{2}\)[1(0) + 3(6) – 1(-6)]
= \(\frac{1}{2}\)[18 + 6] = 12 वर्ग इकाई

प्रश्न 15.
x – 3y = 0 बिन्दुओं (-2, -5) तथा (6, 3) को जोड़ने वाले रेखाखंड को किस अनुपात में विभाजित करती है? इस प्रतिच्छेद बिन्दु के निर्देशांक भी ज्ञात कीजिए।
हल :
माना रेखा x – 3y = 0, बिन्दुओं (- 2, – 5) तथा (6, 3) को जोड़ने वाले रेखाखंड को k : 1 अनुपात में विभाजित करती है।
यहाँ x₁ = – 2, x₂ = 6, y₁ = – 5, y₂ = 3, m = k तथा n = 1
∴ विभाजन सूत्र से,

अतः विभाजन बिन्दु (\(\frac{9}{2}\), \(\frac{3}{2}\)) तथा अभीष्ट अनुपात (\(\frac{13}{3}\) : 1) अथवा (13 : 3) है।

प्रश्न 16.
बिन्दु A, बिन्दुओं X(6, – 6) तथा Y(- 4, – 1) को मिलाने वाले रैखाखण्ड XY इस प्रकार स्थित है कि \(\frac{XA}{XY}=\frac{2}{5}\) है। यदि बिन्दु A रेखा 3x + k(y + 1) = 0 पर भी स्थित है, तो k का मान ज्ञात कीजिए।
हल :
दिया है, \(\frac{XA}{XY}=\frac{2}{5}\)

\(\frac{X A}{X A+A Y}\) = \(\frac{2}{5}\)
5XA = 2XA + 2AY
5XA = 2AY
\(\frac{XA}{XY}=\frac{2}{3}\)
अत: बिन्दु A रेखाखण्ड XY को 2 : 3 अनुपात में विभाजित करता है।
यहाँ m₁ = 2, m₂ = 3
अतः A बिन्दु के निर्देशांक

इसलिए A बिन्दु के निर्देशांक (2, – 4) है।
चूँकि बिन्दु A, रेखा 3x + k(y + 1) = 0 पर भी स्थित है, अतः यह रेखा के समीकरण की संतुष्ट करेगा।
∴ x = 2 तथा y = – 4 रखने पर,
3(2) + k (- 4 + 1) = 0
6 + k (- 3) = 0
– 3k = – 6
k = 2

प्रश्न 17.
यदि A(- 2, 1), B(a, 0), C(4, b) तथा D(1, 2) एक समानान्तर चतुर्भुज ABCD एके शीर्ष बिन्दु हैं, तो a तथा b के मान ज्ञात कीजिए। अतः इस चतुर्भुज की भुजाओं की लम्बाइयाँ ज्ञात कीजिए।
हल :
ज्ञात है, ABCD एक समानान्तर चतुर्भुज है।

विकर्ण AC तथा BD एक-दूसरे को बिन्दु O पर समद्विभाजित करते हैं।


अत: चतुर्भुज ABCD की भुजाओं की लम्बाई
AB = BC = CD = DA = \(\sqrt{10}\) इकाई

प्रश्न 18.
यदि बिन्दु A(k + 1, 2k), B(3k, 2k + 3) तथा C(5k – 1, 5k) सरेख हों, तो k का मान ज्ञात कीजिए।
हल :
बिन्दु A(k + 1, 2k), B (3k, 2k +3) और C(5k – 1, 5k) सरेख हैं। अतः त्रिभुज ABC का क्षेत्रफल = 0
Δ = \(\frac{1}{2}\)[(k + 1)(2k + 3) – 6k² + 15k² – (5k – 1)(2k + 3) + 2k(5k – 1) – (k + 1)(5k)]
0 = \(\frac{1}{2}\)[2k² + 5k + 3 – 6k² + 15k² – 10k² – 13k + 3 + 10k² – 2k – 5k² – 5k]
0 = \(\frac{1}{2}\)[6k² – 15k + 6]
⇒ 6k² – 15k + 6 = 0
⇒ 6k² – 12k – 3k + 6 = 0
⇒ 6k(k – 2) – 3 (k – 2) = 0
⇒ (k – 2) (6k – 3) = 0
⇒ k = 2 या k = \(\frac{1}{2}\)

प्रश्न 19.
दर्शाइए कि ΔABC जहाँ A(- 2, 0), B(2, 0) C(0, 2) तथा ΔPQR जहाँ P(- 4, 0), Q( 4, 0), R (0, 4) हैं, समरूप त्रिभुज है।
हल :
त्रिभुजों के शीर्षों के निर्देशांक है,
A (- 2, 0), B (2, 0), C(0, 2)
P(- 4, 0), Q(4, 0), R(0, 4)

यहाँ ΔPQR की भुजायें ΔABC को भुजाओं की दोगुनी हैं।
अत: दोनों त्रिभुज समरूप है। इति सिद्धम्

प्रश्न 20.
एक त्रिभुज का क्षेत्रफल 5 वर्ग इकाई है। इसके दो शीर्ष (2, 1) तथा (3, – 2) हैं। यदि तीसरा शीर्ष (\(\frac{7}{2}\), y), तो y का मान ज्ञात कीजिए।
हल :
दिया है, 4(2, 1), B(3, – 2) और C(\(\frac{7}{2}\), y)
अब, ΔMBC का क्षेत्रफल

प्रश्न 21.
निम्न आकृति में किसी कक्षा में रखे डेस्कों की व्यवस्था दर्शाइए गई है। आशिमा, भारती तथा आशा क्रमशः बिंदुओं A, B तथा C पर बैठी हैं। निम्न प्रश्नों के उत्तर दीजिए ।
(i) ज्ञात कीजिए कि क्या तीनों लड़कियाँ एक ही रेखा में बैठी हैं।
(ii) यदि A, B तथा C सरेख हैं तो ज्ञात कीजिए कि बिंदु B रेखाखंड AC को किस अनुपात में विभाजित करता है।

हल :
आलेख से बिंदु A, B और C के निर्देशांक हैं :
A = (3, 1), B (6, 4) तथा C = (8, 6)
A, B, C के सरेख के लिए प्रतिबंध
x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂) = 0
∴ x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂)
= 3(4 – 6) + 6(6 – 1) + 8(1 – 4)
= 3(-2) + 6(5) + 8(-3)
= – 6+ 30 – 24
= 0
∴ A, B और C सरेख है।
अतः तीनों लड़कियाँ एक ही रेखा में बैठी है।

(ii) माना B, रेखाखंड AC को m : n अनुपात में विभाजित करता है।

⇒ 8m + 3n = 6m + 6n
⇒ 2m = 3n
⇒ \(\frac{m}{n}=\frac{3}{2}\)
⇒ m : n = 3 : 2
अत: B, AC को 3 : 2 के अनुपात में विभाजित करता है।

प्रश्न 22.
कृष्णा के पास एक सेवों का बाग है जिसके साथ एक 10 मी × 10 मी साइज का एक किचन गार्डन है। उसने उसे एक 10 × 10 ग्रिड के बाँटकर उसमें मिट्टी तथा खाद डाली है। उसने बिंदु पर एक नींबू का पौधा, बिंदु B पर धनिए का पौधा, बिंदु C पर प्याज का पौधा तथा बिंदु D पर एक टमाटर का पौधा लगाया है। उसका पति राम किचन गार्डन को देखकर तारीफ़ करता है तथा कहलाता है A, B, C तथा D को मिलाने पर शायद एक समांतर चतुर्भुज बन जाएं। नीचे दिए गए चित्र को ध्यानपूर्वक देखकर निम्नलिखित के उत्तर दीजिए :

(i) निर्देशांक अक्ष के रूप में 10 × 10 ग्रिड का उपयोग करते हुए बिंदुओं A, B, C तथा D के निर्देशांक ज्ञात कीजिए।
(ii) ज्ञात कीजिए कि क्या ABCD एक समांतर चतुर्भुज है या नहीं।
हल :
(i) चित्र से, A, B, C तथा D के निर्देशांक हैं :
A = (2, 2), B = (5, 4), C = (7, 7), D = (4, 5)

यहाँ AB = BC = CD = DA
लेकिन AC ≠ BD
अत: ABCD एक समांतर चतुर्भुज है।

प्रश्न 23.
यदि A (- 5, 7), B (- 4, – 5), C(- 1, – 6) तथा D(4, 5) एक समानान्तर चतुर्भुज ABCD के शीर्ष बिन्दु हैं, तो चतुर्भुज ABCD का क्षेत्रफल कीजिए ।
हल :
चतुर्भुज ABCD का एक विकर्ण BD को मिलाया।

ΔABD के लिए शीर्ष A(- 5, 7), B(-4, -5), D(4, 5)
ΔABD का क्षेत्रफल
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₃)]
= \(\frac{1}{2}\)[- 5(- 5 – 5) + (- 4)(5 – 7) + 4(7 + 5)]
= \(\frac{1}{2}\)[-5(-10) – 4(-2) + 4(12)]
= \(\frac{1}{2}\)[50 + 8 + 48]
= \(\frac{1}{2}\)[106]
= \(\frac{106}{2}\) = 53 वर्ग इकाई
ΔBCD के लिए शीर्ष
B(- 4, – 5), C(-1, – 6), D(4, 5)
ΔBCD का क्षेत्रफल
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₃)]
= \(\frac{1}{2}\) [-4(- 6 – 5) + (-1)(5 + 5) + 4(-5 + 6)]
= \(\frac{1}{2}\) [-4(- 11) – 1(10) + 4(1)]
= \(\frac{1}{2}\) [44 – 10 + 4]
= \(\frac{1}{2}\) [38] = \(\frac{38}{2}\)
= 19 वर्ग इकाई
चतुर्भुज ABCD का क्षेत्रफल = 53 + 19
= 72 वर्ग इकाई

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क)

1. x निर्देशांक को ……….. भी कहते हैं।
2. y- निर्देशांक को ………….. भी कहते हैं।
3. AOBC एक आयत है जिसके तीन शीर्ष- बिंदु A(0, – 3), O(0, 0) एवं B (4, 0) हैं इसके विकर्ण की लंबाई …………… है।
4. बिन्दुओं (- 3, – 3) तथा (- 3, 3) को जोड़ने वाले रेखाखंड का मध्य बिंदु ………………. है।
5. बिंदुओं (- a, a) तथा (- a, – a) के बीच की दूरी ………………. है।

हल :

1. भुज,
2. कोटि,
3. 5 इकाई,
4. (- 3, 0),
5. 2a इकाई ।

निम्न में सत्य / असत्य ज्ञात कीजिए :

प्रश्न (ख)

1. यदि बिंदु (3, – 6) बिंदुओं (0, 0) तथा (x, y) को जोड़ने वाले रेखाखंड का मध्य-बिंदु है, तो बिन्दु (x, y) का मान (6, – 12) होगा।
2. x – अक्ष पर स्थित वह बिंदु जो (2, 3) तथा (6, – 9) को जोड़ने वाले रेखाखंड को 1 : 3 के अनुपात में विभाजित करता है, के निर्देशांक (0, 3) है।
3. यदि बिंदुओं A(-3, b) तथा B (1, b + 4) को मिलाने वाले रेखाखंड का मध्यबिंदु P(- 1, 1) है, तो b का मान 2 है।
4. यदि एक वृत्त का केंद्र (3, 5) है तथा एक व्यास के अंत बिंदु (4, 7) तथा (2, y) हैं, तो y का मान 3 है।
5. बिंदुओं A(2, – 3) तथा B (5, 6) को मिलाने वाले रेखाखंड x-अक्ष को 1 : 2 के अनुपात में बाँटता है।

हल :

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

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
बिंदुओं (a cos θ + b sin θ, 0) तथा (0, a sin θ – b sin θ) के बीच की दूरी है
(A) a² + b²
(B) a² – b²
(C) \(\sqrt{a^2+b^2}\)
(D) \(\sqrt{a^2-b^2}\)
हल :

अत: सही विकल्प (C) है।

प्रश्न 2.
यदि बिंदु P(k, 0), बिंदुओं A(2, – 2) तथा B(- 7, 4) को मिलाने वाले रेखाखंड को 1 : 2 के अनुपात में विभाजित करता है, तो k का मान है:
(A) 1
(B) 2
(C) – 2
(D) 1
हल :
बिंदु P के निर्देशांक

⇒ P (k, 0) = (- 1, 0)
∴ k = – 1
अत: सही विकल्प (D) है।

प्रश्न 3.
p का वह मान जिसके लिए बिंदु 4 (3, 1), B (5, p) तथा C(7, – 5) सरेख हैं, है :
(A) – 2
(B) 2
(C) – 1
(D) 1
हल :
बिंदु A(3, 1), B(5, p) तथा C(7, – 5) सरेख होंगे यदि,
x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
⇒ 3 (p + 5) + 5 (- 5 – 1) + 7 (1 – p) = 0
⇒ 3p + 15 – 30 + 7 – 7p = 0
⇒ – 4p – 8 = 0
⇒ – 8 = 4p
⇒ p = \(\frac{-8}{4}\) = – 2
अत: सही विकल्प (A) हैं।

प्रश्न 4.
x- अक्ष पर स्थित बिंदु P जो बिंदुओं B(5, 0) से समदूरस्थ है, हैं
(A) (2, 0)
(B) (0, 2)
(C) (3, 0)
(D) (2, 2)
हल :
माना x- अक्ष पद P(x, 0) कोई बिंदु है।
दिया है, PA = PB
⇒ PA² = PB²
(दोनों पक्षों का वर्ग करने पर)
⇒ (x + 1)² + (0 – 0)² = (x + 5)² + (0 – 0)²
⇒ x² + 1 + 2x = x² + 25 – 10x
⇒ 12x = 24
⇒ x = 2
∴ बिंदु (2, 0)
अत: सही विकल्प (A) हैं।

प्रश्न 5.
उस बिंदु के निर्देशांक जो बिंदु (- 3, 5) का x- अक्ष में प्रक्षेप है, है:
(A) (3, 5)
(B) (3, – 5)
(C) (- 3, – 5)
(D) (- 3, 5)
हल :
P(-3, 5) का प्रेक्षप बिंदु P'(- 3, – 5) है ।

अत: सही विकल्प (C) है।

प्रश्न 6.
यदि बिंदुओं A(10, – 6) तथा B (k, 4) को मिलाने वाले रेखाखंड का मध्य-बिन्दु (a, b) है, तथा a – 2b = 18 है, तो k का मान है:
(A) 30
(B) 22
(C) 4
(D) 40
हल :
प्रश्नानुसार,

प्रश्न 7.
उस त्रिभुज जिसके शीर्ष बिंदु (0, 4), (0, 0) तथा (3, 0) है का परिमाप है।
(A) 7 + \(\sqrt{5}\)
(B) 5
(C) 10
(D) 12
हल :
माना ABC एक त्रिभुज है।

त्रिभुज ABC की परिमाप = AB + BC + CA
= 4 + 3 + 5 = 12 इकाई ।
अत: सही विकल्प (D) है।

प्रश्न 8.
बिंदु P(3, 4) की x-अक्ष से दूरी है :
(A) 3 इकाई
(C) 5 इकाई
(B) 4 इकाई
(D) 1 इकाई
हल :
बिंदु P(3, 4) की x अक्ष से दूरी इस बिंदु के y-निर्देशांक के बराबर होती है।
अत: सही विकल्प (B) है।

प्रश्न 9.
यदि बिंदुओं A(4, p) तथा B (1, 0) के बीच की दूरी 5 इकाई है, तो का मान है :
(A) केवल 4
(B) केवल – 4
(c) ± 4
(D) 0
हल :
∵ AB = 5
⇒ \(\sqrt{(4-1)^2+(p-0)^2}\) = 5
⇒ \(\sqrt{3^2+p^2}\) = 5
⇒ 3² + p² = 25
(दोनों पक्षों का वर्ग करने पर)
⇒ p² = 25 – 9
⇒ p² = 16
⇒ p = ± 4
अत: सही विकल्प (C) है।

प्रश्न 10.
आयत AOBC के तीन शीर्ष A(0, 3), O(0, 0) तथा C(5, 0) है। इसके विकर्ण की लम्बाई है:
(A) 5
(B) 3
(C) \(\sqrt{34}\)
(D) 6
हल :
आयत AOBC के तीन शीर्ष 4 (0, 3), O(0, 0) तथा C(5, 0) है।
आयत के विकर्ण (AC) की लम्बाई
= बिन्दु A(0, 3) तथा C(5, 0) के बीच की दूरी (AC)

अत: विकल्प (C) सही है।

प्रश्न 11.
यदि (a, b + c), (b, c + a) और (c, a + b) त्रिभुज के शीर्ष बिन्दु हैं तो त्रिभुज का क्षेत्रफल है :
(A) (a + b + c)²
(B) 0
(C) a + b + c
(D) abc
हल :
(a, b + c), (b, c + a) और (c, a + b) दिए गए त्रिभुज के शीर्ष बिन्दु हैं।
यहाँ
x₁ = a,
x₂ = b,
x₃ = c
y₁ = b + c
y₂ = c + a
y₃ = a + b
त्रिभुज का क्षेत्रफल
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂))]
= \(\frac{1}{2}\)[a(c + a – a – b) + b(a + b – b – c) + c(b + c – c – a)]
= \(\frac{1}{2}\)[a(c – b) + b(a – c) + c(b – a)]
= \(\frac{1}{2}\)[ac – ab + ab – bc + bc – ac]
= \(\frac{1}{2}\) × 0 = 0
अत: विकल्प (B) सही है।

प्रश्न 12.
यदि एक चतुर्भुज के शीर्ष (1, 4), (- 5, 4), (- 5, – 3) और (1, – 3) हों, तो चतुर्भुज का प्रकार है:
(A) वर्ग
(B) आयत
(C) समान्तर चतुर्भुज
(D) समचतुर्भुज
हल :
माना कि चतुर्भुज के शीर्ष 4(1, 4), B(- 5, 4) तथा D(1, – 3) है तब

अतः AB = CD और BC = DA और
विकर्ण AC = विकर्ण BD

अत: दिये गये बिन्दु आयत के शीर्ष हैं।
अत: सही विकल्प (B) है।

प्रश्न 13.
यदि बिन्दु (1, 2), O(0, 0) तथा C (a, b) संरेखी हैं, तब :
(A) a = b
(B) a = 2b
(C) 2a = b
(D) a = – b
हल :
यदि दिए गए बिन्दु A(1, 2), O(0, 0) तथा C(a, b) सरेखी हैं, तो इनसे बने त्रिभुज का क्षेत्रफल शून्य होगा ।
त्रिभुज ABC का क्षेत्रफल = 0
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac{1}{2}\)[1(0 – b) + 0(b – 2) + a(2 – 0)] = 0
⇒ \(\frac{1}{2}\)[- b + 0 + 2a] = 0
⇒ 2a – b = 0
⇒ 2a = b
अत: विकल्प (C) सही है।

प्रश्न 14.
यदि A(- 2, – 1), B (a, 0), C(4, b) और D (1, 2) समान्तर चतुर्भुज के शीर्ष हों, तो a और b का मान होगा :
(A) 1, 3
(B) 2, 4
(C) 2, 3
(D) 1, 4
हल :
समान्तर चतुर्भुज के विकर्ण परस्पर सम- द्विभाजित करते हैं।
∴ विकर्ण AC का मध्य-बिन्दु = विकर्ण BD का मध्य बिन्दु

अत: सही विकल्प (A) है।

प्रश्न 15.
आकृति में बिन्दु P(5, – 3) तथा Q(3, y) बिन्दुओं A (7, – 2) तथा B (1, – 5) को मिलाने वाले रेखाखण्ड को समनिभाजित करते हैं, तो y बराबर है-

(A) 2
(B) 4
(C) – 4
(D) \(\frac{-5}{2}\)
हल :
∵ बिन्दु P तथा Q, AB को समत्रिभाजित करते हैं।

∴ AP = PQ = BQ
AQ = AP + PQ = 2AP
∴ \(\frac{AQ}{BQ}=\frac{2AP}{AP}\) = \(\frac{2}{1}\) = 2 : 1
बिन्दु Q का निर्देशांक है।

अतः विकल्प (C) सही है।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.1

Question 1.
Use Euclid’s division algorithm to find the of (1) 135 and 225 (2) 196 and 38220 (3) 867 and 255.
Solution:
1. 135 and 225
Here, 225 > 135
∴ 225 = 135 × 1 +90
Since remainder ≠ 0, we apply division lemma to 135 and 90.
∴ 135 = 90 × 1 + 45
Since remainder ≠ 0, we apply division lemma to 90 and 45.
∴ 90 = 45 × 2 + 0
Since remainder = 0, the divisor 45 is the HCF.
Hence, HCF (135, 225) = 45.

2. 196 and 38220
Here, 38220 > 196
∴ 38220 = 196 × 195 +0
Since remainder = 0, the divisor 196 is the HCF.
Hence, HCF (196, 38220) = 196.

3. 867 and 255
Here, 867 > 255
∴ 867 = 255 × 3 + 102
Since remainder ≠ 0, we apply division lemma to 255 and 102.
∴ 255 = 102 × 2 + 51
Since remainder ≠ 0. we apply division lemma to 102 and 51.
∴ 102 = 51 × 2 + 0
Since remainder = 0, the divisor 51 is the HCF.
Hence, HCF (867, 255) = 51

Question 2.
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Solution:
Let a be any positive integer and b = 6. Then. by Euclid’s division lemma, a = 6q + r, for some integer q ≥ 0 and r = 0, 1, 2, 3, 4 or 5. because 0 ≤ r < 6.
So, a = 6q or a = 6q + 1 or
a = 6q + 2 = 2(3q + 1) or a = 6q + 3 or
a = 6q + 4 = 2(3q + 2) or a = 6q + 5.
Since a is an odd integer, a cannot be 6q or 6q + 2 or 6q + 4 as they are all divisible by 2.
Therefore, any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.

Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
To arrive at the answer, we have to find the HCF of 616 and 32.
Here, 616 > 32
∴ 616 = 32 × 19 + 8
∴ 32 = 8 × 4 + 0
Thus, HCF (616, 32)=8
Hence, the maximum number of columns in which they can march is 8 columns.

Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [Hint: Let a be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Solution:
Let a be any positive integer and b = 3. Then, by Euclid’s division lemma, a = 3q or a = 3q + 1 or a = 3q + 2; where q is a non- negative integer.
1. If a = 3q, then a² = (39)² = 9q² = 3(3q²) = 3m, where m = 3q² is some integer.
2. If a = 3q + 1, then a² = (3q + 1)² = 9q2 + 6q + 1 = 3(3q² + 2q) + 1 = 3m+ 1. where m = 3q² + 2q is some integer.
3. If a = 3q + 2, then a² = (3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3 (3q² + 4q + 1) + 1 = 3m + 1, where m = 3q² + 4q + 1 is some integer.
Thus, in either case, the square of any positive integer is of the form 3m or 3m + 1.

Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m +8.
Solution:
Let a be any positive integer and b = 3. Then, by Euclid’s division lemma, a = 3q or a = 3q + 1 or a = 3q + 2; where q is negative integer.
1. If a = 3q, then
a³ = (39)³ = 27q³ = 9(3q³) = 9m,
where m = 3q³ is some integer.

2. If a = 3q + 1, then
a³ = (3q + 1)³
= 27q³ + 27q² + 9q + 1
= 9(3q³ + 3q² + q) + 1
= 9m + 1.
where m = 3q³ + 3q² + q is some integer.

3. If a = 3q + 2, then
a³ = (3q + 2)³
= 27q + 54q² + 36q + 8
= 9(3q³ + 6q² + 4q) + 8
= 9m + 8,
where m = 3q³ + 6q² + 4q is some integer.
Thus, in either case, the cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 8 त्रिकोणमिति का परिचय Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 8 त्रिकोणमिति का परिचय

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
यदि tan A = \(\frac{3}{4}\) हो, तो sec A(1 – sin A)(sec A + tan A) का मान ज्ञात कीजिए।
हल:
दिया है कि
tan A = \(\frac{A B}{B C}=\frac{3}{4}\)

माना कि BC = 4k तथा AB = 3k
समकोण त्रिभुज ABC में,
पाइथागोरस प्रमेय से
AC² = AB² + BC²
= (3k)² + (4k)²
= 9k² + 16k²
= 25k²

प्रश्न 2.
cos² 12° + cos² 78° का मान ज्ञात कीजिए।
हल:
cos² 12° + cos² 78°
cos² (90° – 78°) + cos² 78° = 0
[∵ cos (90° – θ) = sin θ]
sin² 78° + cos² 78° = 1.

प्रश्न 3.
मान ज्ञात कीजिए :
\(\frac{2 \cos 65^{\circ}}{\sin 25^{\circ}}-\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-\sin 90^{\circ}\) + tan 5° tan 35° tan 60°.tan 55° tan 85°
हल:
\(\frac{2 \cos 65^{\circ}}{\sin 25^{\circ}}-\frac{\tan 20^{\circ}}{\cot 70^{\circ}}\) – sin 90° + tan 5° tan 35° tan 60°.tan 55° tan 85°
= \(\frac{2 \cos \left(90^{\circ}-25^{\circ}\right)}{\sin 25^{\circ}}-\frac{\tan \left(90^{\circ}-70^{\circ}\right)}{\cot 70^{\circ}}\) – 1 + tan (90° – 85°). tan 85° \(\sqrt{3}\).tan (90° – 55°) tan 55°
[∵ tan 60° = \(\sqrt{3}\) sin 90° = 1]
= \(\frac{2 \sin 25^{\circ}}{\sin 25^{\circ}}-\frac{\cot 70^{\circ}}{\cot 70^{\circ}}\) – 1 + cot 85° tan 85° \(\sqrt{3}\).cot 55°. tan 55°
[∵ cos (90° – θ) = sin θ, tan (90° – θ) = cot θ]
= 2 – 1 – 1 + \(\frac{1}{\tan 85^{\circ}}\).tan 85°\(\sqrt{3}\).\(\frac{1}{\tan 55^{\circ}}\)
= 1 × \(\sqrt{3}\) × 1
= \(\sqrt{3}\)
अत: \(\frac{2 \cos 65^{\circ}}{\sin 25^{\circ}}-\frac{\tan 20^{\circ}}{\cot 70^{\circ}}\) – sin 90° + tan 5° tan 35° tan 60°.tan 55°.tan 85° = \(\sqrt{3}\)

प्रश्न 4.
सिद्ध कीजिए कि :
\(\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}\) = 2 cosec θ.
हल:

= \(\frac{1+1+2 \cos \theta}{\sin \theta(1+\cos \theta)}=\frac{2+2 \cos \theta}{\sin \theta(1+\cos \theta)}\)
= \(\frac{2(1+\cos \theta)}{\sin \theta(1+\cos \theta)}=\frac{2}{\sin \theta}\)
= 2 cosec θ = R.H.S.
∴ L.H.S. = R.H.S.

प्रश्न 5.
यदि sin θ + cos θ = \(\sqrt{3}\), तब सिद्ध कीजिए कि tan θ + cot θ = 1.
हल:
दिया है,
sin θ + cos θ = \(\sqrt{3}\)
(sin θ + cos θ)² = (\(\sqrt{3}\))²
sin² θ + cos² θ + 2 sin θ cos θ = 3
⇒ 1 + 2 sin θ cos θ = 3
⇒ 2 sin θ cos θ = 3 – 1 = 2
⇒ 2 sin θ cos θ = \(\frac{2}{2}\) = 1
⇒ sin θ cos θ = sin² θ + cos² θ [∵ 1 = sin² θ + cos² θ]
⇒ \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}=1\)
⇒ \(\frac{\sin ^2 \theta}{\sin \theta \cos \theta}+\frac{\cos ^2 \theta}{\sin \theta \cos \theta}=1\)
⇒ \(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}=1\)
⇒ tan θ + cot θ = 1.

प्रश्न 6.
यदि tan θ + sec θ, तब सिद्ध कीजिए कि sec θ = \(\frac{l^2+1}{2 l}\)
हल:
दिया है,
tan θ + sec θ = l

प्रश्न 7.
यदि 3 cot A = 4 तो \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\) का मान ज्ञात कीजिए।
हल:
3 cot A = 4
cot A = \(\frac{4}{3}\)

प्रश्न 8.
सिद्ध कीजिए कि :
tan⁴ θ + tan² θ = sec⁴ θ – sec² θ.
हल:
L.H.S. = tan⁴ θ + tan² θ
= tan2 θ + (tan2 θ + 1)
= (sec² θ – 1) (sec² θ – 1 + 1) [∵ tan² θ = sec² θ – 1]
= (sec² θ – 1) (sec² θ)
= sec⁴ θ – sec² θ
= R.H.S.
∴ L.H.S. = R.H.S.

प्रश्न 9.
यदि x = r sin A cos C, y = r sin A sin C तथा z = r cos A है, तो सिद्ध कीजिए कि x² + y² + z² = r² है।
हल:
दिया है,
x = r sin A cos C, y = sin A sin C तथा
z = r cos A
L.H.S. = x² + y² + z²
= (r sin A cos C)² + (r sin A sin C)² + (r cos A)²
= r² sin² A cos² C + r² sin² A sin² C + r² cos A
= r² sin² A (cos² C + sin² C) + r² cos² A
= r² sin² A + r² cos² A [∵ sin² C + cos² C = 1]
= r² (sin² A + cos² A)
= r²
R.H.S. = r²
∴ L.H.S. = R.H.S.

प्रश्न 10.
दिखाइये कि
tan 36° tan 17° tan 54° tan 73° = 1.
हल:
L.H.S. = tan 36° tan 17° tan 54° tan 73°
= tan (90° – 54°) tan (90° – 73°) tan 54° tan 73°
= cot 54° cot 73° tan 54° tan 73°
= 1 = R.H.S.

प्रश्न 11.
यदि cos A = \(\frac{12}{13}\), तो cot A का मान परिकलित कीजिए।
हल:
एक ΔABC की रचना करते हैं, जिसमें ∠B = 90° है।
दिया है, cos A = \(\frac{12}{13}\)
आधार / कर्ण = \(\frac{12}{13}\)
आधार = 12k
तथा कर्ण = 13k

ΔABC में पाइथागोरस प्रमेय से
AC² = AB² + BC²
⇒ BC² = AC² – AB²
⇒ BC² = (13k)² – (12k)²
⇒ BC² = 169k² – 144k²
⇒ BC² = 25k²
⇒ BC ± 5k
cot A = आधार / लम्ब
= \(\frac{12 k}{5 k}\)
= \(\frac{12}{5}\)

प्रश्न 12.
त्रिकोणमितीय अनुपात tan A को sec A के पदों में लिखिए।
हल:
∵ sec² A = 1 + tan² A
⇒ tan² A = sec² A – 1
⇒ tan A = \(\sqrt{\sec ^2 A-1}\)

प्रश्न 13.
(i) यदि cos 3A = sin (A – 34°) हो, जहाँ A एक न्यूनकोण है तो A का मान ज्ञात कीजिए।
(ii) निम्नलिखित सर्वसमिका सिद्ध कीजिए, जहाँ वे कोण, जिनके लिए व्यंजक परिभाषित है, न्यूनकोण है।
\(\frac{1+\cot ^2 A}{1+\tan ^2 A}=\left(\frac{1-\cot A}{1-\tan A}\right)^2\)
हल:
(i) दिया है, cos 3A = sin(A – 34°)
⇒ cos 3A = cos[90° – (A – 34°)]
⇒ 3A = 90° – (A – 34°)
⇒ 3A = 90° – A + 34°
⇒ 4A = 124°
⇒ A = 31°

प्रश्न 14.
(i) (1 + tan θ + sec θ) ( 1 + cot θ – cosec θ) का मान ज्ञात कीजिए।
(ii) सिद्ध कीजिए: \(\frac{\tan A-\sin A}{\tan A+\sin A}=\frac{\sec A-1}{\sec A+1}\)
हल:
(i) (1 + tan θ + sec θ) ( 1 + cot θ – cosec θ)

प्रश्न 15.
सिद्ध कीजिए :
2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1 =0
हल:
L.H.S. = 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1
= 2[(sin² θ)³ + (cos² θ)³] – 3(sin⁴ θ + cos⁴ θ) + 1
= 2[(sin² θ + cos² θ) {sin² θ)² + (cos² θ)² – sin² θ cos² θ}] – 3(sin⁴ θ + cos⁴ θ) + 1
[∵ a³ + b³ = (a + b)(a² + b² – ab)]
= 2[1·(sin⁴ θ + cos⁴ θ – sin² θ cos² θ] – 3(sin⁴ θ + cos⁴ θ) + 1
= 2 sin⁴ θ + 2 cos⁴ θ – 2 sin² θ cos² θ – 3sin⁴ θ – 3 cos⁴ θ + 1
= -sin⁴ θ – cos⁴ θ – 2 sin² θ cos² θ + 1
= -[sin⁴ θ + cos⁴ θ + 2 sin² θ cos² θ] + 1
= -[(sin² θ + cos² θ)²] + 1
= – [(1)²] + 1
= 1 + 1 = 0 = R.H.S.

प्रश्न 16.
सिद्ध कीजिए:

हल:

प्रश्न 17.
मान ज्ञात कीजिए :

हल:

प्रश्न 18.
सिद्ध कीजिए:

हल:


समीकरण (i) व (ii) से,
L.H.S. = R.H.S.

प्रश्न 19.
यदि sin (A + 2B) = \(\frac{\sqrt{3}}{2}\) तथा cos (A + 4B) = 0 है, जहाँ A तथा B न्यूनकोण हैं, तो A तथा B ज्ञात कीजिए।
हल:
दिया है,
sin(A + 2B) = \(\frac{\sqrt{3}}{2}\)
sin(A + 2B) = sin 60° (∵ sin 60° = \(\frac{\sqrt{3}}{2}\))
A + 2B = 60° …..(i)
तथा cos(A + 4B) = 0
cos(A + 4B) = cos 90° (∵ cos 90° = 0)
A + 4B = 90° ……(ii)
समीकरण (i) व (ii) को हल करने पर,
B = 15° तथा A = 30°

प्रश्न 20.
सिद्ध कीजिए :
\(\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{\sec A-\tan A}\)
हल:

प्रश्न 21.
सिद्ध कीजिए:

हल:

प्रश्न 22.
यदि sec θ = x + \(\frac{1}{4 x}\), x ≠ 0, तो (sec θ + tan θ) ज्ञात कीजिए :
हल:
sec θ = x + \(\frac{1}{4 x}\)
sec θ = \(\frac{4 x^2+1}{4 x}\)
दोनों पक्षों का वर्ग भरने पर प्राप्त होता है-

प्रश्न 23.
(1) यदि 4 tan θ = 3 है, तो \(\left(\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}\right)\) का मान ज्ञात कीजिए।
(ii) यदि tan 2A = cot (A – 18°), जहाँ 2A एक न्यून कोण है, तो A का मान ज्ञात कीजिए।
हल:
(i) यदि 4 tan θ = 3
tan θ = \(\frac{3}{4}\)
लम्ब = 3k; आधार = 4k; कर्ण = ?
पाइथागोरस प्रमेय से,
(कर्ण)² = (लम्ब)² + (आधार)²
= (3k)² + (4k)²
= 9k² + 16k²
= 25k²
कर्ण = 5k

(ii) यदि tan 2A = cot(A – 18°)
⇒ cot(90° – 2A) = cot(A – 18°)
⇒ 90° – 2A = A – 18°
-2A – A = – 18° – 90°
-3A = -108°
A = \(\frac{108^{\circ}}{3}\)
A = 36°.

प्रश्न 24.
सिद्ध कीजिए:
\(\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}\) = tan A.
हल:

प्रश्न 25.
सिद्ध कीजिए:
\(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = sec θ – tan θ
हल:

प्रश्न 26.
सिद्ध कीजिए:
\(\frac{\tan ^2 \theta}{1+\tan ^2 \theta}+\frac{\cot ^2 \theta}{1+\cot ^2 \theta}=1\)
हल:

प्रश्न 27.
सिद्ध कीजिए : (1 + tan A – sec A) × (1 + tan A + sec A) = 2 tan A
हल:
L.H.S. = (1 + tan A – sec A) × (1 + tan A + sec A)
= {(1 + tan A) – sec A} {(1 + tan A) + sec A}
= (1 + tan A)² – sec² A
= 1 + tan² A + 2 tan A – sec² A
= sec² A + 2 tan A – sec² A
= 2 tan A = R.H.S.

प्रश्न 28.
सिद्ध कीजिए कि :

हल:

प्रश्न 29.
यदि x = 3 sin θ + 4 cos θ तथा y = 3 cos θ – 4 sin θ, तो ज्ञात कीजिए : x² + y² = 25
हल:
दिया है, x = 3 sin θ + 4 cos θ
x² = (3 sin θ + 4 cos θ)²
= 9 sin² θ + 16 cos² θ + 24 sin θ cos θ
तथा y = 3 cos θ – 4 sin θ
y² = (3 cos θ – 4 sin θ)²
= 9 cos² θ + 16 sin² θ – 24 sin θ cos θ
L.H.S. = x² + y²
= 9 sin² θ + 16 cos² θ + 24 sin θ cos θ + 9 cos² θ + 16 sin² θ – 24 sin θ cos θ
= 25 sin² θ + 25 cos² θ
= 25(sin² θ + cos² θ)
= 25 × 1 = 25 = R.H.S.

प्रश्न 30.
यदि sin θ + sin² θ = 1 है, तो सिद्ध कीजिए : cos² θ + cos⁴ θ = 1
हल:
दिया है,
sin θ + sin² θ = 1
⇒ sin θ = 1 – sin² θ
⇒ sin θ = cos² θ
अब cos² θ + cos⁴ θ = cos² θ + (cos² θ)²
= cos² θ + (sin θ)²
= cos² θ + sin² θ
= 1 = R.H.S.

वस्तुनिष्ठ प्रश्न :

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क).

1. \(\frac{\cos 80^{\circ}}{\sin 10^{\circ}}\) + cos 59° cosec 31° का मान …………… \frac{\cos 80^{\circ}}{\sin 10^{\circ}}है ।
2. \(\left(\sin ^2 \theta+\frac{1}{1+\tan ^2 \theta}\right)\) का मान ………………. है।
3. (1 + tan² θ) (1 – sin θ) (1 + sin θ) का मान ………………….. है।
4. sin 20° cos 70° + sin 70° cos 20° का मान है ……………….. ।
5. sin 65° + sin² 25° का मान है …………………… ।

उत्तर:

1. 2,
2. 1,
3. 1,
4. 1,
5. 1

निम्न में सत्य / असत्य बताइये :

प्रश्न (ख).

1. समकोण त्रिभुज में समकोण की सम्मुख भुजा को लंब कहते हैं।
2. sin व cos के मान सदैव 1 से कम या 1 के बराबर होते हैं।
3. cosee व sec के मान सदैव से अधिक या 1 के बराबर होते हैं।
4. यदि cot θ = \(\frac{12}{5}\) है, sin θ का मान \(\frac{13}{5}\) है।
5. tan² 60° + sin² 45° का मान \(\frac{7}{2}\) है।

उत्तर:

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

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
यदि cos (10° + θ) = sin 30° है, तो θ का मान है:
(A) 50°
(B) 40°
(C) 80°
(D) 20°
हल:
cos (10° + θ) = sin 30°
cos(10° + θ) = sin(90° – 60°)
cos (10° + θ) = cos 60°
10° + θ = 60°
θ = 60° – 10° = 50
अत: सही विकल्प (A) है।

प्रश्न 2.
यदि sin A = cos A, 0 ≤ A ≤ 90° है, तो कोण A बराबर है:
(A) 30°
(B) 60°
(C) 0°
(D) 45°
हल:
sin A = cos A
⇒ sin A = sin (90° – A )
⇒ A = 90° – A
⇒ 2A = 90°
⇒ A = 45°
अतः सही विकल्प (D) है।

प्रश्न 3.
8 cot² A – 8 cosec² A बराबर है:
(A) 8
(B) 1
(C) – 8
(D) – \(\frac{1}{8}\)
हल:
8 cot² A – 8 cosec² A
= 8 cot² A – 8(1 + cot² A)
= 8 cot² A – 8 – 8 cot² A
= -8
अत: सही विकल्प (C) है।

प्रश्न 4.
यदि cos A = \(\frac{\sqrt{3}}{2}\), 0° < A < 90° है, तो A बराबर है:
(A) \(\frac{\sqrt{3}}{2}\)
(B) 30°
(C) 60°
(D) 1
हल :
cos A = \(\frac{\sqrt{3}}{2}\)
⇒ cos A = cos 30°
⇒ A = 30°
अतः सही विकल्प (B) है।

प्रश्न 5.
θ का ऐसा मान जिसके लिए sin (44° + θ) = cos 30° है, होना :
(A) 46°
(B) 60°
(C) 16
(D) 90°
हल:
sin(44° + θ) = cos 30°
⇒ sin(44° + θ) = cos(90° – 60°)
⇒ sin(44° + θ) = sin 60°
⇒ 44° + θ = 60°
⇒ θ = 60° – 44° = 16°
अत: सही विकल्प (C) है।

प्रश्न 6.
2 sin² 60° + 3 cot² 30° – tan 45° का मान होना :
(A) \(\frac{2}{19}\)
(B) \(\frac{12}{19}\)
(C) \(\frac{19}{2}\)
(D) इनमें से कोई नहीं।
हल:
2 sin² 60° + 3 cot² 30° – tan 45°
= 2 × \(\left(\frac{\sqrt{3}}{2}\right)^2\) + 3 × \((\sqrt{3})^2\) – 1
= 2 × \(\frac{3}{4}\) + 3 × 3 – 1
= \(\frac{3}{2}\) + 8 = \(\frac{19}{2}\)
अतः सही विकल्प (C) है।

प्रश्न 7.
यदि ΔABC का ∠C समकोण है, ता cos (∠A + ∠B) का मान है :
(A) 0
(B) 1
(C) \(\frac{1}{2}\)
(D) \(\frac{\sqrt{3}}{2}\)
हल:
दिया है, ΔABC का C समकोण है।

∵ हम जानते हैं कि त्रिभुज के तीनों कोणों का योग 180° होता है।
∴ ∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠B + 90° = 180°
⇒ ∠A + ∠B = 180° – 90°
⇒ ∠A + ∠B = 90°
दोनों तरफ cos लेने पर
⇒ cos (∠A + ∠B) = cos 90°
⇒ cos (∠A + ∠B) = 0
अत: विकल्प (A) सही है।

प्रश्न 8.
यदि sin A + sin² A = 1, तब व्यंजक (cos² A + cos² A) का मान हैं :
(A) 1
(B) \(\frac{1}{2}\)
(C) 2
(D) 3
हल:
दिया है,
sin A + sin² A = 1
⇒ sin A = 1 – sin² A
⇒ sin A = cos² A ……(1)
अब cos² A + cos⁴ A = cos² A + (cos² A)²
= cos² A + sin² A
= 1
[(1) का प्रयोग करने पर]
अत: विकल्प (A) सही है।

प्रश्न 9.
sin (45° + θ) – cos (45° – θ) बराबर है:
(A) 2 cos θ
(B) 0
(C) 2 sin θ
(D) 1
हल:
sin (45° + θ) – cos (45° – θ)
= sin (45° + θ) – cos [90° – (45° + θ)]
[∵ (45° – θ) = {90° – (45° + θ)}]
= sin (45° + θ) – sin (45° + θ)
[∵ cos (90° – θ) = sin θ = 0]
अतः विकल्प (B) सही है।

प्रश्न 10.
दिया गया है कि sin α = \(\frac{1}{2}\) और cos β = \(\frac{1}{2}\) तब (α + β) का मान है:
(A) 0°
(B) 30°
(C) 60°
(D) 90°
हल:
दिया है,
sin α = \(\frac{1}{2}\)
⇒ sin α = sin 30°
⇒ α = 30°
और cos β = \(\frac{1}{2}\)
⇒ cos β = cos 60°
⇒ β = 60°
अतः α + β = 30° + 60° = 90°
अतः विकल्प (D) सही है।

प्रश्न 11.
यदि sin θ – cos θ = θ तब (sin⁴ θ + cos⁴ θ) का मान है:
(A) 1
(B) \(\frac{3}{4}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{4}\)
हल:
दिया है,
sin θ – cos θ = 0 ⇒ sin θ = cos θ
⇒ \(\frac{\sin \theta}{\cos \theta}\) = 1 ⇒ tan θ = tan 45°
⇒ θ = 45°
अब sin⁴ θ + cos⁴ θ = sin⁴ 45° + cos⁴ 45°
= \(\left(\frac{1}{\sqrt{2}}\right)^4+\left(\frac{1}{\sqrt{2}}\right)^4\)
= \(\frac{1}{4}+\frac{1}{4}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
अत: विकल्प (C) सही है।

प्रश्न 12.
बराबर है:
(A) tan θ
(B) cos θ
(C) sin θ
(D) cot θ.
हल:

अतः सही विकल्प (C)।

प्रश्न 13.
यदि sin θ = \(\frac{1}{2}\) हो, तो \(\frac{1-2 \sin ^2 \theta}{\sin \theta}\) का मान ज्ञात कीजिए ।
(A) 1
(B) 0
(C) 2
(D) – 1
हल:

अतः विकल्प (A) सही है।

प्रश्न 14.
यदि sec θ + tan θ = 7 है, तो sec θ – tan θ बराबर है:
(A) \(\frac{1}{7}\)
(B) 7
(C) 6
(D) 49
हल:
∵ हम जानते हैं कि
sec² θ – tan² θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1
⇒ 7 × (sec θ – tan θ) = 1
[दिया है : sec θ + tan θ = 7]
sec θ – tan θ = \(\frac{1}{7}\)
अतः विकल्प (A) सही है।

प्रश्न 15.
यदि 5 tan θ = 12 है, तो \(\frac{13 \sin \theta}{3}\) का मान है :
(A) 2
(B) 4
(C) 12
(D) 1
हल:
दिया है,
5 tan θ = 12
tan θ = \(\frac{12}{5}=\frac{A B}{B C}\)

माना कि AB = 12k तथा BC = 5k
समकोण ΔABC में,
AC² = BC² + AB² (पाइथागोरस प्रमेय से)
⇒ AC² = (5k)² + (12k)²
⇒ AC² = 25k² + 144k²
⇒ AC² = 169k²
⇒ AC = \(\sqrt{169 k^2}\)
⇒ AC = 13k
∴ sin θ = \(\frac{A B}{A C}\)
= \(\frac{12 k}{13 k}=\frac{12}{13}\)
अब \(\frac{13 \sin \theta}{3}\)
= \(\frac{13 \times 12}{3 \times 13}\)
= 4
अतः विकल्प (B) सही है।

Jharkhand Board JAC Class 10 Science Important Questions Chapter 13 Magnetic Effects of Electric Current Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 13 Magnetic Effects of Electric Current

Additional Questions and Answers

Question 1.
State at least three points of difference between the following terms / quantities :
(1) Electric motor and Electric generator
Answer:

Electric motor	Electric generator
1. It converts electrical energy into mechanical energy.	1. It converts mechanical energy into electrical energy.
2. A current-carrying coil placed in a magnetic field experiences force. It works on this principle.	2. Whenever a coil is rotated in a magnetic field, an electric current is induced in the coil. It works on this principle of electromagnetic induction.
3. A current-carrying coil rotates in the magnetic field due to the equal (in magnitude) and opposite (in direction) forces acting on the parallel arms of the coil.	3. The conducting coil is made to rotate in the magnetic field by mechanical means.
4. It is used in fans, washing machines, mixers and many such appliances.	4. It is used in hospitals, shops, banks, etc. when there is a power failure.

(2) Direct Current (DC) and Alternating Current (AC)
Answer:

Direct Current (DC)	Alternating Current (AC)
1. It flows only in one direction.	1. Its direction is reversed at regular intervals of time.
2. It is obtained with a cell/battery or a DC generator.	2. It is obtained with an AC generator.
3. Its production is expensive.	3. Its production is comparatively cheaper.
4. There is large loss of electric power when it is transmitted over long distances.	4. There is comparatively less loss of electric power when it is transmitted over long distances.

(3) Permanent magnet and Electromagnet
Answer:

Permanent magnet	Electromagnet
1. Its strength is fixed.	1. Its strength can be changed by changing the current through the coil.
2. Its polarities are fixed.	2. Its polarities can be reversed by changing the direction of the current.
3. They are usually made of alloys, e.g., Alnico Note: Alnico is an alloy of aluminium, nickel, cobalt and iron.	3. It is made by inserting a soft iron rod into a copper coil / solenoid and connecting the coil / solenoid to a battery.

Question 2.
Give scientific reasons for the following statements:
(1) It is improper to use a copper wire as a fuse wire.
Answer:
The melting point of copper Is very high. Hence, it may not melt even when a current through it is more than the current rating of the circuit.
Hence, it is improper to use a copper wire as a fuse wire.

(2) We use power supply of two different current ratings at our homes.
Answer:
Each electrical appliance has its rating and draws current accordingly for its function-ing. To avoid overloading and short-circuiting, two separate circuits are used in electrical wiring at home, one of 15 A for appliances with higher power ratings and another of 5 A for bulbs, tubes, fans, etc.

(3) We use AC for electric power transmission over long distances, while most of the appliances use DC.
Answer:
Electric power transmission over long distances can result in a lot of loss of electrical energy as the energy gets dissipated as heat energy.

The loss of power is much less in the case of AC relative to DC. Hence, we use AC for electric power transmission over long distances.

Objective Questions and Answers

Question 1.
Answer the following questions in one word / sentence :
(1) Name the instrument used to determine the direction of magnetic field.
Answer:
Magnetic needle.

(2) At which position of a bar magnet is its magnetism maximum?
Answer:
Magnetic pole

(3) Is magnetic field a scalar or a vector?
Answer:
Vector

(4) What is the direction of the magnetic field inside the current-carrying circular ring as shown in the following figure?

Answer:
Perpendicular to the plane of the paper and going into the plane of the paper

(5) As shown in the following figure, the magnetic force is acting in the downward direction on the current-carrying wire placed in the magnetic field. What is the direction of the magnetic field?

Answer:
Prependicular to the plane of the paper and coming out of the plane of the paper

(6) Name the scientist who first suggested that a force would act on a current-carrying wire placed in a magnetic field.
Answer:
Ampere

(7) In given magnetic field, what should the direction of motion of the conductor so that the maximum induced current can be produced?
Answer:
Perpendicular to the magnetic field

(8) While a magnet is moved towards a coil, will the induced current decrease or increase? Why?
Answer:
Increase, because the number of magnetic field lines linked with the coil increases.

(9) Which type of current can be transmitted over long distance without much loss of electrical energy?
Answer:
AC

(10) Which type of connection is done for electric appliances?
Answer:
Parallel

(11) Whose magnetic field resembles the magnetic field of a current-carrying solenoid?
Answer:
A bar magnet

(12) How much current would be induced in a coil while a magnet and the coil both move along the same line with the same velocity?
Answer:
Zero

(13) What is the other name of Right-hand thumb rule?
Answer:
Maxwell’s corkscrew rule

(14) When a wire is moved up and down in a magnetic field, a current is induced in the wire. What is this phenomenon known as?
Answer:
Electromagnetic induction

Question 2.
Fill in the blanks:

1. The magnetic field is ………………. in the region where the magnetic field lines are very close to each other.
2. ………………. had discovered the, magnetic effect of electric current.
3. With the help of a ………………. the presence of electric current can be known.
4. Electric motor converts ………………. energy into ………………. energy.
5. Resistance of a fuse wire is and its melting point is ……………….
6. The colour of the insulation of the earthing wire is ……………….
7. Magnetic field at the centre of a current carrying circular ring is inversly proportional to ……………….
8. In Fleming’s left-hand rule, the thumb denotes the direction of ……………….
9. The appliance with the help of which electricity can be generated is known as ……………….
10. The induced electromotive force produced in a coil is proportional to rate of change of ……………….
11. Electromagnetic induction was discovered by ……………….
12. In India, the direction of AC changes ………………. times in one second.
13. Magnetic force acting on a current-carrying wire kept in a magnetic field is in ………………. direction to the magnetic field.
14. A long wound cylindrical coil of insulated wire is called ……………….
15. Working of an electric motor is based on ………………. rule.
16. Working of an electric generator is based on ………………. rule.
17. When a coil is viewed from one end, if the current flows in an anticlockwise direction, then this end behaves like a ………………. pole.
18. When a coil is viewed from one end, if the current flows in a clockwise direction, then this end behaves like a ………………. pole.
19. In an electric motor, the split ring acts as a ……………….
20. The coil having many turns used in an electric motor or generator is called ……………….

Answer:

1. Strong
2. Oersted
3. galvanometer
4. electrical, mechanical
5. high, low
6. green
7. its radius
8. magnetic force
9. an electric generator
10. number of magnetic field lines with time
11. Michael Faraday
12. 100
13. perpendicular
14. a solenoid
15. Fleming’s left-hand
16. Fleming’s right-hand
17. north
18. south
19. commutator
20. armature

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

1. Magnetic field lines form a closed loop.
2. Magnetism is maximum at the centre of the magnet.
3. The direction of the magnetic field produced by the current-carrying conducting wire is determined by Fleming’s left-hand rule.
4. While going away from the magnet, magnetic field becomes strong.
5. At every point inside the region of a long solenoid, magnetic field is uniform.
6. Electrical generator converts electrical energy into mechanical energy.
7. In India, the frequency of AC is 50 Hz.
8. All electrical appliances in a house are connected in series.
9. The insulated cover on the neutral wire is black in colour.
10. There is a link between electric current and magnetic field.
11. Magnetic field lines intersect each other.
12. Magnitude of the current induced in a circular loop depends on the speed of the magnet moving towards it.
13. The direction of DC remains constant in time.
14. Electromagnet consists of a current-carrying solenoid with Alnico rod (an alloy) put in it.
15. Overloading is caused by connecting too many appliances in a single socket and using them at the same time.
16. A stationary charge is surrounded by a magnetic field.
17. It is safe to touch a live wire.

Answer:

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

Question 4.
Match the following :
(1)

Section ‘A’	Section ‘B’
1. Oersted	p. Force would act on a current-carrying conducting wire when it is placed in a magnetic field.
2. Faraday	q. The rule to find the direction of the force acting on a current-carrying conductor placed in a magnetic field.
3. Ampere	r. When an electric current is passed through a wire, a magnetic field is produced.
4. Fleming	s. Induced current is produced in the coil, while moving a magnet towards the stationary coil.

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

(2)

Section ‘A’	Section ‘B’
1. Right-hand thumb rule	p. Gives the direction of the current induced in a conductor.
2. Fleming’s right-hand rule	q. Gives the direction of the force acting on a current-carrying conductor placed in a magnetic field.
3. Flemintg’s left-hand rule	r. Gives the direction of the magnetic field around a current-carrying straight conductor.

Answer:
(1 – r), (2 – p), (3 – q)

(3)

Section ‘A’	Section ‘B’
1. Current-carrying solenoid	p. Is used to lift heavy objects made of iron.
2. Fuse	q. Converts mechanical energy into electrical energy.
3. Electromagnet	r. Behaves like a bar magnet.
4. Electric motor	s. Converts electrical energy into mechanical energy.
5. Electric generator	t. Is used to prevent damage to the appliances and the circuit due to overloading.

Answer:
(1 – r), (2 – t), (3 – p), (4 – s), (5 – q).

Question 5.
Choose the correct option from those given below each question:
1. The direction of the magnetic field lines in a region outside a bar magnet is …
A. from N-pole to S-pole of the magnet.
B. from S-pole to N-pole of the magnet.
C. coming out from both the poles of the magnet.
D. entering into both the poles of the magnet.
Answer:
A. from N-pole to S-pole of the magnet.

2. Which of the following statements is false?
A. The direction of the magnetic field line in a region outside a bar magnet is from N to S.
B. In the region where the magnetic field lines are at a close distance from each other, there is a strong magnetic field.
C. The magnetic field lines form closed loops.
D. The magnetic field lines can cross each other.
Answer:
D. The magnetic field lines can cross each other.

3. By which instrument can the presence of magnetic field be determined?
A. Voltmeter
B. Ammeter
C. Galvanometer
D. Magnetic needle
Answer:
D. Magnetic needle

4. Who discovered the magnetic effect of an electric current?
A. Faraday
B. Oersted
C. Volta
D. Ampere
Answer:
B. Oersted

5. With the help of which law/rule can the direction of a magnetic field be decided?
A. Faraday’s law
B. Fleming’s right-hand rule
C. Right-hand thumb rule
D. Fleming’s left-hand rule
Answer:
C. Right-hand thumb rule

6. According to the right-hand thumb rule, whose direction is indicated by the thumb?
A. Electric current
B. Magnetic field
C. Magnetic force
D. Motion of a conductor
Answer:
A. Electric current

7. The magnetic field produced by a straight conducting wire on passing the current through it is …
A. in the direction of the current.
B. in the direction opposite to that of the current.
C. circular around the wire.
D. in the direction parallel to the wire.
Answer:
C. circular around the wire.

8. What is the shape of the field line of a magnetic field passing through the centre of a current-carrying circular ring?
A. Circular
B. Straight line
C. Ellipse
D. Magnetic field is zero at the centre of the ring
Answer:
B. Straight line

9. Whose magnetic field is like the magnetic field of a bar magnet?
A. A current-carrying wire
B. A current-carrying ring
C. A current-carrying solenoid
D. A horse-shoe magnet
Answer:
C. A current-carrying solenoid

10. Who discovered electromagnetic induction?
A. Faraday
B. Oersted
C. Ampere
D. Volta
Answer:
A. Faraday

11. What is the direction of the magnetic force acting on a current-carrying wire placed in a magnetic field?
A. Along the magnetic field
B. Along the electric current
C. Perpendicular to the magnetic field
D. Opposite to that of the magnetic field
Answer:
C. Perpendicular to the magnetic field

12. How is a current-carrying wire placed in a magnetic field so that a magnetic field does not act on it?
A. Parallel to the magnetic field
B. Perpendicular to the magnetic field
C. At an angle of 45° with magnetic field
D. At an angle of 120° with magnetic field
Answer:
A. Parallel to the magnetic field

13. State in which of the following cases, the induced current in the loop will not be obtained.
A. The loop is moved towards the magnet.
B. The magnet is moved towards the loop.
C. The loop and the magnet are moved in the opposite directions with the same speed.
D. The loop and the magnet are moved in the same direction with the same speed.
Answer:
C. The loop and the magnet are moved in the opposite directions with the same speed.

14. Which instrument is used to convert electrical energy into mechanical energy?
A. Electric generator
B. Electric motor
C. Electric iron
D. Electric oven
Answer:
B. Electric motor

15. On which principle does the electric generator work?
A. Electrical energy is converted into mechanical energy.
B. Electrical energy is converted into heat.
C. Mechanical energy is converted into electrical energy.
D. Electrical energy is converted into light.
Answer:
C. Mechanical energy is converted into electrical energy.

16. The magnitude of the AC voltage used in India is ……………. and its frequency is …………….
A. 110 V 60 Hz
B. 110 V, 50 Hz
C. 220 V, 50 Hz
D. 220 V, 60 Hz
Answer:
C. 220 V, 50 Hz

17. A wire with ……………. insulation is used for earthing.
A. red
B. black
C. green
D. white
Answer:
C. green

18. Which type of current is obtained from a battery?
A. DC
B. AC
C. AC and DC
D. Depends upon the type of the battery
Answer:
A. DC

19. Which instrument is used to know the presence of an electric current?
A. Fuse
B. Galvanometer
C. Voltmeter
D. Magnetic needle
Answer:
B. Galvanometer

20. A fuse wire is a/an …………….
A. conductor having low melting point
B. insulator having low melting point
C. semiconductor having low melting point
D. conductor having high melting point
Answer:
A. conductor having low melting point

21 is used to find the direction of a current induced in a circuit.
A. Fleming’s left-hand rule
B. Fleming’s right-hand rule
C. Right-hand thumb rule
D. Ampere’s rule
Answer:
B. Fleming’s right-hand rule

22. How many times in one second does an AC with frequency 50 Hz change its direction?
A. 25
B. 50
C. 100
D. 200
Answer:
C. 100

23. At the centre of which of the following four circular rings is the magnetic field maximum while passing equal current through each of them?

A. (a)
B. (b)
C. (c)
D. (d)
Answer:
A. (a)

24. Inside a bar magnet the direction of the magnetic field lines is …………….
A. from N-pole to S-pole of the magnet
B. from S-pole to N-pole of the magnet
C. coming out from both the poles of the magnet
D. entering into both the poles of the magnet
Answer:
B. from S-pole to N-pole of the magnet

25. When the north poles of two different magnets are placed near each other, then the force and when the north pole and the south pole are placed near each other the force of ……………. results.
A. attraction and attraction
B. attraction and repulsion
C. repulsion and attraction
D. repulsion and repulsion
Answer:
C. repulsion and attraction

26. Which figure, out of the following, shows the magnetic field lines of a bar magnet?

A. (a)
B. (b)
C. (c)
D. (d)
Answer:
B. (b)

27. The following figure shows the magnetic field of a magnet. At which point is the intensity of the magnetic field more?

A. D
B. C
C. B
D. A
Answer:
D. A

28. The following figure shows the magnetic field between two magnets. Which magnetic poles are there at points A and B respectively?

A. South pole, north pole
B. North pole, south pole
C. North pole, north pole
D. South pole, south pole
Answer:
B. North pole, south pole

29. On what factor does the direction of the magnetic field produced by a current-carrying straight wire depend?
A. The magnitude of the electric current
B. The direction of the electric current
C. The length of the straight wire
D. The material of the straight wire
Answer:
B. The direction of the electric current

30. The magnetic field at the centre of a current carrying circular loop is ……………. to the current and ……………. to the loop’s radius.
A. inversely proportional, directiy proportional
B. directiy proportional, directiy proportional
C. directly proportional, inversely proportional
D. inversely proportional, inversely proportional
Answer:
C. directly proportional, inversely proportional

31. As we move away from a current-carrying straight wire, the magnetic field of wire varies as ……………. (Where, r is the distance between a given point and the wire.)
A. r²
B. \(\frac { 1 }{ r }\)
C. \(\frac { 1 }{ r² }\)
D. r
Answer:
B. \(\frac { 1 }{ r }\)

32. Out of the following, which statement is wrong for a current-carrying solenoid?
A. The magnetic field inside the solenoid is the same at each point.
B. Placing an iron object inside the solenoid increases the magnetic field.
C. The pattern of the magnetic field lines of the solenoid is different from that of the magnetic field lines of a bar magnet.
D. On reversing the direction of the current through the solenoid, the positions of N-pole and S-pole of the solenoid also get reversed.
Answer:
C. The pattern of the magnetic field lines of the solenoid is different from that of the magnetic field lines of a bar magnet.

33. When an iron piece is introduced inside a current-carrying solenoid the magnetic field of the solenoid …………….
A. increases
B. decreases
C. first increases then decreases
D. does not change
Answer:
A. increases

34. The magnetic force acting on a current-carrying wire placed in a magnetic field is directiy. proportional to …………….
A. the electric current
B. the magnetic field
C. the length of the wire in the magnetic field
D. all the given
Answer:
D. all the given

35. When electric current I flows through a straight wire of length l placed in a magnetic field, a magnetic force of 1 N acts on it. If magnitude of the electric current is halved and the magnetic field is doubled, what will be the magnetic force acting on the wire?
A. 0.5 N
B. 1 N
C. 2N
D. 4 N
Answer:
B. 1 N

36. As shown in the following figure, a current-carrying coil ABCD is placed in a magnetic field. Which portion of the coil does not experience any magnetic force?

A. Portion AB
B. Portion BC
C. Portion CD
D. Both A and B
Answer:
B. Portion BC

37. What happens when a bar magnet is quickly moved towards a coil?
A. The resistance of the coil increases.
B. The magnetic field linked with the coil decreases.
C. The magnetic field linked with the coil increases.
D. The coil moves towards the magnet.
Answer:
C. The magnetic field linked with the coil increases.

38. In which of the following cases will the electric current induced in the loop be maximum?
A. The magnetic field linked with the loop changes rapidly.
B. The magnetic field linked with the loop changes slowly.
C. The magnetic field linked with the loop remains constant.
D. The loop is placed in an electric field.
Answer:
A. The magnetic field linked with the loop changes rapidly.

39. What is the frequency of 220 V DC voltage?
A. 50 Hz
B. 60 Hz
C. 220 Hz
D. Zero
Answer:
D. Zero

40. What is the colour of insulation on the neutral wire coming from the electric board?
A. Red
B. Black
C. Green
D. White
Answer:
B. Black

41. What is the red coloured wire coming from the electric main board called?
A. The earthing wire
B. The neutral wire
C. The live wire
D. The fuse wire
Answer:
C. The live wire

42. On which effect of electric current does the fuse work?
A. Magnetic effect of electric current
B. Chemical effect of electric current
C. Heating effect of electric current
D. Electric effect of electric current
Answer:
C. Heating effect of electric current

43. At the time of short circuit, the electric current ……………….
A. suddenly decreases
B. suddenly increases
C. instantaneously becomes zero
D. does not change
Answer:
B. suddenly increases

44. The direction of the current in the coil of an electric motor changes per ………………. rotation.
A. two
B. one
C. half
D. one-fourth
Answer:
C. half

45. In the following figure, a magnetic needle is placed at different positions near a bar magnet. Which position of the magnetic needle represents the correct direction of the magnetic field?

A. A
B. B
C. C
D. D
Answer:
C. C

46. On which metallic piece is the conducting wire wound to prepare an electromagnet?
A. Copper
B. Soft iron
C. Aluminium
D. Nichrome
Answer:
B. Soft iron

47. To double the voltage generated in a generator, ……………….
A. the rotational speed of the coil should be halvened.
B. the number of turns of the coil should be doubled.
C. the cross-sectional area of the coil should be halvened.
D. none of the given
Answer:
B. the number of turns of the coil should be doubled.

48. In the following figure, magnetic field of different current-carrying wires is shown. Which figure represents the correct direction of the magnetic field?

A. Position A
B. Position B
C. Position C
D. Position D
Answer:
B. Position B

49. A current-carrying conducting wire kept in a magnetic field should experience a force. This was suggested by ……………….
A. Oersted
B. Faraday
C. Ampere
D. Ohm
Answer:
C. Ampere

50. The magnitude of the induced current in electromagnetic induction depends on ……………….
A. the motion of the coil only
B. the motion of the magnet only
C. the relative speed of the coil and magnet
D. the deflection of the pointer in the galvanometer
Answer:
C. the relative speed of the coil and magnet

51. In Fleming’s left-hand rule, ………………. denotes the direction of the magnetic force.
A. the thumb
B. the forefinger
C. the middle finger
D. the palm
Answer:
A. the thumb

52. In the following figure, three bar magnets are shown. Magnets (1) and (3) are fixed. In which direction will the magnet (2) move?

A. Towards magnet (1)
B. Towards magnet (3)
C. Upward
D. Magnet (2) will remain at rest
Answer:
B. Towards magnet (3)

53. To find the direction of the magnetic field produced by a current-carrying wire, a magnetic needle is placed at different positions as shown in the following figure. Which position of the magnetic needle corresponds to the correct direction of the magnetic field?

A. Position A
B. Position B
C. Position C
D. Position D
Answer:
D. Position D

54. In the following figure, current-carrying loop ABCD is placed in a uniform magnetic field. The magnetic forces acting on wires AB and CD are in ………………. and ………………. direction respectively.

A. Downward, upward
B. Downward, downward
C. Upward, upward
D. Upward, downward
Answer:
D. Upward, downward

Question 6.
Answer the following questions in very- short as directed (miscellaneous):
(1) Name the physical qualities which are indicated by the direction of the thumb and forefinger in Fleming’s right-hand rule.
Answer:
The thumb denotes direction of motion of the conductor and the forefinger indicates the direction of the magnetic field.

( 2) State the effect on the strength of the magnetic field produced at a point near a straight conductor if the electric current flowing through it increases.
Answer:
The strength of the magnetic field increases.

(3) Mention the angle between a current carrying straight conductor and magnetic field for which the force experienced by the conductor placed in a magnetic field is maximum.
Answer:
90°

(4) List two sources of magnetic fields.
Answer:
Magnet, moving charges in space, electric current through conductor.

(5) There are two wires. One of them carries a current. How will you find out which one carries a current?
Answer:
Bring a magnetic needle near the wire. The current-carrying wire will produce a deflection in the needle whereas the wire without a current will not.

(6) How will the magnetic field intensity B at the centre of a circular coil carrying current change, if the current through the coil is doubled and the radius of the coil is halved?
Answer:
The magnetic field at the centre of the coil, B ∝ \(\frac { I }{ R }\).
When current I is doubled and radius R is halved, the magnetic field at the centre of the coil becomes four times the original / initial magnetic field.

(7) PQ is a current-carrying long straight conductor in the plane of the paper as shown in the following figure. A and B are two points at distances r₁ and r₂ from it.

If r₂ > r₁ at which point will the strength of the magnetic field be greater? Justify your answer.
Answer:
The magnetic field at a point near the current-carrying straight conductor is B ∝ \(\frac { 1 }{ r }\).
Hence, at point A B₁ ∝ \(\frac{1}{r_1}\) and at point B B₂ ∝ \(\frac{1}{r_2}\)
As r₁ < r₂, B₁ > B₂.

(8) PQ is a current- carrying long straight conductor in the plane of the paper as shown in the following figure. A and B are two points at distances r₁ and r₂ from it.

Mention the direction of the magnetic field produced by it at points A and B.
Answer:
Using the Right-hand thumb rule and S properties of magnetic fields lines we find that at point A magnetic field intensity would be perpendicular to the plane of paper in the inward direction (x) going into the plane and at point B it would be perpendicular to the plane of paper in the outward direction (.) [coming out of the plane.]

(9) In the experiment to show that a current-carrying conductor (rod) when placed in a uniform? magnetic field experiences a force. What will happen if you interchange the terminals of the battery connected in the circuit?
Answer:
The conductor (rod) will be deflected in the opposite direction as the direction of the current is reversed.

(10) A compass needle is placed near a current-carrying straight wire. State your observation when the current in the wire is increased. Give reason.
Answer:
The deflection of the compass needle increase.
Reason: Magnetic field strength produced s near a current-carrying straight wire is directly proportional to the current.

(11) The magnetic field lines of two bar s magnets A and B are shown below:

Name the poles of the magnets facing each other.
Answer:
North poles

(12) Name any two sources of direct current.
Answer:
Electric cell and DC generator.

(13) Name the device used to prevent damage to the electrical appliances and domestic circuit due to overloading or short-circuiting.
Answer:
Electric fuse

(14) State value of potential difference between l the live wire and neutral wire of power supply in our country.
Answer:
220 V

(15) State one difference between the wires used in the element of an electric heater and in a fuse.
Answer:
The melting point of the material of the wire used in a heater element is high while a fuse wire has low melting point.

(16) In a domestic electric circuit-wiring, with which wire do we connect a fuse?
Answer:
The live wire

(17) Mention two ways to increase the strength of the magnetic field of a current-carrying solenoid.
Answer:
Two ways to increase the strength of the magnetic field of a current-carrying solenoid :

• Increasing the number of turns of the solenoid.
• Increasing the current through the solenoid.

(18) Give the significance of the electric meter in a domestic circuit.
Answer:
Electric meter is used to record the consumption of electrical energy in kWh [unit] (commercial unit of electric energy) in the domestic circuit.

(19) A domestic circuit has a line of 5 A rating. How many lamps of ratings 40 W, 220 V can be run simultaneously on this line safely?
Answer:
The current rating of one lamp,
I = \(\frac { P }{ V }\) = \(\frac { 40 W }{ 220 V }\) = \(\frac { 2 }{ 11 }\) A
Thus, current of \(\frac { 2 }{ 11 }\)A is required by one lamp to glow completely.
Now, the current rating (i.e., Maximum current in the line) = 5 A.
∴ The number of lamps that can be run simultaneously on this line safely.
= \(\frac{\text { current rating of the line }}{\text { current rating of the lamp }}\)
= \(\frac{5 \mathrm{~A}}{\frac{2}{11} \mathrm{~A}}\)
= \(\frac { 55 }{ 2 }\)
= 27.5
≈ 27

(20) Give any two uses of an electromagnet.
Answer:

1. In electric bells
2. For separating magnetic substances such as iron and other substances having magnetic property from a metallic scrap.
3. In hospitals, doctors use it to remove particles of iron or steel from a patient’s eye.

(21) A constant current I flows in a horizontal wire in the plane of paper from east to west as shown in the following figure :

At which point will the direction of the magnetic field be from north to south?
Answer:
According to the Right-hand thumb rule, when a current-carrying wire is held in the right hand, with the stretched thumb from east to west in the direction of the current, the fingers wrapped around the wire will be from north to south at the point lying directly below the wire. Hence, the magnetic field at that point would be from north to south.

(22) How do you magnetise a piece of magnetic material?
Answer:
A piece of magnetic material can be magnetised by keeping it inside a current-carrying solenoid.

(23) How does the current in a domestic circuit differ from that used to run a clock?
Answer:
The current in a domestic circuit is an alternating current (AC), while the current used to run a clock is a direct current (DC).

(24) State what would happen to the direction of rotation of the coil of an electric motor, if the direction of the current flowing through the coil is reversed.
Answer:
The direction of rotation of the coil would be reversed.

(25) State the use of Maxwell’s corckscrew rule.
Answer:
Maxwell’s corkscrew rule is used to find the magnetic polarities of a current-carrying coil or a solenoid.

(26) What type of dynamo (a machine for changing other power into electric power or a machine to produce electric current) is used in (i) a bicycle (ii) industry?
Answer:
In a bicycle DC dynamo is used while in industry AC dynamo is used.

(27) Give one reason for not using a series arrangement for domestic circuits.
Answer:
If one of the appliances does not work and consequently the circuit becomes incomplete other appliances in the domestic circuit will stop working.

(28) What does the following symbol indicate / represent?

Answer:
Electric fuse

(29) What is the function of the brushes in an AC generator?
Answer:
In 311 AC generator brushes are used to transfer current from the simple coil / armature of the generator to the load in (extermal) circuit.

(30) What determines the frequency of AC produced by a generator?
Answer:
The frequency of AC produced by a generator is determined by the rotation of the coil and is equal to the frequency of rotation of the coil.

(31) What is the role of each of the three pins in a power plug?
Answer:
The thick pin at the top of a power plug is for earthing. The live pin is on the left and the neutral pin is on the right of the power plug.

(32) State two serious hazards of electricity.
Answer:

1. If a person accidently touches the live wire, he gets a severe shock which may prove fatal.
2. Short-circuiting or overloading can cause a spark which may lead to fire in a building.

(33) Name an instrument used in navigation.
Answer:
Magnetic compass needle

(34) What is the effect on the resistance in the circuit at the time of short-circuit?
Answer:
At the time of short-circuit, the resistance in the circuit becomes very small (nearly zero).

(35) Name the current-carrying conductor, whose magnetic field is represented by the following diagram :

Answer:
A Circular coil.

(36) What is a commutator?
Answer:
A device that reverses the direction of s flow of current through the circuit of an electric? motor is called a commutator.

(37) In an electric motor, which of the s following remains fixed and which rotates with the coil?
(i) Commutator
(ii) Brush
Answer:
(i) Commutator rotates with the coil.
(ii) Brush remains fixed.

(38) What change should be made in an AC generator, so that it may become a DC generator?
Answer:
If we replace rings R₁ and R₂ of an AC generator by a split ring type commutator, then it will become a DC generator.

(39) If fuses of 250 mA, 500 mA, 1 A, 5 A and 10 A were available, which one would be the most suitable for protecting an amplifier rated as 240 V, 180 W?
Answer:
1 A
Because, the current rating of an amplifier is
I = \(\frac { P }{ V }\) = \(\frac { 180 W }{ 240 V }\) = \(\frac { 3 }{ 4 }\) A = 0.75 A
Hence, the fuse of 1 A rating is most suitable compared to other given fuses for safety purpose.

(40) What is a galvanometer?
Answer:
A galvanometer is an instrument used to detect the presence of an electric current in a circuit.

(41) What are the two most commonly used domestic circuits?
Answer:

1. 5 A for low power rating appliances.
2. 15 A for high power rating appliances.

(42) What do you mean by earthing?
Answer:
Earthing means connecting the metal case of an electrical appliance to the earth (at zero potential) by means of a metal wire called the ‘earth wire’.

Value Based Questions With Answers

Question 1.
Shalini and his brother Amey observed a water-pump working. Shalini inquired how an electric motor works. Amey collected different parts of an electric motor e.g., field – magnets, cabon – brushes, armature, battery, switch, connecting wire, split-rings, etc. and showed them to Shalini. Amey then explained how an armature could rotate.

1. State the principle how an electric motor works.
2. Is an electric motor needed for water-pump to work?
3. What values do you learn from Amey’s activities?

Answer:

1. When a current-carrying coil is placed in a magnetic field, it experiences a force. The direction of this force is found by using the Fleming’s left-hand rule.
2. Yes. An electric motor is needed for water-pump to work.
3. Aptitude for practical demonstration and enthusiasm to learn.

Question 2.
Jay’s friend Dev was playing and that too with a magnet. Jay asked him not to play with a magnet. Even then Dev continued playing and brought the magnet very near to the TV screen. That TV then got damaged and there was a dark patch on that area of TV due to effect of magnetism. Jay informed all of his friends to be aware of such things in future.

1. How was TV damaged to get dark patch, when the magnet was brought closer?
2. Name any two domestic devices having electromagnetic effect.
3. What values do you learn from Jay?

Answer:

1. TV set has an electromagnet installed in it. When a magnet is brought closer to the TV screen, the two magnetic fields interfered with each other and spoiled the functioning of the TV.
2. Mobile phones and CD players have electromagnetic effect.
3. We learn values of helpfulness, kindness and caring of others.

Question 3.
Dharmendra is a welder, working at Amit’s house. After electric welding done with naked eyes, he used a grinder to smoothen the welding joints. Accidentally, some particle fell into his eye. He was then crying due to pain. Amit took him to an eye hospital by a taxi. The doctor used a device connected to two electric wires to remove the particle from his eye.

Dharmendra inquired what happened to his eye and what device did the doctor use to remove the particle from his eye. Amit as a science student of class X explained to Dharmendra and advised him to be careful in future as he had a responsible risky job of welding.

1. Which particle fell into Dharmendra’s eye while welding? What precautions should Dharmendra take while doing electric welding and grinding work?
2. What device was used by the doctor to remove the particle?
3. What values do you learn from Amit during this episode?

Answer:

1. A tiny iron particle fell into Dharmendra’s eye while welding. Dharmendra should use some eye-protecting device e.g., welding goggles or handheld shield while doing electric welding and grinding work.
2. The doctor used an electromagnet to remove the tiny iron particle.
3. We learn the values such as (a) will to help others (b) ability to handle a serious situation with calmness.

Question 4.
Bharat had an electric iron. He connected it into two-pin plug. Obviously the green wire was left unconnected. After few days, once his sister got a severe electric shock while ironing the clothes. Bharat called the electrician. He told that this situation could have been avoided if she would have connected the green wire also by using a three-pin plug. Bharat learnt a lesson.

1. Which terminal gets connected by using a green wire?
2. What care should Bharat take to avoid such mistakes?
3. If you were the electrician, what else would you do apart from mere explaining to Bharat?

Answer:

1. Earthing terminal gets connected using a green wire.
2. Bharat should be more careful in future regarding safety measures and put them into practice.
3. As an electrician, I would check all other devices at his house regarding proper earthing. I would also check overloading, if any in the house-hold circuits.

Question 5.
Atul built a small house. He used well-insulated copper wires of good quality for wiring and switches, sockets and plugs too. He however, had a room-heater of poor quality. Once on a winter night .he was sleeping with his family in a closed room with the heater ON.

During sleep he was suffocated. Hence he woke up and found flames of heater on fire. His son Raghav got up quickly and switched off the main switch. He made a call to the fire brigade. The fire crew of fire brigade reached quickly and put off fire. Atul and his family thanked the firemen.

1. How Atul’s heater was on fire?
2. Name the safety device missing in the electric wiring of Atul’s house.
3. What values do you learn from Raghav in this episode?

Answer:

1. The cause of fire was short-circuiting due to defective heater of poor quality.
2. Electric fuse is the safety device which was missing in wiring of Atul’s house.
3. We learn the values like – (a) Awareness (b) Presence of mind

Memory Map:

Jharkhand Board JAC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current

Jharkhand Board Class 10 Science Magnetic Effects of Electric Current Textbook Questions and Answers

Question 1.
Which of the following correctly describes the magnetic field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.
Answer:
(d) The field consists of concentric circles centred on the wire.
Hint: This is an experimental fact. Which is only exhibited by the right-hand thumb rule.

Question 2.
The phenomenon of electromagnetic induction is…
(a) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coll.
(c) producing induced current in a coil due to relative motion between a magnet and the coil.
(d) the process of rotating a coil of an electric motor.
Answer:
(c) producing induced current in a coil due to relative motion between a magnet and the coil.

Question 3.
The device used for producing electric current is called a…
(a) generator.
(b) galvanometer.
(c) ammeter.
(d) motor
Answer:
(a) generator.
Hint: A generator converts mechanical energy into electrical energy.

Question 4.
The essential difference between an AC generator and a DC generator is that…
(a) an AC generator has an electromagnet while a DC generator has permanent magnet.
(b) a DC generator will generate a higher voltage.
(c) an AC generator will generate a higher voltage.
(d) an AC generator has slip rings while the DC generator has a commutator.
Answer:
(d) an AC generator has slip rings while the DC generator has a commutator.
[Due to the slip rings, the direction of the current produced by an AC generator is reversed at regular interval of time, while the current produced by a DC generator always flows in one direction.]

Question 5.
At the time of short circuit, the current in the circuit…
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) varies continuously.
Answer:
(c) increases heavily.

Question 6.
State whether the following statements are true or false :
(a) An electric motor converts mechanical energy into electrical energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The magnetic field lines at the centre of a long circular coil carrying current are parallel straight lines.
(d) The wire with green insulation is usually the live wire of an electric supply.
Answer:
(a) False
[An electric motor converts electrical energy into mechanical energy.]
(b) True
(c) True
(d) False
[The wire with green insulation is the earth wire, whereas the wire with red insulation is the live wire of an electric supply.]

Question 7.
List two methods of producing magnetic fields.
Answer:
Magnetic fields can be produced by passing an electric current through:

• a straight conductor
• a circular loop and
• a solenoid.

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:

• A current-carrying solenoid behaves like a bar magnet and the polarities of its ends depend upon the direction of the current through it.
• Yes, we can use a bar magnet to determine the north and south poles of a current-carrying solenoid.
• In order to determine the magnetic poles of a current-carrying solenoid, place it in a brass hook and suspend it with a long thread so that it can moves freely in a horizontal plane.

Bring the north pole of a bar magnet near one of its ends. If that end of the solenoid moves towards the bar magnet, then it must be its south pole and the other end its north pole, On the contrary if that end of the solenoid moves away from the magnet, then it must be its north pole and the other end its south pole.

Question 9.
When is the force experienced by a current carrying conductor placed in a magnetic field largest?
Answer:
The force experienced by a current-carrying conductor placed in a magnetic field is the largest, when the current in the conductor is perpendicular to the magnetic field.

For reference see the following figure :

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:

• For reference, see figure (a) and (b).
• The direction of the conventional current is opposite to the direction of motion of the electron.
• Applying Fleming’s left-hand rule, we find that when the middle finger points in the direction of the current and the thumb points in the direction of the force, the forefinger points downward giving the direction of the magnetic field.

[Note : IMg indicates a vector (arrow) going into the plane of figure, while IMG indicates a vector (arrow) coming out of the plane of the figure.]

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of the split ring in an electric motor?
Answer:
Construction: The construction of an electric motor is as shown in figure 13.24.

• Electric motor consists of a rectangular coil ABCD of insulated copper wire.
• The coil is placed between the two poles of a permanent magnet such that the arms AB and CD are perpendicular to the direction of the magnetic field.
• The ends of the coil are connected to two halves P and Q of a split ring.
• The inner sides of these halves are insulated and attached to an axle such that they can rotate easily on it.
• The external conducting edges of P and Q touch two conducting stationary brushes (i.e., carbon strips) X and Y respectively.
• These brushes are connected to a plug key and battery as shown in figure 13.24.

Working:
(1) Current in coil ABCD enters from the source – battery through conducting brush X and flows back to the battery through brush Y.

(2) Current in arm AB of the coil flows from A to B while in arm CD it flows from C to D, i.e., opposite to the direction of the current through AB. Both the currents flowing in AB and CD are perpendicular to the magnetic field.

(3) On applying Fleming’s left hand rule for finding the direction of the force on a current carrying conductor in a magnetic field (see figure 13.24), we find that the force acting or arm AB pushes it downwards while the force acting on arm CD pushes it upwards. These forces are also equal in magnitude and perpendicular to the respective lengths of arms AB and CD.

(4) So, the coil and the axle, mounted free to turn about an axis, rotate anticlockwise.

(5) At half rotation, Q makes contact with brush X and P with brush Y. Therefore, the current in the coil gets reversed and flows along the path DCBA.

(6) A device that reverses the direction of flow of current through a circuit is called a commutator.
In an electric motor the split rings act as a commutator.

(7) Now, the reversal of current also reverses the direction of the force acting on the arms AB and CD. Thus arm AB that was earlier pushed down, is now pushed up and arm CD previously pushed up, is now pushed down.

(8) Therefore, the coil and the axle rotate half a turn more in the same direction.

(9) The reversing of the current is repeated at each half rotation, giving rise to a continuous rotation of the coil and the axle.

Important note : When currents through arms BC and DA are either parallel or antiparallel to the magnetic field, magnetic force does not act on them.

When a rectangular coil carrying a current is placed in a magnetic field, then forces, equal in magnitude and opposite in direction, act on its two parallel sides, perpendicular to their lengths, due to which the coil rotates continuously.

The function of the split ring : Due to the split ring commutator in an electric motor, the direction of the current through the coil is reversed after every half rotation of the coil. This reverses the direction of the force acting on the two arms of the coil. The arm that was earlier pushed downward is now pushed upward, while the other arm is now pushed downward. Therefore, the coil continues to rotate in the same direction.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric motors are used in electric fans, refrigerators, mixers and grinders, washing machines, water-pumps, coolers, etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is (i) pushed into the coil, (ii) withdrawn from inside the coil, (iii) held stationary inside the coil?
Answer:
(i) The galvanometer will show a momentary deflection in one direction. It means a current is induced in the coil in one direction due to the relative motion between the coil and the magnet.

(ii) The galvanometer will show a momentary deflection in the opposite direction. It means a current is induced in the coil in the opposite direction due to the relative motion between the coil and the magnet.

(iii) There will be no deflection in the galvanometer. It means no current is induced in the coil, as there is no relative motion between the coil and the magnet.

[Note: The greater the speed of the magnet, the greater is the deflection of the pointer in the galvanometer.]

Question 14.
Two circular coils A and B are placed close to each other. If the current in coil A is changed, will some current be induced in coil B? Give reason.
Answer:
Yes. Current will be induced in coil B. Reason : When the current in coil A is changed, the magnetic field around it changes.

As the two circular coils (with their planes parallel to each other) are placed close to each other, the magnetic field lines linked with coil B also change. Therefore, some current is induced in coil B.

Question 15.
State the rule to determine the direction of a (i) magnetic field produced around a straight conductor-carrying current, (ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and (iii) current induced in a coil due to its rotation in a magnetic field.
Answer:
(i) Right-hand thumb rule : This rule is also called ‘Maxwell’s corkscrew rule.
Rule : Imagine that you are holding a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your fingers of right hand wrap around the conductor in the direction of the field lines of the magnetic field.

This is known as the Right-hand thumb rule.

(ii) Fleming’s left-hand rule : In this case, Fleming’s left hand rule is used to find the direction of the force acting on the conductor.
Fleming left-hand rule:

Stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular (figure 13.22). If the first (fore) finger points in the direction of magnetic field and second (middle) finger points in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Note : Experimentally it is found that, when direction of current through the rod and direction of the magnetic field are perpendicular to each other, the force exerted on the rod is perpendicular to both of them.

(iii) Fleming’s right-hand rule : Direction of induced current in a conductor can be known with the help of Fleming’s right-hand rule.

Stretch the thumb, forefinger and middle finger of right hand so that they are perpendicular to each other. If forefinger indicates the direction of the magnetic field and thumb shows the direction of motion of conductor, then the middle finger will show the direction of induced current.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of the brushes?
Answer:
Principle : When a coil is rotated properly in a magnetic field, electromotive force is induced in it. As a result, a current flows in the circuit containing the coil.
OR
When a coil is rotated in a magnetic field, electric current is induced in the circuit containing the coil.

Labelled diagram and working : An electric generator is a device that converts mechanical energy into electrical energy. Its working is based on electromagnetic induction.

Construction :

• Figure 13.30 (a) shows the construction of an AC generator. It consists of a rectangular coil ABCD placed between the two poles of a permanent magnet.
• Two ends of this coil are connected to the two metal (copper) rings R₁ and R₂. The inner side of these rings are insulated.
• The two conducting stationary brushes R₁ and R₂ are kept pressed separately on the rings R₁ and R₂ respectively.
• The two rings R₁ and R₂ are internally attached to an axle. The axle may be mechanically rotated from outside to rotate the coil inside the magnetic field.
  
• Outer ends of the two brushes are s connected to the galvanometer to show the flow of current in the given external circuit.

Working:
(1) Suppose, the axle attached to the two rings is rotated such that the arm AB moves up and the arm CD moves down in the magnetic field produced by the permanent magnet. Then the coil ABCD rotates clockwise in the arrangement shown in the figure. By applying Fleming’s right- hand rule, we find that the induced currents are set up in these arms along the directions AB and CD. This means that the current in external circuit flows from brush B₂ to B₁.

[If the coil is made of a larger number of turns, the current generated in each turn adds up to give a large current through the coil.]

(2) After half a rotation, arm CD starts moving up and AB moving down. As a result, the directions of the induced currents in both the arms change, giving rise to the net induced current in the direction DCBA. This means that the current in the external circuit now flows from brush B₁ to B₂.

(3) Thus, after every half rotation the direction of the current in the respective arms changes. Such a current, which changes direction after equal intervals of time, is called an alternating current (abbreviated as AC).

The function of the brushes in an electric generator : Brushes are used to transmit current induced externally from coil ABCD to the external circuit.

Question 17.
When does an electric short circuit occur?
Answer:
An electric short circuit takes place when the live wire and neutral wire of the electric supply line touch each other directly or indirectly via a conducting wire. This occurs when the plastic insulation of the wires gets torn or there is a fault in the electrical appliance.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
Function of an earth wire : The function of an earth wire is to provide a low-resistance conducting path to the current leaking from an electrical appliance to the earth and thereby prevent electric shock to the user of the electrical appliance.

Therefore, it is necessary to earth metallic appliances such as an electric press, toaster, table fan, refrigerator, etc.

Sometimes accidently, the insulation of the wires connected to an electrical appliances may melt resulting in a current through the metallic casing of the appliance. A person touching such an appliance may get a severe electric shock which can be fatal. To avoid this, the current should be diverted to the earth.

Moreover, due to very low resistance (almost nil) offered by the earth wire, the current in the circuit rises to a very high value, thereby melting fuse in that circuit and cutting off its electric supply.

Thus, earthing the electrical appliances properly, saves us from possible electric shock.

Jharkhand Board Class 10 Science Magnetic Effects of Electric Current InText Questions and Answers

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
A compass needle is a small bar magnet with one north pole and the other south pole. When a compass is brought near a bar magnet, compass needle gets deflected due to the forces acting on its pole due to the magnetic field of the bar magnet.

Note: When compass needle is brought near a bar magnet it is acted upon by the force due to the magnetic field of the earth as well as that due to the bar magnet. Therefore, it gets deflected and finally comes to rest in the direction of the net force.

Question 2.
Draw magnetic field lines around a bar magnet.
Answer:

Question 3.
List the properties of magnetic field lines.
Answer:
(1) The magnetic field lines emerge from north pole and merge at the south pole outside the magnet, while inside the magnet the direction of field lines is from its south pole to its north pole.
Thus, the magnetic field lines are closed and continuous curve.

(2) The magnetic field lines are crowded near the pole where the magnetic field is strong and are far apart near the middle of the magnet and far from the magnet where the magnetic field is weak.

(3) The magnetic field lines never intersect each other because if they do so, there would be two directions of magnetic field at that point which is absurd.

(4) In case the field lines are parallel and equidistant, these represent a uniform magnetic field.

Important note : The relative strength of the magnetic field is shown by the degree of closeness of the field lines.

Question 4.
Why don’t two magnetic field lines intersect each other?
Answer:
The direction of the magnetic field B at a point is obtained by drawing a tangent to the magnetic field line at that point.

If two magnetic field lines intersect each other, it would mean that there are two directions of the magnetic field at the point of intersection, which is not possible because magnetic field is a vector, at a given point in space, it can have only one direction, (or the resultant force on a pole (north / south) at any point can only be in one direction.)

Question 5.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the Right-hand thumb rule to And out the direction of the magnetic field inside and outside the loop.
Answer:
Using the Right-hand thumb rule, the direction of the magnetic field inside and outside the circular loop of wire carrying an electric current can be found. This is shown in the following figure:

The dotted magnetic field lines are perpendicular to the plane of the paper.
The front face of the loop behaves as the south pole and the back face, i.e., the face touching the plane of the table behaves as the north pole.

Question 6.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
A uniform magnetic field in a given region can be represented by straight, parallel equally spaced field lines.

Question 7.
Choose the correct option :
The magnetic field inside a long straight solenoid carrying current…
(a) is zero.
(b) decreases as we move towards its end.
(c) increases as we move towards its end.
(d) is the same at all points.
Answer:
(d) is the same at all points.
Hint: The magnetic field inside a long straight current-carrying solenoid is uniform which is represented by straight parallel equally spaced lines. Thus, it is the same at all the points.

Question 8.
Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:
(c) velocity and
(d) momentum
Hint: The proton is a charged particle. Whenever a charged particle moves in a magnetic field except in the direction of the magnetic field and opposite to the direction of the magnetic field, a magnetic force exerts on it. As a result its velocity and momentum (mass x velocity) both change.

Note: The magnetic force is perpendicular to the velocity of the charged particle (here, the proton). Therefore, only the direction of the velocity of the particle changes, i.e., there is no change in magnitude of the velocity (speed) of the particle. The mass does not change.

Question 9.
In Activity 13.7. how do we think the displacement of the rod AB will be affected if (i) the current in rod AB is increased; (ii) a stronger horse-shoe magnet is used; and (iii) the length of the rod AB is increased?
Answer:
Experiments have shown that, when a current-carrying conductor is placed in a magnetic field such that its length is perpendicular to the magnetic field, the force acting on the conductor and hence its displacement (as the conductor is initially at rest) is proportional to…

• the current in the conductor
• the strength of the magnetic field
• the length of the conductor

(i) When the current in the rod AB is increased, then more force will act on the rod and hence the displacement of the rod will also be more (in the same proportion).

(ii) If a stronger horse-shoe magnet is used, then the strength of magnetic field will increase leading to a greater force on the rod. Due to the greater force, the displacement of the rod will also be more (in the same proportion.).

(iii) If the length of the rod AB is increased, then more force will act on the rod and hence the displacement of the rod will also be more (in the same proportion).

Question 10.
A positively charged particle (alpha particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is…
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(d) upward
Hint:

• Here, the positively charged particle (alpha particle) is initially moving towards the west, so the direction of the corresponding conventional current (current opposite to electron-current) is towards the west.
• The deflection of alpha particle is towards north. This shows that the magnetic force acts on it towards the north (direction).
• Hence, here (a) direction of current I is towards west and (b) direction of force F is towards north
  
• Let us now hold the first (fore) finger, middle finger and thumb of your left hand at right angles to one another.
• Adjust the left hand in such a way that your middle finger points towards the west (in the direction of the current) and the thumb points towards the north (in the direction of the force.)

In this case, your first (fore) finger points upwards [shown by the symbol G], Since the direction of the forefinger gives the direction of the magnetic field, the magnetic field would be in the upward direction, [see figure 13.23 (a) or 13.23 (b)].

Question 11.
State Fleming’s left-hand rule.
Answer:
In this case, Fleming’s left hand rule is used to find the direction of the force acting on the conductor.

Fleming left-hand rule:

Stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular (figure 13.22). If the first (fore) finger points in the direction of magnetic field and second (middle) finger points in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Note : Experimentally it is found that, when direction of current through the rod and direction of the magnetic field are perpendicular to each other, the force exerted on the rod is perpendicular to both of them.

Question 12.
What is the principle of an electric motor?
Answer:
When a rectangular coil carrying a current is placed in a magnetic field, then forces, equal in magnitude and opposite in direction, act on its two parallel sides, perpendicular to their lengths, due to which the coil rotates continuously.

Question 13.
What is the role of the split rings s in an electric motor?
Answer:
The split rings act as a commutator and its function is to reverse the direction of the current flowing through the coil after every half rotation of the coil.

Due to reversing of the current, the direction of rotating couple (i.e., torque) remains unchanged and the coil continues to rotate in the same direction.

Question 14.
Explain different ways to induce a current in a coil.
Answer:

• A current can be induced in a coil by moving a magnet towards or away from it or by moving the coil towards or away from the magnet.
• A current can be induced in a coil by changing the current in the coil placed near it.
• A current can be induced in a coil by moving a coil properly in a non-uniform magnetic field or by changing a magnetic field around steady coil by some means.
• A current can be induced in a coil by rotating it properly in a magnetic field or by rotating a magnet properly placed near the coil.

[Note : The magnet used may be a bar magnet or a current-carrying conductor.]

Question 15.
State the principle of an electric generator.
Answer:
The working of an electric generator is based upon the principle of electromagnetic induction.
Principle : An electric current produced in a closed circuit or coil by a changing magnetic field is called an induced current. This phenomenon is called electromagnetic induction.

Question 16.
Name some sources of direct current.
Answer:
Electrochemical dry cells, batteries, DC generators, solar cells, etc. are sources of direct current (DC).

Question 17.
Which sources produce alternating current?
Answer:
AC generators (or powerhouse generators), car alternators, bicycle dynamos are sources of alternating current (AC).

Question 18.
Choose the correct option :
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each…
(a) two rotations
(b) one rotation
(c) half rotation
(d) one-fourth rotation
Answer:
(c) half rotation
Hint: After each half rotation, the direction of motion of the two parallel sides of the coil (AB and CD) is reversed and so the direction of the induced current changes once in each half rotation.

Question 19.
Name two safety measures commonly used in electric circuits and appliances.
Answer:

1. A safety fuse of proper rating : It prevents damage to the appliances and circuit due to overloading.
2. An earth wire : It prevents possible electric shock when the live wire accidentally touches the metallic body of an appliance.

Question 20.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
The current drawn by this electric oven,
I = \(\frac { P }{ V }\)
= \(\frac { 2000 W }{ 220 V }\)
= 9.09 A
Since the current rating of the circuit is 5 A. the fuse used in this circuit is only of 5 A capacity.

Hence, when the oven is switch ON, the fuse wire (rated 5 A) will get heated too much, melt and break the circuit. This will save the electric oven from getting damaged.

[If fuse of current rating more than 9.09 A is used in the circuit or if no fuse has been put in the circuit, there may be a fire.]

Question 21.
What precaution should be taken to avoid the overloading of domestic electric circuits?
Answer:
To avoid overloading of domestic electric circuits, the following precautions should be taken :

• Wires used for carrying current should be of proper current rating.
• Two separate circuits should be used, one of 5 A current for lighting electric bulbs, tubes, TV, etc. and another of 15 A for higher current rating appliances such as heating appliances, AC, etc.
• Parallel circuits should be used and each circuit should have a fuse of proper rating.
• Too many higher power rating electrical appliances such as electric iron, geyser, air-conditioner, etc. should not be switched on at the same time.
• Too many electrical appliances should not be operated on a single socket simultaneously.
• Wires should be replaced by new wires of proper rating and good insulation after every 5-6 years.
• PVC of good quality should be used.

Activity 13.1 [T. B. Pg. 223]

To show that the magnetic field is produced s due to electric current.

Procedure:
1. Take a straight thick copper wire and place it between points X and Y in an electric circuit, as shown in figure 13.1. Wire XY is kept perpendicular to the plane of the paper.

2. Horizontally place a small compass near this copper wire.
See the position of its needle.

3. Pass the current through the circuit by inserting the key into the plug.

• Observe the change in the position of the compass needle.
• What does it indicate? (or What does it mean?)

Observation:
1. When no current flows in the straight thick copper wire (i.e., when plug key K is open), the compass needle (i.e., magnetic needle) remains stationary in the geographical north- south direction of the earth.

2. On passing the current through the copper wire XY (i.e., conducting wire), by inserting the key into the plug, compass needle is deflected.
[Any change in the direction of current through copper wire will show a variation in deflection.]

3. This indicates (shows) that the electric current through copper wire (the conductor) has produced a magnetic effect. It means that magnetic field is produced around the (copper) wire.

Conclusion :
Magnetic field is produced due to electric current. Thus, we can say that +electricity and ++magnetism are linked to each other.

Note: The ‘magnetic effect of electric current’ means that an electric current flowing in a conducting wire produces a magnetic field around it.

Activity 13.2 [T. B. Pg. 224]

To understand the magnetic field produced by a bar magnet.
OR
To obtain the magnetic field lines around a bar magnet.

Procedure :
1. Fix a sheet of white paper on a drawing board using some adhesive material.

2. Place a bar magnet in the centre of it.

3. Sprinkle some iron filings uniformly around a bar magnet as shown in figure 13.3. A salt- sprinkler may be used for this purpose.

4. Now tap the board gently.

• What do you observe?
• Why do the iron filings arrange in such a pattern?
• What does this pattern demonstrate?

Observation :
1. The iron filings arrange themselves in a pattern as shown in figure 13.3.

2. A bar magnet exerts its influence in the region surrounding it. Therefore, the iron filings experience a magnetic force. This force makes iron filings arrange in a specific (particular) pattern.

3. The region surrounding a magnet, in which the force of the magnet can be detected, is said to have a magnetic field.
The lines along which the iron filings align themselves represent magnetic field lines.

Conclusion:
A bar manget possesses a magnetic field which can be detected by sprinkling iron filings around it.

Activity 13.3 [T. B. Pg. 224]

To draw magnetic field lines of a bar magnet using a compass.

Procedure:
1. Take a small compass and a bar magnet.

2. Place the magnet on a sheet of white paper fixed on a drawing board, using some adhesive material.

3. Mark the boundary of the magnet.

4. Place a compass near the north pole of the magnet. How does it behave?

5. Mark the positions of two ends of the needle.

6. Now move the needle to a new position such that its south pole occupies the position previously occupied by its north pole.
In this way, proceed step by step till you reach the south pole of the magnet as shown in figure 13.4.

7. Join the points marked on the paper by a smooth curve. This curve represents a magnetic field line.

8. Repeat the above procedure and draw as many lines as you can. You will get a pattern shown in figure 13.5.
These lines represent the magnetic field around the magnet.
These are known as magnetic field lines.

• Observe the deflection in the compass needle as you move it along a field line. The deflection increases as the needle is moved towards the poles.

Observation:
1. The south pole of the needle points towards the north pole of the magnet. The north pole s of the compass needle is directed away from the north pole of the magnet.

2. Magnetic field is a quantity that has both direction and magnitude. The direction of the magnetic field is taken to be direction in which a north pole of the compass needle moves inside it.

Therefore, the field lines emerge from the north pole and merge at the south pole of a magnet. Inside the magnet, the direction of the field lines is from its south pole to its north pole. Thus, the magnetic field lines are closed curves.

Conclusion:

1. The region surrounding a magnet has a magnetic field, i.e., a bar magnet possesses magnetic field.
2. The direction of magnetic field is the direction in which the north pole of the compass needle move / deflect.
3. The field is stronger where the field lines are crowded and weaker where the field lines are farther away from each other.
4. The deflection of the needle increases as the needle is moved towards the poles.

Activity 13.4 [T. B. Pg. 226]

To show that the direction of the magnetic field due to a current depends on the direction of the current.

Procedure:
1. Take a long straight copper wire, two or three cells of 1.5 V each, and a plug key. Connect all of them in series as shown in figure 13.6 (a).

2. Place the straight wire parallel to and over a compass needle.

3. Plug the key in the circuit.
Observe the direction of deflection of the north pole of the needle.

4. Replace the cell connections in the circuit as shown in figure 13.6 (b). This would result in the change of the direction of current through the copper wire, that is, from south to north.
Observe the change in the direction of deflection of the needle.

Observation:
1. If the current flows from north to south as shown in figure 13.6 (a), the north pole of the compass needle moves towards the east.

2. Now, when the current flowing through the copper wire is reversed, the compass needle aligns / moves in the opposite direction i.e., towards the west as shown in figure 13.6 (b).

It means that the direction of the magnetic field due to a current is reversed, when the direction of the current is reversed.

Conclusion :
The direction of the magnetic field due to a current depends on the direction of the current.

Activity 13.5 [T. B. Pg. 226]

To study the pattern of the magnetic field (lines) due to a current through a straight conductor.

Procedure:
1. Take a battery (12 V), a variable resistance (or a rheostat), an ammeter (0-5 A), a plug key, connecting wires and a long straight thick copper wire.

2. Insert the thick wire through the centre, normal to the plane of a rectangular cardboard. Take care that the cardboard is fixed and does not slide up or down.

3. Connect the copper wire vertically between the points X and Y, as shown in figure 13.7 (a), in series with the battery and a plug key.

4. Sprinkle some iron filings uniformly on the cardboard. (You may use a salt-sprinkler for this purpose.)

5. Close the key so that a current flows through the wire. Ensure that the copper wire placed between the points X and Y remains vertically straight.

6. Keep the variable of the variable resistance (or rheostat) at a fixed position and note the current through the ammeter.

• Gently tap the cardboard a few times. Observe the pattern of the iron filings.
• What do these concentric circles represent?
• How can the direction of the magnetic field be found?
  (Place a compass at a point (say P) over a circle. Observe the direction of the needle.)
• Does the direction of the magnetic field lines get reversed if the direction of the current through the straight copper wire is reversed? Check it.
  
• What happens to the deflection of the compass needle placed at a given point if the current in the copper wire is changed?
• (To see this, vary the current in the wire using the variable resistance.)
• What happens to the deflection of the needle if the compass is moved away from the copper wire keeping the current through the wire remains tire same?
• (To see this, now place the compass needle at a farther point from the conducting wire [say at point Q (figure 13.7 (b)]).
• What change do you observe?

Observation:

• It is found that the iron filings align themselves forming a pattern of concentric circles around the copper wire (figure 13.7 (c)).
• These concentric circles represent the (pattern of) magnetic field lines.
• The direction shown by the north pole of the compass needle gives the direction of the magnetic field lines at point P. In figure 13.7 (a), it is shown by an arrow.
• Yes. The direction of the magnetic field lines get reversed if the direction of the current through the straight copper wire is reversed (see figure 13.7 (b)).
• We find that the deflection of the needle changes, when the current is changed.

If the current is increased, the deflection increases and if the current is decreased, the deflection decreases.

• When the compass is moved away from the wire, the deflection of the needle decreases.
• When the compass is moved away from the wire, it is observed that the concentric circles representing the magnetic field around a current-carrying straight wire become larger and larger.
• In other words, as we move away from the wire, the radius of a circle representing the magnetic field line increases.

Conclusion :
(1) Pattern of the magnetic field produced due to a straight conductor carrying current is concentric circles, which are nothing but the magnetic field lines of the produced magnetic field.

(2) Direction of magnetic field at a point is in the direction of tangent drawn at that point on particular circle, which is nothing but the direction of north pole of compass needle at that point.

(3) Direction of magnetic field lines get reversed if direction of current through the straight copper wire is reversed, i.e., it depends on the direction of the current.

(4) Magnitude of the magnetic field produced at a given point increases as the current through the wire increases.
i.e., Magnetic field B ∝ Current I.

(5) Magnitude of the magnetic field produced by a given current in the conductor decreases as the distance from wire increases.
i.e., Magnetic field B ∝ \(\frac{1}{\text { Distance from the wire }}\)

Activity 13.6 [T. B. Pg. 229]

To study the magnetic field produced near a circular coil carrying current.

Procedure:

• Take a rectangular cardboard having two holes. Insert a circular coil having a large number of turns through them, normal to the plane of the cardboard.
• Connect the ends of the coil in series with a battery, a plug key and a rheostat, as shown in figure 13.13.
• Sprinkle iron filings uniformly on the cardboard.
• Plug the key.
• Tap the cardboard gently a few times.
• Note the pattern of the iron filings that emerges on the cardboard.
  

Observation :
The iron filings arrange themselves in the pattern as shown in figure 13.13.

Conclusion:

• The magnetic field at the centre of the coil is nearly uniform.
• The magnetic field lines near the coil are circular and concentric.
• At the centre, as the field lines are crowded, the strength of the magnetic field is maximum.

Activity 13.7 [T. B. Pg. 230]

To understand the force experienced by a current-carrying small aluminium (conducting) rod placed in a magnetic field.

Procedure:
1. Take a small aluminium rod AB (of length about 5 cm.) Using two connecting wires suspend it horizontally from a stand, as shown in figure 13.20.

2. Place a strong horse-shoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. For this put the north pole of the magnet vertically below and the south pole vertically above the aluminium rod (see figure 13.20).

3. Connect the aluminium rod in series with a battery, a plug key and a rheostat.

4. Now pass a current through the aluminium rod from end B to end A.

• What do you observe?
• Reverse the direction of current flowing through the rod and observe the direction of its displacement.
  
• Why does the rod get displaced?
• Change the direction of magnetic field by interchanging the two poles of the magnet to vertically downwards and observe the direction of the force / displacement of the current-? carrying rod.

Observation :
1. It is observed that the rod is displaced to the left.

2. The direction of the displacement of the rod is reversed, i.e., now it is to the right, on reversing the direction of the current flowing through the rod.

It suggests that the direction of the force is also reversed when the direction of the current through the conductor (rod) is reversed.

3. The displacement of the rod is due to the force exerted on the current- carrying aluminium rod when it is placed in a magnetic field.

4. The direction of the force acting on the current-carrying rod and hence that of the displacement of the rod (as the rod is initially ; at rest) gets reversed on changing the direction of the magnetic field to vertically downwards.

Conclusion :

• A current-carrying conductor (here a rod) experiences a force when placed in a magnetic field.
• The direction of the force and hence that of the displacement of the conductor (rod) depends upon the direction of the current and the direction of the magnetic field.

Notes :

• Here, the displacement of the rod has the same direction as that of the force acting on the rod because the rod is initially at rest. Recall the kinematical equations.
• The magnetic force also depends upon the length of the rod in the magnetic field.

Activity 13.8 [T. B. Pg. 233]

To study the phenomenon of electromagnetic induction.

Procedure:
1. Take a coil of wire AB having a large number of turns.

2. Connect the ends of the coil to a galvanometer as shown in figure 13.25.

3. Take a strong bar magnet and move its north pole towards the end B of the coil.
Do you find any change in the galvanometer needle?

4. Now withdraw the north pole of the magnet away from the coil.
What do you observe?

5. Place the magnet stationary at a point near the coil, keeping its north pole towards the end B of the coil.

6. Move the coil to right and later to left.
What do you observe?

7. When the coil is kept stationary with respect to the magnet.
What do you observe?

8. Now, move the south pole of the magnet towards the end B of the coil.

• What do you observe?
• When the coil and the magnet both are stationary what do you observe?
• What do you conclude from this activity?
  

Observation :
1. There Is a momentary deflection in the needle of the galvanometer, say to the right.
This indicates the presence of a current in the coil AB. The deflection becomes zero the moment the motion of the magnet stops.

2. When the north pole of the magnet is moved away from the coil, the galvanometer needle is deflected to the left, showing that the current is now set up in the direction opposite to the first.

3. We see that the galvanometer needle deflects to the right when the coil is moved towards the north pole of the magnet. Similarly, the needle moves to the left when the coil is moved away.

4. When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer needle drops to zero.

5. If you move the south pole of the magnet ‘c towards the end B of the coil, the deflections in the galvanometer would just be opposite > to those in the previous case.

6. When the coil and magnet both are stationary, there is no deflection in the galvanometer.

Conclusion :
Whenever there is a relative motion between s a magnet and coil, potential difference is produced between two ends of the coil, which sets up an electric current in the circuit.

This potential difference is called the induced potential difference and the corresponding current is called the induced current.

Activity 13.9 [T. B. Pg. 235]

To study the phenomenon of electromagnetic induction.

Procedure:
1. Take two different coils of copper wire having large number of turns (say 50 and loo turns respectively). Insert them over a non-conducting cylindrical roll, as shown in figure 13.27 (You may use a thick paper roll for this purpose.)

2. Connect the coil 1. having larger number of turns, in series with a battery and a plug key. Also connect the other coil 2 wIth a galvanometer as shown.

3. Plug in the key. Observe the galvanometer.
Is there a deflection in its needle?

4. Disconnect coil 1 from the battery.
What do you observe?

Observation :
1. When the key is plugged, the needle of the galvanometer is instantly deflected to one side and then quickly returns to zero, indicating a momentary current in coil 2.

2. When coil 1 is disconnected from the battery, the needle of the galvanometer momentarily moves, but to the opposite side, i.e., now the current flows in the opposite direction in coil 2.

In short in this activity, we observe that as soon as the current in coil 1 reaches either a steady value or becomes zero, the galvanometer connected to coil 2 shows no deflection.

Conclusion :
(a) A potential difference is induced in coil 2 whenever the electric current through the coil 1 changes with time.

(b) As the current in coil 1 (which is called Primary coil) changes, the magnetic field associated with it also changes. Then the magnetic field lines around coil 2 (which s is called Secondary coil) also change.

Thus, the change in magnetic field lines associated with the secondary coil is the cause of induced electric current in it. s

This process, by which a changing magnetic field in a conductor induces a current in another conductor, is called electromagnetic induction.

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 9 त्रिकोणमिति का अनुप्रयोग Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 9 त्रिकोणमिति का अनुप्रयोग

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
क्षैतिज तल पर किसी निश्चित बिन्दु से एक मीनार के शिखर का उन्नयन कोण 30° है। यदि प्रेक्षक 20 मीटर मीनार की ओर चलता है, तो मीनार के शिखर का उन्नयन कोण 15° बढ़ जाता है। मीनार की ऊँचाई ज्ञात करो ।
हल:
माना AB एक मीनार है जिसकी ऊँचाई h मीटर है। बिन्दु C से मीनार के शिखर का उन्नयन कोण 30° है। बिन्दु C से 20 मीटर मीनार की तरफ चलने पर मीनार के शिखर का उन्नयन कोण 15° बढ़ जाता है।
अर्थात् ∠ACB = 30° तथा ∠ADB = 45°

समकोण ΔABD में,
tan 45° = \(\frac{A B}{B D}\)
⇒ 1 = \(\frac{h}{B D}\)
⇒ BD = h मी. …..(1)
समकोण ΔABC में tan 30° = \(\frac{A B}{B C}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{20+B D}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{20+h}\)
[ समी. (1) से BD = h]
20 + h = \(\sqrt{3}\)h
20 = \(\sqrt{3}\)h – h
20 = h(\(\sqrt{3}\) – 1)
h = \(\frac{20}{\sqrt{3}-1}\)
h = \(\frac{20(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\)
h = \(\frac{20(\sqrt{3}+1)}{3-1}\)
h = 10(\(\sqrt{3}\) + 1) मीटर
अतः मीनार की ऊँचाई = 10 (\(\sqrt{3}\) + 1) मीटर ।

प्रश्न 2.
एक झील में पानी के तल से 20 मीटर ऊंचे बिन्दु A से, एक बादल का उन्नयन कोण 30° है। झील में बादल के प्रतिबिम्ब का A से अवनमन कोण 60° हैं। A से बादल की दूरी ज्ञात कीजिए।
हल:
माना कि BD पानी का तल है। B से 20 मीटर ऊँचे बिन्दु A से एक बादल (P) का उन्नयन कोण 30° है। माना झील में बादल के प्रतिबिम्ब की स्थिति C है तथा प्रतिबिम्ब इस प्रकार है कि झील में बादल के प्रतिबिम्ब का अवनमन कोण 60° है। माना कि PQ = h मी है।

अतः ∠PAQ = 30° तथा ∠QAC = 60°
QD = AB = 20 मी
CD = PD = (20 + h) मी
QC = 20 + h + 20
= (40 + h) मी
BD = AQ
समकोण ΔPAQ में
tan 30° = \(\frac{P Q}{A Q}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{A Q}\)
⇒ AQ = h\(\sqrt{3}\) मीटर …..(1)
समकोण ΔAQC में
tan 60° = \(\frac{Q C}{A Q}\)
⇒ \(\sqrt{3}=\frac{40+h}{h \sqrt{3}}\)
[समीकरण (1) का प्रयोग करने पर]
⇒ 3h = 40 + h
⇒ 3h – h=40
⇒ 2h = 40
⇒ h = \(\frac{40}{2}\) = 20 मी
पुन: समकोण ΔPAQ में,
sin 30° = \(\frac{P Q}{A P}\)
⇒ \(\frac{1}{2}=\frac{h}{A Q}\)
⇒ \(\frac{1}{2}=\frac{20}{A Q}\) मीटर
AQ = 20 × 2
अतः A से बादल की दूरी = 40 मीटर ।

प्रश्न 3.
4000 मीटर की ऊँचाई पर उड़ते हुए वायुयान के ठीक नीचे जिस क्षण दूसरा वायुयान आता है, उसी क्षण क्षैतिज तल पर किसी बिन्दु से इन वायुयानों के उन्नयन कोण क्रमश: 60° और 45° हैं। उस क्षण पर दोनों वायुयानों के बीच की ऊर्ध्वाधर दूरी ज्ञात कीजिए।
हल:
माना दो वायुयान A और B हैं जिनके बीच की ऊर्ध्वाधर दूरी AB = h मीटर
क्षैतिज तल पर स्थित बिन्दु D से इन वायुयानों A और B के उन्नयन कोण क्रमश: 60° और 450 हैं।
अर्थात् ∠ADC = 60° तथा ∠BDC = 45°
AC = 4000 मीटर
BC = AC – AB = (4000 – h) मीटर
समकोण ΔBCD में,
tan 45° = \(\frac{B C}{C D}\)

1 = \(\frac{4000-h}{x}\)
x = 4000 – h …..(i)
पुन: समकोण ΔACD में,
tan 60° = \(\frac{A C}{C D}\)
\(\sqrt{3}\) = \(\frac{4000}{x}\)
⇒ x\(\sqrt{3}\) = 4000
⇒ x = \(\frac{4000}{\sqrt{3}}\) …..(ii)
समीकरण (i) से x का मान रखने पर,
4000 – h = \(\frac{4000}{\sqrt{3}}\)
⇒ h = 4000 – \(\frac{4000}{\sqrt{3}}\)
⇒ h = 4000\(\left(1-\frac{1}{\sqrt{3}}\right)\)
⇒ h = 4000\(\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)\)
⇒ h = 4000\(\frac{4000 \times 0.732}{1.732}\)
⇒ h = \(\frac{2928}{1.732}\) = 1690.53 मीटर
अतः दोनों वायुयानों के बीच ऊर्ध्वाधर दूरी 1690.53 मीटर होगी।

प्रश्न 4.
धरातल के एक बिन्दु से एक हवाई जहाज का उन्नयन कोण 60° है। 15 सेकण्ड की उड़ान के पश्चात्, उन्नयन कोण 30° का हो जाता है। यदि हवाई जहाज एक निश्चित ऊँचाई 1500\(\sqrt{3}\) मीटर पर उड़ रहा हो, तो हवाई-जहाज की गति किमी / घंटा में ज्ञात कीजिए ।
हल:
माना P और Q हवाई जहाज की दो स्थितियाँ है। माना ABC एक क्षैतिज रेखा है जो A से जाती है।
दिया है कि स्थिति P और Q से, A बिन्दु से हवाई जहाज द्वारा बने उन्नयन कोण 30° तथा 60° है।
∴ ∠PAB = 60° और ∠QAC = 30°
इसी तरह, PB = QC = 1500\(\sqrt{3}\) मीटर

ΔABP में,
tan 60° = \(\frac{B P}{A B}\)
\(\sqrt{3}\) = \(\frac{1500 \sqrt{3}}{A B}\)
AB = 1500 मीटर
ΔACQ में,
tan 30° = \(\frac{Q C}{A C}\)
\(\frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{A C}\)
⇒ AC = 1500 × 3 = 4500 मीटर
∴ BC = AC – AB = 4500 – 1500
= 3000 मीटर
इस प्रकार हवाई जहाज 15 सेकण्ड में 3000 मीटर की दूरी तय करता है।
∴ हवाई जहाज की चाल = \(\frac{3000}{15}\) = 200 मी/से
= \(\frac{200 \times 60 \times 60}{1000}\)
= 720 किमी/ घण्टा
अतः हवाई जहाज की चाल 720 किमी/घण्टा है।

प्रश्न 5.
10 मीटर ऊँचे भवन के शिखर से एक टॉवर के शिखर का उन्नयन कोण 60° है और इसके पाद का अवनमन कोण 45° है। टावर की ऊँचाई ज्ञात कीजिए।
हल:
माना AB एक टॉवर है उसी धरातल में एक भवन CD है जिसकी ऊँचाई 10 मीटर है।
टॉवर के शिखर का उन्नयन कोण और पाद का अवनमन कोण क्रमश: 60° और 45° है।
अर्थात् ∠ACE = 60°
और ∠ECB = 45°
BD || CE, CD || BE
CD = BE = 10 मी.
अब समकोण त्रिभुज CBD में,
tan 45° = \(\frac{C D}{D B}\)
\(1=\frac{10}{D B}\)
DB = 10 मी.
CE = DB = 10 मी.
पुनः समकोण त्रिभुज AEC में,

tan 60° = \(\frac{A E}{C E}=\frac{A E}{10}\)
AE = 10\(\sqrt{3}\) मी
अतः टॉवर AB की ऊँचाई
= AE + EB = 10\(\sqrt{3}\) + 10
= 10 (\(\sqrt{3}\) + 1) मीटर
अतः टॉवर AB की ऊँचाई = 10(\(\sqrt{3}\) + 1) मीटर

प्रश्न 6.
एक नदी के पुल के एक बिन्दु से नदी के सम्मुख किनारों के अवनमन कोण क्रमश: 30° और 45° है। यदि पुल किनारों से 4 मी की ऊँचाई पर है तो नदी की चौड़ाई ज्ञात कीजिए।
हल:
दिया है, नदी से पुल की ऊँचाई
AC = 4 मी
BC = x, CD = y

∴ समकोण ΔABC में,
\(\frac{A C}{B C}\) = tan 45°
⇒ \(\frac{4}{x}\) = 1 ⇒ x = 4 मी
पुन: समकोण ΔACD में,
\(\frac{A C}{C D}\) = tan 30°
⇒ \(\frac{4}{y}=\frac{1}{\sqrt{3}}\)
y = 4\(\sqrt{3}\) मी …..(ii)
समी. (i) व (ii) से,
∴ नदी की चौड़ाई (x + y)
= 4\(\sqrt{3}\) + 4
= 4(\(\sqrt{3}\) + 1)
= 10.92 मी ।

प्रश्न 7.
एक मीनार के पाद से एक भवन के शिखर का उन्नयन कोण 30° है और भवन के पाद से मीनार के शिखर का उन्नयन कोण 60° है। यदि मीनार के ऊँचाई 48 मी है तो भवन की ऊंचाई ज्ञात कीजिए।
हल:
माना AB एक मीनार है जिसके ऊँचाई 48 मी है तथा CD एक भवन है जिसकी ऊँचाई h मी है। दिया है, मीनार के पाद से भवन के शिखर का उन्नयन कोण 30° तथा भवन के पाद से मीनार के शिखर का उन्नयन कोण 60° है।
अर्थात् ∠CBD = 30°
तथा ∠ADB = 60°

समकोण ΔCDB में,
tan 30° = \(\frac{C D}{B D}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{B D}\)
⇒ BD = h\(\sqrt{3}\) मी
पुन: समकोण ΔABD में,
tan 60° = \(\frac{A B}{B D}\)
\(\sqrt{3}\) = \(\frac{48}{h \sqrt{3}}\)
h = \(\frac{48}{\sqrt{3} \times \sqrt{3}}=\frac{48}{3}\)
= 16 मी
अतः भवन की ऊँचाई 16 मी है।

प्रश्न 8.
आँधी आने पर एक पेड़ टूट जाता है और टूटा हुआ भाग इस तरह मुड़ जाता है कि पेड़ का शिखर जमीन को छूने लगता है और इसके साथ 60° का कोण बनाता है। पेड़ के पाद- बिंदु की दूरी जहाँ पेड़ का शिखर जमीन को छूता है, 3 मी है। पेड़ की ऊंचाई ज्ञात कीजिए।
हल:
माना आँधी से पहले पेड़ की लम्बाई BD है। आँधी के बाद पेड़ A स्थान से टूटकर पेड़ का शिखर जमीन पर C बिन्दु पर पड़ता है। टूटा हुआ भाग जमीन से 30° का कोण बनाता है।

∴ ∠ACB = 30°
समकोण ΔABC में,
tan 30° = \(\frac{A B}{B C}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{A B}{3}\)
⇒ AB = \(\frac{3}{\sqrt{3}}\) मी
पुन: समकोण ΔABC में,
cos 30° = \(\frac{B C}{A C}\)
⇒ \(\frac{\sqrt{3}}{2}=\frac{3}{A C}\)
⇒ AC = \(\frac{6}{\sqrt{3}}\) मी
पेड़ की कुल ऊँचाई = BD
= AB + AD
= AB + AC [∵ AD = AC]
= \(\left(\frac{3}{\sqrt{3}}+\frac{6}{\sqrt{3}}\right)\) मी
= \(\frac{9}{\sqrt{3}}\) मी = \(\frac{9 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
= \(\frac{9 \sqrt{3}}{3}\) मी
= 3\(\sqrt{3}\) मी

प्रश्न 9.
अमित जो कि समतल जमीन पर खड़ा है, अपने से 200 मी दूर उड़ते हुए पक्षी का उन्नयन कोण 30° पाता है। दीपक जो कि 50 मी ऊँचे भवन की छत पर खड़ा है, उसी पक्षी का उन्नयन कोण 45° पाता है। अमित और दीपक पक्षी के विपरीत दिशा में हैं। दीपक से पक्षी की दूरी ज्ञात कीजिए।
हल:
माना अमित बिन्दु C पर खड़ा है तथा AB चिड़िया की धरातल पर स्थित बिन्दु B से ऊँचाई है तथा माना दीपक बिन्दु D पर स्थित है, जहाँ DE भवन की ऊँचाई है।
दिया है, ∠ACB = 30°, ∠ADF = 45° तथा DE = 50 मी

अब, समकोण ΔABC में,
sin 30° = लम्ब / कर्ण
= \(\frac{A B}{A C}\)
\(\frac{1}{2}=\frac{A B}{200}\)
AB = 100 मी
समकोण ΔAFD में,
sin 45° = लम्ब / कर्ण
= \(\frac{A F}{A D}\)
(AB = AF + BF
100 = AF + 50
AF = 50 मी)
= \(\frac{1}{\sqrt{2}}=\frac{50}{A D}\)
AD = 50\(\sqrt{2}\) मी
अतः दीपक की चिड़िया से दूरी 50\(\sqrt{2}\) मी है।

प्रश्न 10.
एक ऊर्ध्वार मीनार क्षैतिज तल पर खड़ी है तथा उसके ऊपर एक 6 मी ऊँचा झंडा लगा है। तल के किसी बिन्दु से झंडे के पाद तथा शिखर के उन्नयन कोण क्रमश: 30° तथा 45° है। मीनार की ऊँचाई ज्ञात कीजिए। (\(\sqrt{3}\) = 1.732 लीजिए)
हल:
माना AB एक मीनार है तथा BC झंडा है। अब माना कि P भूमि पर एक ऐसा बिन्दु है, जिसका मीनार के शिखर का उन्नयन कोण ∠APB = 30° तथा मीनार पर स्थित झंडा का उन्नयन कोण ∠APC = 45° है तथा BC = 6 मी अब माना AB = h मी तथा PA = x मी

समकोण ΔPAB से,
cos 30° = \(\frac{P A}{P B}=\frac{x}{h}\)
\(\sqrt{3}=\frac{x}{h}\) (∵ cot 30° = \(\sqrt{3}\))
x = \(\sqrt{3}\)h …..(i)
समकोण ΔPAC से,
cot 45° = \(\frac{x}{h+6}=\frac{P A}{A C}\) (∵ cot 45° = 1)
x = h + 6 ….. (ii)
समीकरण (i) व (ii) से,
\(\sqrt{3}\)h = h + 6
\(\sqrt{3}\)h – h = 6
= \(\frac{6}{\sqrt{3}-1}\)
= \(\frac{6}{1.732-1}\)
= \(\frac{6}{0.732}\)
= \(\frac{6000}{732}\)
= 8.19 मी
अतः मीनार की ऊँचाई 8.19 मी है।

प्रश्न 11.
100 मी ऊंचे एक लाइट हाउस से दूर एक नाव को ले जाता हुआ व्यक्ति 2 मिनट में लाइट हाउस में शिखर का उन्नयन कोण को 60° से 30° बढ़लता हुआ पाता है। मीटर प्रति मिनट में नाव की चाल ज्ञात कीजिए। (\(\sqrt{3}\) = 1.732 लीजिए)
हल:
माना AB एक 100 मी ऊँचाई का लाइट हाउस है प्रारम्भ में व्यक्ति C बिन्दु पर था तथा 2 मिनट बाद D बिन्दु पर आता है।

यहाँ, ∠ACB = 60° तथा ∠ADB = 30° है। माना BC = x मी तथा BD = y मी है।
समकोण ΔABC में,
tan 60° = \(\frac{A B}{B C}\)
\(\sqrt{3}=\frac{100}{x}\)
x = \(\frac{100}{\sqrt{3}}\) ……..(i)
पुन: समकोण ΔABD में,
tan 30° = \(\frac{A B}{B D}\)
\(\frac{1}{\sqrt{3}}=\frac{100}{B D}\)
BD = 100\(\sqrt{3}\)
BC + CD = 100\(\sqrt{3}\)
x + y = 100\(\sqrt{3}\) …..(ii)
समीकरण (i) से, x = \(\frac{100}{\sqrt{3}}\) समीकरण (ii) में रखने पर,
\(\frac{100}{\sqrt{3}}\) + y = 100\(\sqrt{3}\)
y = \(100 \sqrt{3}-\frac{100}{\sqrt{3}}\)
= \(\frac{300-100}{\sqrt{3}}\)
y = \(\frac{200}{\sqrt{3}}\) मी
बिन्दु C से D तक जाने मैं नाव द्वारा लगा समय 2 मिनट है।
तथा CD = \(\frac{200}{\sqrt{3}}\)
अतः नाव को चाल = समय / दूरी
\(\frac{C D}{2}\) ⇒ \(\frac{y}{2}\)
= \(=\frac{200}{\sqrt{3} \times 2}\) (∵ y = \(\frac{200}{\sqrt{3}}\) मी)
= \(\frac{100}{\sqrt{3}}=\frac{100}{1.732}\)
= 57.73 मीटर / मिनट

प्रश्न 12.
एक मीनार के पाद से गुजरने वाली सीधी रेखा पर पाद से क्रमशः 4 मी तथा 16 मी की दूरियों पर दो बिंदु C व D स्थित हैं। यदि C व D से मीनार के शिखर के उन्नयन कोण एक-दूसरे के पूरक हों, तो मीनार की ऊँचाई ज्ञात कीजिए।
हल:
माना मीनार की ऊँचाई = h मीटर
ΔABC में, \(\frac{A B}{B C}\) = tan(90° – θ)
\(\frac{h}{4}\) = cot θ …..(i)

ΔABD में,
\(\frac{A B}{B D}\) = tan θ
\(\frac{h}{16}\) = tan θ …..(ii)
समीकरण (i) और (ii) का गुणा करने पर,
\(\frac{h}{4} \times \frac{h}{16}\) = cot θ × tan θ
\(\frac{h^2}{64}=1\)
[∵ cot θ × tan θ = \(\frac{1}{\tan \theta}\) × tan θ = 1]
⇒ h² = 64
⇒ h = 8 मी
अतः मीनार की ऊँचाई 8 मीटर है।

प्रश्न 13.
एक हवाई जहाज भूतल से ऊपर 300 मी की ऊँचाई पर उड़ रहा है। इस ऊँचाई पर उड़ते हुए हवाई जहाज से एक नदी के दोनों किनारों पर परस्पर विपरीत दिशाओं में स्थित दो बिंदुओं के अवनमन कोण क्रमशः 45° तथा 60° हैं। नदी की चौड़ाई ज्ञात कीजिए। [\(\sqrt{3}\) = 1.732 प्रयोग कीजिए]
हल:
माना हवाई जहाज A बिंदु पर नदी से 300 मीटर ऊँचाई पर है। C व D नहीं के विपरीत किनारों पर है।

समकोण ΔABC में,
\(\frac{B C}{A B}\) = cot 60°
⇒ \(\frac{x}{300}=\frac{1}{\sqrt{3}}\)
⇒ x = \(\frac{300}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= 100\(\sqrt{3}\) मी
= 100 × 1.732 = 173.2 मी
समकोण ΔABD में,
⇒ \(\frac{B D}{A B}\) = cot 45°
⇒ \(\frac{y}{300}\) = 1
⇒ y = 300
नदी को चौड़ाई = x + y
= 173.2 + 300
= 473.2 मी

प्रश्न 14.
समुद्र तल से 75 मी. ऊँचे लाइट हाउस के शिखर से देखने पर दो समुद्री जहाजों के अवनमन कोण 30° तथा 45° है। यदि दोनों जहाज लाइट हाउस की विपरीत दिशाओं में हो, तो दोनों जहाजों के बीच की दूरी ज्ञात कीजिए।
हल:
माना AB लाइट हाउस है।
जहाज क्रमश: बिन्दु C व D पर है।

समकोण ΔABC में,
⇒ \(\frac{A B}{B C}\) = tan 30°
⇒ \(\frac{75}{x}=\frac{1}{\sqrt{3}}\)
⇒ x = 75\(\sqrt{3}\) मी
समकोण ΔABD में,
⇒ \(\frac{A B}{B D}\) = tan 30°
⇒ \(\frac{75}{y}\) = 1
y = 75 मी
जहाजों के बीच की दूरी = x + y
= (75\(\sqrt{3}\) + 75) मी
= 75(\(\sqrt{3}\) + 1) मी

प्रश्न 15.
एक झील के पानी की सतह में 60 मी ऊँचाई पर स्थित एक बिन्दु से बादल का उन्नयन कोण 30° है, तथा झील के पानी में बादल का परछाई का अवनमन कोण 60° है। बादल की झील के पानी की सतह से ऊँचाई प्राप्त कीजिए।
हल:
ΔCMP में,
tan 30° = \(\frac{C M}{P M}\)
\(\frac{1}{\sqrt{3}}=\frac{h}{P M}\) या PM = \(\sqrt{3}\)h …….(i)

ΔPMC में,
tan 60° = \(\frac{C M}{P M}\)
= \(\frac{h+60+60}{P M}=\sqrt{3}\)
या PM = \(\frac{h+120}{\sqrt{3}}\) …..(ii)
समीकरण (i) और (ii) से,
\(\sqrt{3}\)h = \(\frac{h-120}{\sqrt{3}}\)
3h = h + 120
2h = 120 ⇒ h = 60 मी
पानी के तल से बादल की ऊँचाई = h + 60
= 60 + 60 = 120 मी

प्रश्न 16.
एक मीनार के एक ही ओर तथा इसके आध र से एक ही सरल रेखा में दो बिंदु A तथा B हैं। मीनार के शिखर से इन बिंदुओं अवनमन कोण क्रमश: 60° व 45° हैं। यदि मीनार की ऊँचाई 15 मी हो, तो इन बिंदुओं के बीच की दूरी ज्ञात कीजिए।
हल:
माना PT एक मीनार है।

समकोण ΔPTA में,
tan 60° = \(\frac{P T}{T A}\)
⇒ \(\sqrt{3}=\frac{15}{T A}\)
⇒ TA = \(\frac{15}{\sqrt{3}}\)
पुन: समकोण ΔPTB में,
tan 45° = \(\frac{P T}{T B}\)
⇒ 1 = \(\frac{15}{T B}\)
⇒ TB = 15 मी
बिन्दुओं A व B के बीच की दूरी
AB = TB – TA
= 15 – \(\frac{15}{\sqrt{3}}\) = 15\(\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)\) मी

प्रश्न 17.
एक नदी के एक किनारें पर खड़ा एक व्यक्ति, नदी के दूसरे किनारे पर खड़े एक वृक्ष के शिखर का उन्नयन कोण 60° पाता है जब वह किनारे से 30 मी दूर जाता है, तो वह उन्नयन कोण 30° पाता है। वृक्ष की ऊँचाई तथा नदी की चौड़ाई ज्ञात कीजिए। [\(\sqrt{3}\) = 1.732 प्रयोग कीजिए]
हल:
माना नदी के एक किनारे पर एक वृक्ष AB है तथा नदी के दूसरे किनारे पर व्यक्ति P बिंदु पर है। यहाँ AP नदी की चौड़ाई है।
जब व्यक्ति P से बिंदु M पर जाता है, तो उन्नयन कोण 60° से 30° हो जाता है।

समकोण ΔPAB में,
⇒ tan 60° = \(\frac{A B}{P A}\)
⇒ \(\sqrt{3}\) = \(\frac{A B}{P A}\)
⇒ AB = \(\sqrt{3}\)PA …..(i)
पुन: समकोण ΔMAB में,
tan 30° = \(\frac{A B}{M A}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{A B}{30+A P}\)
⇒ 30 + AP = \(\sqrt{3}\)AB
⇒ 30 + AP = \(\sqrt{3}\)(\(\sqrt{3}\)AP) [समी (i) से]
⇒ 30 + AP = 3AP
⇒ 2AP = 30
⇒ AP = 15 मी.
समीकरण (i) से,
AB = \(\sqrt{3}\) × 15 = 15\(\sqrt{3}\) मी
अतः नदी की चौड़ाई = 15 मी
तथा पेड़ की ऊँचाई = 15\(\sqrt{3}\) मी

प्रश्न 18.
भूमि पर स्थित बिंदु A से एक हवाई जहाज का उन्नयन कोण 60° है। 10 सैकंड की उड़ान के बाद उसी ऊँचाई पर उड़ते हुए हवाई जहाज का उन्नयन कोण बिंदु A से 30° हो जाता है। यदि हवाई जाहज की औसत चाल 720 किमी / घंटा हो, तो हवाई जाहज की धरती से स्थिर ऊँचाई ज्ञात कीजिए।
हल:
माना P और Q हवाई जहाज की दो स्थितियाँ है। माना ABC एक क्षैतिज रेखा है जो A से जाती है।
∵ हवाई जाहज द्वारा 1 घंटे में तय दूरी,
PQ = 720 किमी

∵ हवाई जहाज द्वारा 1 सकण्ड में तय दूरी
= \(\frac{720 \times 1000}{60 \times 60}\) मी
= 200 मी
और 10 सेकंड में तय दूरी = 10 × 200 मी = 2000 मी
∴ PQ = 2000 मी
समकोण ΔABP में
tan 60° = \(\frac{P B}{A B}\)
⇒ \(\sqrt{3}=\frac{P B}{A B}\)
⇒ PB = \(\sqrt{3}\)AB …..(i)
⇒ tan 30° = \(\frac{Q C}{A C}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{Q C}{A C}\)
⇒ AC = \(\sqrt{3}\)QC
⇒ AB + BC = \(\sqrt{3}\)PB [∵ QC = PB]
⇒ AB + 200 = \(\sqrt{3}\) × \(\sqrt{3}\)AB
[∵ PQ = 2000
[तथा PB = \(\sqrt{3}\)AB]
⇒ AB + 2000 = 3AB
⇒ 2AB = 2000
⇒ AB = 1000 मी
समीकरण (i) से
PB = \(\sqrt{3}\) × 1000
= 1000\(\sqrt{3}\) मी
अतः हवाई जहाज की धरती से स्थिर ऊँचाई 1000\(\sqrt{3}\) मी है।

वस्तुनिष्ठ प्रश्न :

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क).

1. यदि कोई प्रेक्षक किसी वस्तु को देख रहा है, तो प्रेक्षक की आँख को उस वस्तु से मिलाने वाली क्षैतिज रेखा को ……………. रेखा कहते हैं।
2. वह रेखा, जो प्रेक्षक की आँख से सीधी भूमि के समांतर जाती है, ……………… रेखा कहलाती है।
3. जब प्रेक्षक किसी वस्तु को देखने के लिए अपने सिर को ऊपर उठाता है, तो वस्तु प्रेक्षक की आँख पर ………………. कोण बनाती है।
4. जब प्रेक्षक किसी वस्तु को देखने के लिए अपना सिर नीचे झुकता है, तो वस्तु की आँख पर कोण ……………… बनाती है।
5. उन्नयन कोण एवं अवनमन कोण सदैव बराबर और ……………… कोण होते हैं।

उत्तर:

1. दृष्टि,
2. क्षैतिज,
3. उन्नयन,
4. अवनमन,
5. न्यून ।

निम्न में सत्य / असत्य ज्ञात कीजिए :

प्रश्न (ख).

1. अवनमन कोण को अवनति कोण भी कहते हैं।
2. उन्नयन कोण एवं अवनमन कोण एकांतर कोण होते हैं।
3. उन्नयन कोण सदैव अधिक कोण होता है।
4. अवनमन कोण सदैव समकोण होता है।
5. त्रिकोणमिति की सहायता से दूरियों तथा ऊँचाइयों की गणना सरलता से की जा सकती है।

उत्तर:

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

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
एक सीधी खड़ी छड़ की लंबाई तथा उसकी परछाई में 1 : \(\sqrt{3}\) का अनुपात है। उस समय सूर्य का उन्नयन कोण ज्ञात कीजिए:
(A) 30°
(B) 60°
(C) 45°
(D) 90°
हल:
माना छड़ की लंबाई AB तथ उसकी परछाई BC है।
माना उन्नयन कोण θ है।
दिया है, BA : BC = 1 : \(\sqrt{3}\)
⇒ \(\frac{B A}{B C}=\frac{1}{\sqrt{3}}\)

समकोण ΔCAB में
sin θ = \(\frac{A B}{B C}\)
⇒ sin θ = \(\frac{1}{\sqrt{3}}\)
⇒ sin θ = sin 60°
⇒ θ = 60°
अत: सही विकल्प (B) है।

प्रश्न 2.
निम्न आकृति में वस्तु 1 को बिंदुओं O₁ तथा O₂ से देखने पर बने अवनमन कोण क्रमश: हैं:
(A) 45°, 75°
(B) 60°, 90°
(C) 30°, 60°
(D) 45°, 30°
हल:
एक रेखा PO₁ इस प्रकार खींची कि PO₁ || AC
यहाँ ∠PO₁A + ∠AO₁C = 90°
⇒ ∠PO₁A + 60° = 90°

⇒ ∠PO₁A = 90° – 60° = 30°
अब ∠PO₂A = ∠O₂AB = 45° (एकान्तर कोण)
O₁ से अवनमन कोण = 30°
O₂ से अवनमन कोण = 45°
अत: सही विकल्प (D) है।

प्रश्न 3.
निम्न आकृति में अच्छी तरह से तनी हुई एक 20 मी लम्बी रस्सी भूमि पर सीधे लगे खंभे के शिखर से बंधी है। यदि भूमि स्तर के साथ रस्सी द्वारा बनाया गया कोण 30° का है, तो खंभे की ऊँचाई ज्ञात कीजिए।
(A) 10 मी
(B) 20 मी
(C) 40 मी
(D) 50 मी
हल:
समकोण ΔBAC में.

sin 30° = \(\frac{A B}{B C}\)
⇒ \(\frac{1}{2}=\frac{A B}{20}\)
⇒ \(\frac{20}{2}\) मी = 10 मी
अत: सही विकल्प (A) है।

प्रश्न 4.
निम्न आकृति में, भूमि के एक बिन्दु C से, जो मीनार के पाद बिन्दु से 60 मी की दूरी पर है, मीनार AB के शिखर का उन्नयन कोण 30° है। मीनार की ऊँचाई है :
(A) 60\(\sqrt{3}\) मी.
(B) 60 मी
(C) 20\(\sqrt{3}\) मी
(D) 20 मी
हल:
समकोण ΔABC में,

tan 30° = \(\frac{A B}{B C}\)
\(\frac{1}{\sqrt{3}}=\frac{A B}{60}\)
AB = \(\frac{60}{\sqrt{3}}=\frac{60 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
\(\frac{60 \sqrt{3}}{3}\) = 20\(\sqrt{3}\) मी
अत: सही विकल्प (C) है।

प्रश्न 5.
एक मीनार के आधार से 100 मीटर की दूरी पर स्थित बिन्दु से उसके शिखर का उन्नयन कोण 45° है। मीनार की ऊँचाई है:
(A) 50 मीटर
(B) 100 मीटर
(C) \(\frac{100}{\sqrt{2}}\) मीटर
(D) \(\frac{100 \times \sqrt{3}}{2}\) मीटर
हल:
माना मीनार की ऊँचाई (BC)h मीटर है।
मीनार के आधार से 100 मीटर दूरी पर स्थित बिन्दु उसके शिखर का उन्नयन कोण 45° है। अर्थात् AB = 100 मी. तथा ∠CAB = 45°
समकोण ΔABC में,

tan 45° = \(\frac{B C}{A B}\)
⇒ 1 = \(\frac{h}{100}\)
∴ h = 100 मीटर
अतः मीनार की ऊँचाई = 100 मी.
सही विकल्प (B) है।

प्रश्न 6.
15 मी लम्बी एक सीढ़ी एक ऊर्ध्वाधर दीवार के शिखर तक पहुँचती है। यदि यह सीढ़ी दीवार के साथ 60° का कोण बनाती है, तो दीवार की ऊँचाई है :
(A) 15\(\sqrt{3}\) मी.
(B) \(\frac{15 \sqrt{3}}{2}\) मी.
(C) \(\frac{15}{2}\) मी.
(D) 15 मी.
हल:
माना कि AB एक ऊर्ध्वाधर दीवार है जिसकी ऊँचाई 1⁄2 मी. है। माना कि AC एक सीढ़ी है जो दीवार के शिखर तक पहुँचती है तथा सीढ़ी की लम्बाई 15 मी. है। सीढ़ी दीवार के साथ 60° का कोण बनाती है, तब
∠ACB = 60° तथा AC = 15 मी.
समकोण ΔABC में,

sin 60° = \(\frac{A B}{A C}\)
\(\frac{\sqrt{3}}{2}=\frac{h}{15}\)
h = \(\frac{15 \sqrt{3}}{2}\) मी.
अत: विकल्प (B) सही है।

प्रश्न 7.
6 मीटर ऊँचे एक खम्भे की छाया 2\(\sqrt{3}\) मीटर लम्बी हो तो सूर्य का उन्नतांश कोण है:
(A) 60°
(B) 45°
(C) 30°
(D) 90°
हल:
माना कि AB एक खम्भा है जिसकी ऊँचाई 6 मीटर है।
खम्भे की छाया की लम्बाई (BC) = 2\(\sqrt{3}\) मीटर
माना कि सूर्य का उन्नतांश कोण (∠ACB) = θ
अत: समकोण ΔABC में

tan θ = \(\frac{A B}{B C}\)
⇒ tan θ = \(\frac{6}{2 \sqrt{3}}\)
⇒ tan θ = \(\sqrt{3}\)
⇒ tan θ = tan 60°
⇒ θ = 60°
अत: सही विकल्प (A) है।

प्रश्न 8.
किसी मीनार की छाया उसकी ऊँचाई के बराबर हो तो सूर्य का उन्नयन कोण है:
(A) 90°
(B) 60°
(C) 45°
(D) 30°
हल:
माना BC कोई मीनार है, जिसकी ऊँचाई / मीटर है।

इसकी छाया AB, h मीटर होगी।
पुनः माना सूर्य का उन्नयन कोण ∠CAB = θ
समकोण ΔABC में,
tan θ = \(\frac{B C}{A B}=\frac{h}{h}\)
⇒ tan θ = 1
⇒ tan θ = tan 45°
∴ θ = 45°
अतः सही विकल्प (C) है।

प्रश्न 9.
10 मीटर ऊँची एक मीनार के शिखर से पृथ्वी पर एक बिन्दु का अवनमन कोण 30° है। बिन्दु की मीनार के आधार से दूरी है:
(A) 10 \(\sqrt{3}\) मीटर
(B) \(\frac{10}{\sqrt{3}}\) मीटर
(C) 10 मीटर
(D) 5\(\sqrt{3}\) मीटर
हल:
माना AC कोई मीनार है जिसकी ऊँचाई 10 मीटर है।

माना मीनार के आधार से बिन्दु की दूरी (BC) = x मीटर
मीनार के शिखर से पृथ्वी पर एक बिन्दु का अवनमन कोण 30° है।
∴ ∠XAB = 30°
∠XAB = ∠ABC = 30° (एकान्तर कोण)
समकोण ΔACB में,
tan 30° = \(\frac{A C}{B C}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{10}{x}\)
∴ x = 10\(\sqrt{3}\) मीटर
अतः बिन्दु की मीनार के आधार से दूरी 10\(\sqrt{3}\) मीटर होगी सही विकल्प (A) है।

प्रश्न 10.
एक पतंग भूमि से 30 मी की ऊँचाई पर 60 मी लंबी डोरी की सहायता से उड़ रही है। यह मानते हुए कि डोरी में कोई ढील नहीं है, पतंग का भूमि पर उन्नयन कोण है:
(A) 45°
(B) 30°
(C) 60°
(D) 90°
हल:
माना कि भूमि से 30 मीटर की ऊँचाई पर पतंग की स्थिति है जोकि 60 मीटर लंबी डोरी (AC) की सहायता से उड़ रही है। माना कि पतंग की डोरी का क्षैतिज के साथ कोण θ है।
अर्थात् ∠ACB = θ
समकोण ΔABC में,

sin θ = \(\frac{A B}{A C}\)
⇒ sin θ = \(\frac{30}{60}\)
⇒ sin θ = \(\frac{1}{2}\)
⇒ sin θ = sin 30°
⇒ θ = 30°
अतः विकल्प (B) सही है।

प्रश्न 11.
एक नदी के ऊपर एक पुल नदी के तट के साथ 45° का कोण बनाता है। यदि नदी के ऊपर पुल की लम्बाई 150 मीटर है, तो नदी की चौड़ाई क्या होगी:
(A) 75\(\sqrt{2}\) मीटर
(B) 50\(\sqrt{2}\) मीटर
(C) 75 मीटर
(D) 150 मीटर
हल:
माना AC पुल है जिसकी लम्बाई 150 मीटर है
तथा BC नदी की चौड़ाई है। पुल नदी के साथ 45° का कोण बनाता है।
अर्थात् ∠CAB = 45°
समकोण ΔABC में,
sin 45° = \(\frac{B C}{A C}\)

\(\frac{1}{\sqrt{2}}=\frac{B C}{150}\)
BC = \(\frac{150}{\sqrt{2}}\)
BC = \(\frac{150 \sqrt{2}}{\sqrt{2} \times \sqrt{2}}\)
BC = 75\(\sqrt{2}\)
अतः नदी की चौड़ाई 75\(\sqrt{2}\) मीटर होगी।
सही विकल्प (A) है।

Jharkhand Board JAC Class 10 Science Important Questions Chapter 12 Electricity Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 12 Electricity

Additional Questions and Answers

Question 1.
Solve the following examples :
1. 240 joules of work Is done in moving 20 coulomb electric charge from one pole to the other pole of a battery. Calculate the voltage of the battery.
Answer:
12 V

2. 5 V electric potential difference is applied between two ends of a conducting wire. If 600 C of electric charge passes through it in 10minutes, calculate the resistance of the wire.
Answer:
5 Ω

3. TWo copper wires A and B have the same mass. The resistance of wire A is 0.5 Ω and the length of wire B is double that of wire
A. Find the resistance of wire B.
Answer:
2 Ω

4. If the length of a given conducting wire is kept constant and its diameter is doubled, what will be the resistance of the new wire?
Answer:
\(\frac { R }{ 4 }\)

5. Two resistors of 1 kΩ and 200Ω are connected in series with a 12 V battery. Calculate the current flowing in the circuit and the voltage developed across the 200 Ω resistor.
Answer:
I = 0.01 A, V = 2 V

6. For the circuit shown in figure given below, calculate
(a) the total resistance of the circuit.
(b) the total current flowing in the circuit.
(c) the voltage developed across the two ends of R₁.

Answer:
(a) 12 Ω
(b) 0.5 A
(c) 3.6 V

7. Two 60W bulbs are used for 4hours every day and five 100W bulbs are used for 5 hours everyday. How many units of electricity will be consumed in 30 days?
Answer:
89.4 units

8. 24 W power is consumed by a bulb when it is connected with 12 V battery. How much power will be consumed if it is connected with 6V battery?
Answer:
6 W

9. Find the equivalent resistance of the circuit shown in the following figure :

Also find the total current flowing in this circuit.
Answer:
R = 4Ω, I = 3 A

10. In the following circuit A, B and C are three ammeters. 0.5A current is shown by ammeter B.

(a) Find the currents passing through ammeters A and C.
(b) Find the total resistance of the circuit.
Answer:
(a) current passing through ammeter C = 1 A, current passing through ammeter A= 1.5 A
(b) 4Q

11. If an electric bulb gives light for lhour while carrying 0.5A current, how much electric charge passes through it and how many electrons pass through it during this time? (e = 1.6 x 10^-19C)
Answer:
Q = 1800 C,
n = 1.125 x 10²²

12. An electric current of 64mA flows through a bulb for 10minutes. How many electrons pass through the bulb during this time?
(e = 1.6 x 10^-19C)
Answer:
n = 2.4 x 10²⁰

13. In order to get a current of 0.5A in a circuit by connecting a bulb of resistance 20 Ω with 12 V battery, what should be the resistance to be connected in series? What will be the voltage drop across the bulb?
Answer:
R = 4Ω, V = 10 V

14. Three resistors are connected in parallel with a 30V battery. A current of 7.5A flows through the circuit. If two out of the three resistors are of 10Ω and 12Ω, determine the resistance of the third resistor.
Answer:
15 Ω

15. For the circuit shown in the following figure, determine the equivalent resistance between points A and B. Also find the current I flowing through the circuit.

Answer:
10 Ω, 1 A

16. Determine the equivalent resistance between point A and B in the following circuit:

Answer:
2 Ω

17. In a house, if three bulbs of 100W, 60W and 40 W respectively are used 2 hours per day, how many units of electrical energy will be consumed in 30 days?
Answer:
12 units

18. Find the electric current in the following circuit:

Answer:
0.1 A

19. Determine the equivalent resistance between points X and Y in the following circuit:

Answer:
15 Ω

20. An electric heater consumes 4.4 kW power when connected with a 220V line voltage.
(1) Calculate the current flowing in the heater.
(2) Calculate the resistance of the heater.
(3) Calculate the energy consumed in 2 hours.
Answer:
(1) 20 A
(2) 11 Ω
(3) 3.168 x 10⁷ J

Question 2.
State at least three points of difference between the following terms / quantities:
(1) Resistance and Resistivity
Answer:

Resistance	Resistivity
1. Resistance is the ratio of the voltage applied between the two ends of a conductor to the electric current flowing through the conductor. (R = \(\frac { V }{ I }\))	1. Resistivity of a substance is the resistance offered by a conducting wire of unit length and unit cross-sectional area. (ρ = R. \(\frac { A }{ l }\))
2. It depends on the type of substance, physical conditions such as temperature and pressure and dimensions of the conductor (length and area of cross-section).	2. It depends on the type of substance, temperature of the conductor and pressure exerted on the conductor.
3. Its SI unit is Ω.	3. Its SI unit is Ω m.

(2) Series combination of resistors and Parallel combination of resistors
Answer:

Series combination of resistors	Parallel combination of resistors
1. In a series combination of resistors, the resistors are connected across two points in the circuit such that the current flowing through each resistor is the same and only one path is available for the current to flow.	1. In a parallel combination of resistors, the resistors are connected across two points in the circuit such that the voltage drop across two ends of each resistor remains the same and more than one path are available for the current to flow.
2. In this case, the total voltage drop is equal to the sum of the voltage drops across each resistor.	2. In this case, the total current flowing through the circuit is equal to the sum of the currents flowing through each resistor.
3. The equivalent resistance of a series combination of resistors is equal to the sum of the resistances of the individual resistors.	3. The reciprocal of the equivalent resistance of a parallel combination of resistors is equal to the sum of the reciprocal of the resistances of the  individual resistors.
4. In this case, the equivalent resistance is greater than the largest of all the individual resistances.	4. In this case, the equivalent resistance is smaller than the smallest of all the individual resistances.
5. This combination is used in order to increase the resistance of the circuit.	5. This combination is used in order to decrease the resistance of the circuit.
6. In this case, the total current in the circuit decreases.	6. In this case, the total current in the circuit increases.

Question 3.
Give scientific reasons for the following statements:
(1) It is not advisable to connect an electric bulb and electric heater in series.
Answer:
An electric bulb and electric heater have different resistances and the requirement of current for their proper functioning is different.
Moreover, sometimes the devices may get damaged. If one of the devices stops and the circuit breaks, the other device cannot be used.

(2) For domestic purposes, different electrical devices are connected in parallel instead of connecting them in series.
Answer:
Each device gets the full and same voltage as that of the electric supply.

Also each device gets proper current depending on its resistance.
If one of the devices is switched OFF / ON, other electrical devices remain unaffected.

(3) Fairy decorative lights are always connected in parallel.
Answer:
If the fairy lights are connected in series, the equivalent resistance offered will be greater and the brightness of the bulbs will be affected. But in parallel connection all the bulbs will glow with same intensity and if any one bulb gets fused the other bulbs will continue to glow.

(4) In a tungsten electric-bulb, a coil of wire, rather than a straight wire, is used.
Answer:
A coil of wire has much greater surface area than a straight wire if enclosed in the same bulb. This results in more emission of light.
Hence, in a tungsten electric-bulb, a coil of wire, rather than a straight wire, is used.

Objective Questions and Answers

Question 1.
Answer the following questions in one word / sentence:
(1) Name the scientist in whose honour the SI unit of electric current is named.
Answer:
Andre Ampere

(2) Which type of electric force acts between the proton and the electron?
Answer:
Attractive electric force

(3) Write the relation between the joule, coulomb and the volt.
Answer:
1 volt = \(\frac { 1 joule }{ 1 coulomb }\)

(4) How is the device measuring electric current connected in the circuit?
Answer:
In series

(5) You are given aluminium wire and copper wire having the same dimensions. Which wire will carry more electric current if the applied potential difference is the same for the two wires?
Answer:
Copper wire

(6) Name one electric appliance working on the principle of heating effect of electric current.
Answer:
Electric iron

(7) If a resistive wire is uniformly stretched, what will be a change in its resistance?
Answer:
Its resistance will increase

(8) Name a physical quantity which can be expressed in Ws.
Answer:
Electric energy

(9) Name the particle responsible for flow of electric current in a metallic conductor.
Answer:
The electron

(10) Which instrument is used to measure electric potential difference ?
Answer:
Voltmeter

(11) Draw a symbol for variable resistor.
Answer:

(12) How much energy in joule is consumed when 100 units electricity are used?
Answer:
3.6 x 10⁸ J

(13) Write the Ohm’s law in the form of a formula.
Answer:
V = IR
Where
V = Potential difference across the ends of metallic wire of resistance R at constant temperature
I = Current flowing through the metallic wire

Question 2.
Fill in the blanks:

1. Electric charge on …………………… is taken negative.
2. In a substance, the direction of conventional current is opposite to the direction of …………………… current.
3. Electric component, …………………… has symbol
   
4. The SI unit of electric potential is ……………………
5. In a circuit, a voltmeter is always connected …………………… to the electrical component.
6. When a battery is connected in a circuit, in the external circuit, electric current flows from the …………………… terminal to the ………………….. terminal of the battery.
7. The SI unit of electric power is ……………………
8. According to Joule’s law, heat energy produced in a resistor is directly proportional to the square of ……………………
9. 1 unit (commercial unit of electric energy) = …………………… J
10. The equivalent resistance between points X and Y in the circuit shown in the following figure is ……………………
    
11. The rate of consumption of electric energy with time is called ……………………
12. 1 A = …………………… µA.

Answer:

1. the electron
2. electron
3. resistor
4. the volt or the joule per coulomb
5. in parallel
6. positive, negative
7. the watt
8. the electric current flowing through the resistor
9. 3.6 x 10⁶
10. 5 Ω
11. electric power
12. 10⁶

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

1. The SI unit of electric current is the coulomb.
2. To measure electric current, an ammeter is connected in series with a resistor.
3. W = VQ.
4. The current flowing through each resistor is the same when resistors having different! resistances are connected in parallel.
5. The equivalent resistance (R_s) of a series combination of resistors is given by \(\frac{1}{R_{\mathrm{s}}}=\frac{1}{R_1}+\frac{1}{R_2}\) + ……………
6. Nichrome wire is used to make resistors.
7. When the temperature of a metallic substance increases up to a certain limit, its resistance decreases.
8. For two wires of the same material and having the same length, the resistance of the thicker wire is less than that of the other wire.
9. Electric charge is measured by using an ammeter.
10. Work can be expressed in coulomb-volt.
11. Two bulbs of 60 W, 220 V when connected in series with a supply voltage of 220 V, light up with maximum intensity.

Answer:

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

Question 4.
Match the following :
(1)

Column I	Column II	Column III
1. Electric current	a. VI	p. The volt / ampere
2. Electric potential difference	b. Q/t	q. The joule
3. Resistance	c. W/Q	r. The coulomb / second
4. Electric energy	d. V/I	s. The joule / second
5. Electric power	e. V I t	t. The volt

Answer:
(1 – b – r), (2 – c – t), (3 – d – p), (4 – e – q), (5 – a – s).

(2)

Column I	Column II	Column III
1. Copper	a. Insulator	p. It is used to make resistance.
2. Nichrome	b. Semi-conductor	q. It stops electric current.
3. Wood	c. Conductor	r. It is used to make conduc-ting wire.
4. Silicon	d. Resistor	s. It is used to make elec-tronic components.

Answer:
(1 – c – r), (2 – d – p), (3 – a – q ), (4 – b – s).

(3)

Answer:
(1 – e), (2 – c), (3 – f), (4 – b), (5 – d), (6 – a).

Question 5.
Choose the correct option from those given below each question:
1. The SI unit of electric charge is the ……………….
A. ampere
B. volt
C. watt
D. coulomb
Answer:
D. coulomb

2. How many electrons will be there in 1.6 C charge?
A. 10¹⁷
B. 10¹⁸
C. 10¹⁹
D. 10²⁰
Answer:
C. 10¹⁹

3. 1 µA = …………………. mA
A. 10^-16
B. 10^-3
C. 10¹⁹
D. 10²⁰
Hint IµA = 10^-3 x 10^-3A = 10^-3 mA
Answer:
B. 10^-3

4. Which of the following materials has more number of free electrons in a given volume?
A. Copper
B. Glass
C. Rubber
D. Iron
Answer:
A. Copper

5. According to Ohm’s law ………………….
A. the resistance increases with an increase in current.
B. the resistance increases with an increase in voltage.
C. the current increases with an increase in voltage.
D. both the resistance and current increase with an increase in voltage.
Answer:
C. the current increases with an increase in voltage.

6. The formula for electric current is
A. I = Qt
B. I = Y
C. I = \(\frac { Q }{ t }\)
D. I = W.t
Answer:
C. I = \(\frac { Q }{ t }\)

7. 2 A electric current is passed for 1 minute through a conducting wire. How much electric charge will pass through this wire?
A. 2C
B. 30 C
C. 60 C
D. 120 C
Answer:
D. 120 C
Hint : I = \(\frac { Q }{ t }\)
∴ Q = I x t = (2 A)(1 x 60s)
= 120 C

8. If 4.8 A current is passed through an electrical appliance, the number of electrons passing through it in 1 second will be ………………..
A. 0.33 x 10¹⁹
B. 3.3 x 10¹⁹
C. 3 x 10¹⁹
D. 4.8 x 10¹⁹
Answer:
C. 3 x 10¹⁹
Hint: I = \(\frac { Q }{ t }\) = \(\frac { ne }{ t }\)
∴ n = \(\frac { I × t }{ e }\)
= \(\frac{4.8 \times 1}{1.6 \times 10^{-19}}\)
= 3 x 10¹⁹

9. Which of the following formulae represents voltage?
A. \(\frac{\text { Work }}{\text { current } \times \text { time }}\)
B. \(\frac{\text { Work } \times \text { time }}{\text{ Current} }\)
C. Work x electric charge
D. Work x electric charge x time
Answer:
A. \(\frac{\text { Work }}{\text { current } \times \text { time }}\)
Hint: V = \(\frac { W }{ Q }\) = \(\frac { W }{ I.t }\)

10. The unit of electric potential difference is ………………..
A. J
B. J/C
C. JC
D. C/J
Answer:
B. J/C

11. If the work done to take 3C electric charge from one point to another point is 15 J, what will be the potential difference between these two points?
A. 3V
B. 15 V
C. 5V
D. 45 V
Answer:
C. 5V
Hint: Potential difference V = \(\frac { W }{ Q }\)
= \(\frac { W }{ Q }\)

12. The resistance of a conducting wire is 10 Ω. If a battery of 1.5 V is connected to it, the electric current flowing through it will be
A. 0.15 mA
B. 1.5 mA
C. 15 mA
D. 150 mA
Answer:
D. 150 mA
Hint:
R = \(\frac { V }{ I }\) = \(\frac { 1.5 }{ 10 }\) = 0.15
= 150 x 10^-3
= 150 mA

13. On which factors does the resistivity of a conducting wire depend?
A. The length of the wire
B. The area of cross-section of the wire
C. The volume of the wire
D. The material of the wire
Answer:
D. The material of the wire

14. If five equal pieces of a resistance wire having 5 Ω resistance are connected in parallel, their equivalent resistance will be ……………………….
A. \(\frac { 1 }{ 5 }\) Ω
B. 1 Ω
C. 5 Ω
D. 25 Ω
Answer:
A. \(\frac { 1 }{ 5 }\) Ω
Hint: When a resistance wire having 5 Ω resistance is cut into five equal pieces, each piece will have resistance 1 Ω, i.e., R₁ = R₂ = R₃ = R₄ = R₅ = 1 Ω.
The equivalent resistance (R_p) of their parallel combination is given by,
\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\)
∴ \(\frac{1}{R_{\mathrm{p}}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}\)
∴ \(\frac{1}{R_{\mathrm{p}}}=\frac{5}{1}\)
∴ R_p = \(\frac { 1 }{ 5 }\) Ω

15. The SI unit of resistivity is …………………
A. Ω
B. Ω m
C. \(\frac { Ω }{ m }\)
D. \(\frac { m }{ Ω }\)
Answer:
B. Ω m

16. What will be the equivalent resistance between points A and B of the following electric circuit?

A. 1 Ω
B. 2 Ω
C. 5 Ω
D. 10 Ω
Answer:
D. 10 Ω
Hint: The given electric circuit can be drawn as follows:

As all resistors are connected in series, their equivalent resistance,
R_s = 2 + 2 + 2 + 2 + 2 = 10 Ω

17. The equivalent resistance between points A and B of the following electric circuit will be ………………..

A. 4 Ω
B. 8 Ω
C. 2 Ω
D. 16 Ω
Answer:
A. 4 Ω
Hint: The given electric circuit can be drawn as follows:

It is clear from the circuit that resistors R₁ and R₂ are connected in series.
∴ R’ = R₁ + R₂ = 4 + 4 = 8 Ω
Now, R’ is connected with R₃ in parallel combination.
∴\(\frac{1}{R}=\frac{1}{R^{\prime}}+\frac{1}{R_3}\)
= \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\)
= \(\frac { 2 }{ 8 }\) = \(\frac { 1 }{ 4 }\)
∴ R = 4 Ω

Question 18.
The equivalent resistance between points A and B in the following electric circuit is ……………..

A. 2.5 Ω
B. 5 Ω
C. 12.5 Ω
D. 20 Ω
Answer:
C. 12.5 Ω
Hint: Resistors R₁ and R₂ are connected in parallel.
∴ R’ = \(\frac{R_1 R_2}{R_1+R_2}=\frac{5 \times 5}{5+5}\) = 2.5 Ω
Now, R’, R₃ and R₅ are connected in series
∴R = R’ + R₃ + R₅
= 2.5 + 5 + 5 = 12.5 Ω

19. Which physical quantity is expressed in kWh?
A. Work
B. Electric power
C. Electric Current
D. Electric potential
Answer:
A. Work

20. 1 kWh = ………………. J
A. 3.6 x 10⁶
B. 3.6 x 10³
C. 3.6 x 10^-6
D. 3.6 x 10^-3
Answer:
A. 3.6 x 10⁶
Hint : 1 kWh = (1000 W) (3600 s)
= 3.6 x 10⁶ J

21. An electric heater consumes 1.1 kW power when 220 V voltage is applied to it. The current flowing through it must be ……………….
A. 1.1 A
B. 2.2 A
C. 4 A
D. 5 A
Answer:
D. 5 A
Hint: P = VI
∴ I = \(\frac{P}{V}=\frac{1.1 \times 10^3}{220}=\frac{1100}{220}\)

22. 1 A = ……………. mA
A. 100
B. 10³
C. 10^-3
D. 10^-6
Answer:
B. 10³

23. In an electric field, electric potential at a point is 10 V How much work has to be done to bring 0.5 C electric charge from infinite distance to that point?
A. 0.5 J
B. 2 J
C. 5 J
D. 10 J
Answer:
C. 5 J
Hint: W = VQ = (10) (0.5) = 5 J

24. ……………… is the work done in moving a unit positive charge from point A to B in an electric field.
A. Electric potential at point A.
B. Electric potential at point B.
C. Electric potential difference between points A and B.
D. Electric current from point A to point B.
Answer:
C. Electric potential difference between points A and B.

25. In an electric field, electric potential at point A is 40 V and that at point B is 90 V The work to be done in taking 2 C charge from point A to point B is ………………
A. 25 J
B. 50 J
C. 90J
D. 100 J
Answer:
D. 100 J
Hint: V_B – V_A = \(\frac { W }{ Q }\)
∴ W = Q (V_B – V_A)
= (2) (90 – 40) = 100 J

26. Sonika is working with the circuit shown in the following figure :

The circuit has two gaps : X and Y. She has wires of five different materials – I, II, III, IV and V She knows that the bulb will light up only when both gaps are filled with conducting materials.
She records her observations in a table. After completing the experiment, ink fell on the paper and she lost entries in row 3.

No.	Material in X	Material in Y	Bulb (On/Off)
1	I	II	Off
2	I	IV	On
3			Off
4	III	V	On

Based on the rest of the information in the table, out of the following what could be the materials in row 3?
A. II and III
B. III and IV
C. IV and V
D. I and III
Answer:
A. II and III
Hint: The entries in rows 2 and 4 show that
I : conducting material, IV : conducting material, III : conducting material, V : conducting material,
This information, combined with the entries in row 1, shows that
II : Non-conducting material Row 3 must include II.

27. ‘Ω m’ is the SI unit of ……………..
A. resistance
B. resistivity
C. conductivity
D. resistance per unit length
Answer:
B. resistivity

28. I – V graph for a resistance wire is shown in the figure given below:

What will be resistance of this wire?
A. 50 Ω
B. 20 Ω
C. 7.2 Ω
D. \(\frac { 1 }{ 120 }\)Ω
Answer:
B. 20 Ω
Hint: The slope of I-V graph represents the reciprocal of resistance.
∴ Slope = \(\frac { 1 }{ R }\) = \(\frac { ∆I }{ ∆V }\)
∴R = \(\frac{\Delta V}{\Delta I}=\frac{12-0}{0.6-0}=\frac{12}{0.6}\) = 2.0 Ω

29. What will be the change in the resistivity of a resistance wire, when its length is doubled by stretching it uniformly?
A. Will be doubled
B. Will be half
C. Will be one-fourth
D. Will not change
Answer:
D. Will not change

30. Observe the circuit in the following figure and select the correct statement:

A. Resistors with resistances R₁ and R₂ are connected in series.
B. Resistors with resistances R₁ and R₂ are connected in parallel.
C. Resistors with resistances R₂ and R₃ are connected in series.
D. Resistors with resistances R₂ and R₃ are connected in parallel.
Answer:
D. Resistors with resistances R₂ and R₃ are connected in parallel.

31. Resistors with resistances 5 Ω, 10 Ω and 15 Ω are connected in parallel to each other. What will be the equivalent resistance of the circuit? A. Less than 5 Ω
B. More than 15 Ω
C. More than 30 Ω
D. Equal to 30 Ω
Answer:
A. Less than 5 Ω

32. The resistance of a wire is R. When this wire is stretched so that its length becomes double, its cross-sectional area becomes half. What will be the resistance of this stretched wire?
A. \(\frac { R }{ 2 }\)
B. R
C. 2R
D. 4R
Answer:
D. 4R
Hint: Original resistance of the wire R = ρ.\(\frac { 1 }{ A }\)
Resistance of the stretched wire R’ = ρ.\(\frac { l’ }{ A’ }\)
But l’ = 2l and A’ = \(\frac { A }{ 2 }\)
∴ R’ = ρ\(\frac{2 l}{\frac{A}{2}}\) = (4)ρ\(\frac { 1 }{ A }\) = 4R

33. When three resistors, each of resistance R, are connected in parallel, the equivalent resistance is found out to be 10 Ω. R must be …………..
A. 10 Ω
B. 20 Ω
C. 30 Ω
D. 5 Ω
Answer:
C. 30 Ω
Hint :
\(\frac{1}{R^{\prime}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
Now, R₁ = R₂ = R₃ = R
∴ \(\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}\)
∴ R = 3R’ = 3 (10) = 30 Ω

34. In the following figure, what will be the equivalent resistance between points A and B?

A. 1 Ω
B. 2 Ω
C. 4 Ω
D. 8 Ω
Answer:
B. 2 Ω
Hint: Here, resistors with resistances R₁ and R₂ are in parallel.
∴ R’ = \(\frac{R_1 R_2}{R_1+R_2}=\frac{2 \times 2}{2+2}\) = 1 Ω
Also, resistors with resistances R₃ and R₄ are in parallel.
∴ R” = \(\frac{R_3 R_4}{R_3+R_4}=\frac{2 \times 2}{2+2}\) = 1 Ω
The equivalent resistance of the resulting series combination will be
R = R’ + R”
= 1 + 1 = 2 Ω

35. A wire having resistance 20 Ω is bent in the form of a circle as shown in the following figure :

What will be the resistance between two end points A and B located on the diameter?
A. 5 Ω
B. 10 Ω
C. 20 Ω
D. 40 Ω
Answer:
A. 5 Ω
Hint: The resistance of the wire is R = 20 Ω. Hence, the resistance of the upper part of the circle = \(\frac { R }{ 2 }\) = 10 Ω
Similarly, the resistance of the lower part of the circle = \(\frac { R }{ 2 }\) = 10 Ω.

As shown in the figure above, resistors with resistances 10 Ω and 10 Ω are in parallel. So, the equivalent resistance of the combination will be R’ = \(\frac { (10)(10) }{ 10+10 }\) = 5 Ω

36. Which of the following options is not the formula for heat energy (in joule) produced when a current is passed through a conductor?
A. H = I²Rt
B. H = \(\frac { V²t }{ R }\)t
C. H = IVt
D. H = \(\frac { V² }{ I² }\)t
Answer:
D. H = \(\frac { V² }{ I² }\)t

37. Which option is correct for the following statements A and B?
Statement A : The current passing through each resistor is the same in a series combination of resistors.
Statement B : The voltage between two ends of each resistor is the same in a parallel combination of resistors.
A. Statement A is true but statement B is false.
B. Statement A is false but statement B is true.
C. Statement A and statement B both are true.
D. Statement A and statement B both are false.
Answer:
C. Statement A and statement B both are true.

38. In which of the following circuits the voltmeter and ammeter are connected properly to verify the Ohm’s law?

Answer:

Hint: An ammeter should be connected in series to measure electric current and a voltmeter should be connected in parallel to the resistor to measure potential difference. While joining these two instruments their positive and negative terminals should be taken into consideration.

39. Maximum 1A current can pass through a bulb having resistance 100 Q. What will be the power of this bulb?
A. 10W
B. 100 W
C. 1000 W
D. 0.01 W
Answer:
B. 100 W
Hint: P = I²R = (1)²(100) = 100 W

40. Which of the following is not a unit of electric energy?
A. The watt-second
B. The kilowatt hour
C. The joule
D. The watt
Answer:
D. The watt

41. Of which physical quantity is the unit, ‘VA’?
A. Electric energy
B. Electric power
C. Heat energy
D. Electric potential
Answer:
B. Electric power

42. Which of the following circuit symbols is used to represent an electric cell?

Answer:

43. The resistance of 300 m long wire with cross-sectional area 10^-6m² and resistivity 10^-7 Ω m is
A. 30 Ω
B. 3 Ω
C. 0.3 Ω
D. 300 Ω
Answer:
A. 30 Ω
Hint: R = ρ\(\frac { l }{ A }\)
= 10^-7 x \(\frac{300}{10^{-6}}\)
= 30 Ω

44. What will be the equivalent resistance of the following circuit?

A. 10 Ω
B. 5 Ω
C. 2.5 Ω
D. 1 Ω
Answer:
C. 2.5 Ω
Hint: The resistance of the upper arm of the circuit is 2 + 3 = 5Ω.
Now, the upper arm and lower arm are connected in parallel to each other. So, the equivalent resistance of the circuit
R = \(\frac { 5×5 }{ 5+5 }\) = 2.5 Ω

45. Three resistors of resistances 2 Ω, 3 Ω and 5 Ω are connected in series with a 10 V battery.
The voltage across 2 Ω resistor will be
A. 10 V
B. 5V
C. 3V
D. 2V
Answer:
D. 2V
Hint: The equivalent resistance of the circuit R = 2 + 3 + 5= 10 Ω.
The current flowing through the circuit
I = \(\frac { V }{ R }\) = \(\frac { 10 }{ 10 }\) = 1 A
The voltage across 2 Ω resistor
= 2 x I = 2 Ω x 1 A = 2 V

46. An electric iron of power 2 kW is used for 3 hours. At ₹ 5 per unit, the electricity bill will be
A. ₹ 45
B. ₹ 30
C. ₹ 15
D. ₹ 10
Answer:
B. ₹ 30
Hint: Electric energy consumed in 3 hours W = P x t
= 2 kW x 3 h = 6 kWh = 6 unit
Cost of electric power consumed (electricity bill)
= 5 x 6 = ₹ 30

47. The working of electric fuse is based on the ……………
A. chemical effect of electric current.
B. heating effect of electric current.
C. voltage regulation in the circuit.
D. current regulation in the circuit.
Answer:
B. heating effect of electric current.

48. Resistivity of a material depends on …………………. of the material.
A. the length of the wire
B. the area of cross-section of the wire
C. the temperature
D. the volume of the wire
Answer:
C. the temperature

49. When resistors of resistances R₁ and R₂ (R₂ < R₁) are connected in parallel and the currents flowing through them are I₁ and I₂ respectively, then
A. I₁ = I₂
B. I₁ > I₂
C. I₁ < I₂
D. nothing can be said about the currents
Answer:
C. I₁ < I₂

50. When resistors of resistances R₁ and R₂ (R₂ > R₁) are connected in series and the currents flowing through them are I₁ and I₂ respectively, then
A. I₁ = I₂
B. I₁ > I₂
C. I₁ < I₂
D. nothing can be said about the currents
Answer:
A. I₁ = I₂

51. In the following circuit, which electric component is connected in a wrong manner?

A. Voltmeter
B. Ammeter
C. Plug key
D. Cell
Answer:
A. Voltmeter

52. 1 C = ……………. µC
A. 10^-6
B. 10^-3
C. 1
D. 10⁶
Answer:
D. 10⁶

53. Connection of two cells of 1.5 V is shown in the following figure :

What is the voltage between the points A and B?
A. 1.5 V
B. 3 V
C. 0.75 V
D. 0 V
Answer:
B. 3 V

54. The following figure shows the V-I graph for four different resistors :

Which resistor has the maximum resistance?
A. A
B. B
C. C
D. D
Answer:
D. D

55. A potential difference of 100 V is applied across an electric bulb marked 40 W, 200 V. The power consumed in the bulb is …………………
A. 100 W
B. 40 W
C. 20 W
D. 10W
Answer:
D. 10W
Hint: P = \(\frac{V^2}{R}\)
∴ Resistance of the filament of the bulb,
R = \(\frac{V^2}{P}=\frac{(200)^2}{40}\) = 1000 Ω
Now, power consumed in the bulb,
P = \(\frac{V^2}{R}=\frac{(100)^2}{1000}\) = 10 W

Question 6.
Answer the following questions as directed (Miscellaneous):
(1) The voltage – current (V – I) graphs for a metallic conductor at two different temperatures T₁ and T₂ are shown below:

At which temperature is the resistance higher?
Answer:
At temperature T₁.
Because, the slope \(\left(\frac{\Delta V}{\Delta I}\right)\) of the graph i.e., its resistance is greater at T₁.

(2) A wire of given length is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire change?
Answer:
As the length of the wire become y- of the original length its area of cross-section become 4 times.
So, new resistance
R’ = ρ\(\frac { l’}{ A’}\) = ρ\(\frac{(1 / 4)}{4 A}=\frac{1}{4 \times 4}\left(\rho \frac{l}{A}\right)\) = \(\frac { R}{ 16}\) (∵ R = ρ\(\frac { l}{ A’}\))
Thus, the new resistance is \(\frac { R}{ 16}\) of the original resistance.

(3) Through which of the following two wires, does the electric current flow more easily?
(a) A thick wire
(b) a thin wire of the same material and of the same length when connected to the same source. State the reason.
Answer:
The thick wire because its resistance would be less compared to that of the thin wire.

(4) 400 J of heat is produced in 4 s in a 4 Q resistor. Find the potential difference across the resistor.
Answer:
Here, H = 400 J
∴ \(\frac { V²t }{ R }\) = 400
∴ V² = 400 x \(\frac { R }{ t }\) = 400 x \(\frac { 4 }{ 4 }\) = 400
∴ V = 20 volts

(5) What is the commercial unit of electrical energy?
Answer:
The kilowatt hour, i.e., kWh

(6) On what principle is an electric bulb based?
Answer:
Heating effect of current

(7) In a circuit, two resistors of resistances 5 Ω and 10 Ω are connected in series. Compare the current passing through the two resistors.
Answer:
In a series combination of resistors, the same current passes through all the resistors and it is same as the total current of the circuit.
Hence, here current will be the same in two resistors.
So, the ratio of the current will be 1:1.

(8) Write the relation between resistance R of the filament of a bulb, its power and a constant S voltage V applied across it.
Answer:
R = \(\frac { V² }{ P }\)

(9) Which of the following bulbs has more resistance?
(a) A 220 V, 100 W bulb
(b) a 220 V 60 W bulb.
Answer:
60 W bulb
Here, R \(\frac { 1 }{ p }\) (∵ as both the bulbs have the j same voltage rating.)

(10) Find the minimum resistance that can be made using five resistors, each of 5 Ω.
Answer:
By connecting given resistors in parallel, resistance 1 Ω is obtained which is the minimum.
R_p = \(\frac { R }{ n }\) = \(\frac { 5 Ω }{ 5 }\) = 1 Ω

(11) How does the resistance (R) of a wire depend upon its radius (r)?
Answer:
As R ∝ \(\frac{1}{A} \propto \frac{1}{\pi r^2}\) (∵A = πr²)
∴ R ∝ \(\frac { 1 }{ r² }\) (∵π = constant)

(12) An ammeter has a range (0 – 3A) and there are 30 divisions on its scale. Calculate the least count of the ammeter.
Answer:
Here, range of ammeter = 3 ampere
(i.e., The maximum current that can be measured by the given ammeter is 3A.)
No. of divisions on scale of ammeter = 30
The least count of the ammeter
= \(\frac{\text { Range of the ammeter }}{\text { Total no. of divisions }}\)
= \(\frac { 3 A }{ 30 }\)
= 0.1 A
(i.e., Minimum current that can be accurately measured by a given ammeter is 0.1 A.)

(13) In a voltmeter there are 20 divisions between 0 mark and 0.5 V mark. Calculate the least count of the voltmeter.
Answer:
Range of voltmeter = 0.5 V
No. of divisions = 20
∴The least count of the voltmeter
= \(\frac{\text { Range }}{\text { Total no. of divisions }}\)
= \(\frac { 0.5 V }{ 20 }\)
= 0.025 V

(14) To verify the Ohm’s law a circuit diagram was drawn by a student as shown below:

What do K, L, M, N stand for?
Answer:
K = voltmeter, L = rheostat, M = ammeter and N = plug key

(15) Four students connect 4 cells of 1.5 V each to get a battery of voltage 6 V. State the incorrect connection / connections. Justify your answer.

Answer:
A, B and C are incorrect connections (D is correct).
Because for a proper series connection of four cells, each of 1.5 V to obtain a battery of 6 V, the negative terminal of each cell should be connected to the positive terminal of the next cell or the positive terminal of each cell should be connected to the negative terminal of the next cell (connection D).

(16) Which of the following statements is correct?
(a) one volt is one joule per ampere.
(b) one volt is one joule per coulomb.
Answer:
(b) one volt is one joule per coulomb.

(17) Keeping the resistance constant, the potential difference applied across the ends of a component is halved. How does the current change?
Answer:
The current becomes half the initial current.

(18) Keeping the potential difference constant, the resistance of a circuit is halved. How does the current change?
Answer:
The current becomes double the initial S current.

(19) A potential difference of 10 V is needed i to make a current of 0.02 A flow through a wire. What potential difference is needed to make a current of 250 mA to flow through the same wire?
Answer:
According to the Ohm’s law,
R = \(\frac { V }{ I }\)
So, R = \(\frac { 10 }{ 0.02 }\)
= 500 Ω
Now, to make a current of
250 mA = 250 x 10^-3A to flow through the ; same wire, the required potential difference is
V = IR = 250 x 10^-3 x 500 = 125 V

(20) A current of 250 mA flows through a 4 kΩ resistor. What is the potential difference across the resistor?
Answer:
The potential difference across the resistor
V = IR
= 200 x 10^-3 x 4 x 10³ = 800 V

(21) The electrical resistivities of four materials ; A, B, C and D are given below:
A : 110 x 10^-8 Ω m
B : 1.0 x 10¹⁰ Ω m
C : 10.0 x 10^-8 Ω m
D : 2.3 x 10³ Ω m
Which material is a (a) good conductor (b) resistor (c) insulator (d) semi-conductor?
Answer:
(a) C
(b) A
(c) B
(d) D

(22) The equivalent resistance of a series combination of two resistors is 9 Ω and the equivalent resistance of a parallel combination of the same two resistors is 2Ω. Find the resistances of the resistors.
Answer:
Suppose two unknown resistances are x and y.
When the resistors are connected in parallel \(\frac { xy }{ x+y }\) = 2Ω and when they are connected in series x + y = 9 Ω.
So, \(\frac { xy }{ 9 }\) = 2 ∴ xy = 18 Ω
Now,
From x + y = 9, y = 9 – x
∴ x(9 – x) = 18
∴ 9x – x² = 18
∴ x² – 9x + 18 = 0
∴ x² – 6x – 3x + 18 = 0
∴ x (x – 6) – 3 (x – 6) = 0
∴ (x – 3) (x – 6) = 0
∴ x = 3 Ω or x = 6 Ω
Hence y = 9 – x = 9 – 3 = 6 Ω or y = 9 – 6 = 3 Ω
Thus, x = 3 Ω and y = 6 Ω or x = 6Ω and y = 3 Ω.

(23) An electric lamp is labelled 12 V, 36 W. This indicate that it should be used with a 12 V supply. What other information does the label provide?
Answer:
The electric lamp consumes electric energy at the rate of 36 J/s.

(24) Name two devices whose working is based on the heating effect of electric current.
Answer:
An electric iron, electric toaster

(25) Name the gases which are filled into the filament-type electric light bulbs.
Answer:
Argon, nitrogen

(26) Why are the filament-type electric light bulbs not power efficient?
Answer:
When electric power is supplied to the filament-type electric light bulb, most of the electric power consumed by the filament appears as heat (due to which the bulb becomes hot), only a small amount of electric power is converted into light. So, the filament-type electric light bulbs are not power efficient.
[Note : Tube-lights are much more power efficient because they have no filaments.]

(27) Under what conditions is the Ohm’s law applicable?
Answer:
The Ohm’s law is applicable when the physical conditions like temperature, pressure, etc. of a conductor do not change when a potential difference is applied across it.

(28) Why is Nichrome used as a heating element?
Answer:
Nichrome (alloy) has high resistivity and high melting point. Moreover, it does not react with air when it is red-hot (800 °C).

(29) Why are Constantan and Manganin used for making standard resistors?
Answer:
The resistivity of Constantan and Manganin is moderate and almost independent of temperature.
Hence, these materials are used for making standard resistors.

(30) Why is lead-tin alloy used for making a fuse?
Answer:
Lead-tin alloy has low melting point. Hence it is used for making a fuse.

(31) What is the difference between a resistor and resistance?
Answer:
A resistor is a component whereas resistance is its property due to which it opposes the flow of electrons through it.

(32) What do you mean by the term load in an electric circuit?
Answer:
Electric lamps, heaters, air conditioners, motors and other devices which work on electricity are called loads because they consume electric energy.

(33) How does the current divide itself in a parallel combination of resistors?
Answer:
In a parallel combination of resistors, the current divides in the inverse ratio of the resistances.

(34) How does the potential difference divide itself in a series combination of resistors?
Answer:
In a series combination of resistors, the potential difference across any of the resistors is directly proportional to its resistance.

(35) How is heat produced in a conductor by an electric current?
Answer:
Heat produced in a conductor by an electric current is due to collisions of free electrons with the ions and atoms in the conductor.

(36) If a number of bulbs of different wattages? are joined in parallel with a voltage source, which s bulb will glow with maximum brightness?
Answer:
The bulb with highest wattage (here, s lowest resistance) will glow with maximum brightness as P = \(\frac { V² }{ R }\) and here P ∝ \(\frac { 1 }{ R }\).

(37) Which bulb will glow with maximum brightness in case bulbs of different wattages are : joined in series with a voltage source?
Answer:
The bulb with highest wattage (here S highest resistance) will glow with maximum brightness as P = I²R and here P ∝ R.

(38) What can you say about the resistance s of an ammeter?
Answer:
The resistance of an ammeter should s be very low, ideally zero.

(39) What can you say about the resistance ; of a voltmeter?
Answer:
The resistance of a voltmeter should be very high, ideally infinite.

(40) How does the resistance of a metallic wire depend on its temperature?
Answer:
The resistance of a metallic wire increases with temperature.

(41) Two wires have the same length, same s radius but one of them is made of copper and l the other is made of iron. Which will have more resistance?
Answer:
As resistance R = ρ\(\frac { 1 }{ A }\) and resistivity ρ for iron is more than that for copper, the iron wire will have more resistance.

(42) Name a substance whose resistance decreases with temperature.
Answer:
Germanium (a semi-conductor)

(43) Name a substance whose resistance almost? remains unchanged with a change in temperature.
Answer:
Manganin (an alloy of Cu, Mn and Ni)

(44) Distinguish between the kilowatt and the kilowatt hour.
Answer:
The kilowatt (kW) is a unit of power and the kilowatt hour (kWh) is a unit of electric energy. .
1 kWh = 1 kW x 1 h

(45) Name two characteristics of a heater coil.
Answer:
A heater coil should have (i) high resistivity and (ii) high melting point.

(46) The power-voltage rating of an electric appliance is 100W-250V What does it signify?
Answer:
The electric appliance consumes a power of 100 W (i.e., an electric energy of 100 J per second) when connected to a 250 V electric supply.

(47) Why is much less heat generated in long electric cables than in the filament of an electric bulb?
Answer:
Electric cables are made of thick copper i.e., their area of cross-section is large, hence they have much less resistance as compared to the thin tungsten filament of an electric bulb.

(48) How much energy is consumed by 2 kW AC when used for 2 hour?
Answer:
Electric energy E = P x t
= 2 kW x 2 hour = 4 kWh
Where, 1 kWh = 3.6 x 10⁶ J

(49) A wire of resistivity p is pulled to double its length. What will be its new resistivity?
Answer:
Resistivity of wire remains the same as resistivity p depends on the nature of material and temperature.

(50) How does the resistance of an ohmic < conductor depend on the applied voltage?
Answer:
Resistance R is a constant, so it does not depend upon the applied voltage.

(51) What causes the potential difference between the two terminals of a cell?
Answer:
Chemical reactions taking place within a cell result in excess of electrons at the one terminal and deficiency of electrons at the other terminal.

(52) Name and define the smallest commercial unit of electricity.
Answer:
The smallest commercial unit of electricity is the watt-hour (Wh).
It is defined as the amount of electric energy consumed by an electrical device of power rating 1 W in one hour.

(53) If the current I through a resistor is increased by 100 % (assume that the temperature of the resistor remains unchanged), find the percentage increase in power dissipated.
Answer:
Power dissipated P = I²R
Current after increased by 100% = I’
= I + \(\frac { 100 I }{ 100 }\)
= 2 I
So, new power P’ = I’²R = (2I)²R = 4I²R
∴ Percentage increase in power dissipation
= \(\frac{P^{\prime}-P}{P}\) x 100 %
= \(\frac{4 I^2 R-I^2 R}{I^2 R}\) x 100 %
= \(\frac{I^2 R(4-1)}{I^2 R}\) x 100 %
= 3 x 100%
= 300 %

Value Based Questions With Answers

Question 1.
Medha observed that the tubelights in the corridor of her school were always switched on the whole day. She brought the matter to the notice of her class teacher who talked to the Principal about it. The Principal took immediate action.

1. Medha helped this way to reduce air pollution. Explain how.
2. What values do you learn from Medha in this episode?
3. What steps can the school take to get electricity consumption reduced?

Answer:

1. Power / Electricity production involves burning of fossil fuel. Electricity saved is fuel saved, less fuel burnt means less air pollution.
2. Courage, appreciation and protection of environment.
3. (i) Students should switch off the lights while going out of their class, (ii) they should not waste water as it also involves the use of electricity many a times.

Question 2.
Shalini is a student of class X. Her mother was preparing tea in an old electric kettle having metal case. When she switched on the electric kettle, she got a severe electric shock. Shalini put off the main switch quickly and found that the connecting cord was torn, where her mother touched the metal case of the kettle. She also found that the red and black wires of the connecting cord were firmly connected to the two lower terminals of the power plug but S the green wire of the cord was not connected to the upper terminal of the plug.

Shalini replaced the torn connecting cord and also connected the three wires of the cord firmly to the power plug terminals.

On the basis of the above passage, answer the following questions :

1. Why did Shalini switch off the main switch quickly?
2. Which wire red, black or green, touched the metal case of electric kettle when Shalini’s s mother got electric shock?
3. What values are displayed by Shalini in this incident?

Answer:

1. Shalini switched off the main switch to cut-off the electricity supply to the faulty s’ kettle in order to save her mother.
2. Red wire which is at high potential of 220 V was touching the metal case of electric kettle.
3. The values displayed by Shalini are :
   • • Presence of mind.
     • Concern for her mother.
     • Knowledge of household wiring and daily life activities.

Question 3.
Bharat was doing an experiment by using an ammeter. Suddenly, it fell from his hand and broke. He was afraid, that he might be scolded and threaten by his teacher. His classmate advised him not to tell the teacher, but he refused and told his teacher. On listening to him patiently, the teacher did not scold him and threaten as it was just an accident and used the opportunity to show the internal structure of the ammeter to the whole class.

On the basis of above passage, answer the following questions :

1. What values do you learn from Bharat?
2. What is the use of ammeter? How is it connected in the circuit?
3. State the aim of any one experiment, where Bharat could have used the ammeter.

Answer:

1. The values learnt from Bharat are honesty, confidence and courage.
2. An ammeter is used to measure the electric current flowing in a circuit. It is always connected in series in a circuit.
3. Bharat could have used the ammeter to verify the Ohm’s law.

Question 4.
Mitali’s mother was cooking in the kitchen for guests. Mitali saw her mother had plugged in microwave, hot plate and food processor on the same plug point. She immediately switched off the plug and removed all the plugs and re-plugged them in separate individual plugs.

1. What happens when we use too many electrical devices plugged in one power point?
2. What is the power of a device?
3. What value did Mitali display in the above act?

Answer:

1. Plugging too many devices in one power point causes overloading which can lead to electrical fire.
2. The power of an electrical device is the amount of electric energy used by it per unit time.
3. Mitali showed the values of awareness and responsibility.

Question 5.
Vishva noticed in a dhaba around 50 bulbs, each of 100 W all were glowing. She calculated the cost of electricity consumed in one hour and told the dhaba owner to reduce the expenses and at the same time save electricity by using CFL bulbs instead of ordinary filament-type bulbs.
(1) What would be the cost of electric energy if 50 bulbs, each of 100 W, are used for one hour? 1 unit costs ₹ 5.
(2) What is a CFL?
(3) What value did Vishva display in the above case?
Answer:
(1) The cost of electricity consumed when 50 bulbs are used for 1 hour
= (number of units consumed) x ₹ 5 per unit
= (50 x \(\frac { 100 }{ 1000 }\) kWh x 1 h) x \(\frac { ₹ 5 }{ 1 unit }\)
= (5 k Wh) x \(\frac { ₹ 5 }{ 1 unit }\)
= ₹ 25

(2) A CFL is a compact fluorescent lamp.

(3) Vishva displayed responsible behaviour, awareness as a citizen, critical thinking and analysis.

Practical Skill Based Questions With Answers

Question 1.
The scales of an ammeter and a voltmeter are shown below:

(1) What is the range of the ammeter and the range of the voltmeter?
(2) Find the least count of the ammeter and the voltmeter.
Answer:
(1) The range of the ammeter is 6 A and the range of the voltmeter is 1.5 V.

(2) The least count of the ammeter

Question 2.
See the figure given below and find the readings of ideal ammeter and ideal voltmeter. (There is no zero error.)

Answer:
The reading of the ammeter = 1.15 A. The value of 1 division on the scale of the ammeter (the least count) is \(\frac { 1.5 }{ 30 }\) = 0.05 A and its pointer shows 23 divisions on its scale. Hence, the ammeter reading = 23 x 0.05 A = 1.15 A.
The reading of the voltmeter = 2.2 V
The value of 1 division on the scale of the voltmeter (the least count) is \(\frac { 3 }{ 15 }\) = 0.2 V and its pointer shows 11 divisions on its scale. Hence the voltmeter reading 11 x 0.2 = 2.2V

Question 3.
In a given ammeter, a student observes that the needle indicates 17 divisions in the ammeter when performing an experiment to verify Ohm’s law. If the ammeter has 10 divisions between 0 and 0.5 A, then what is the value of 17 divisions?
Answer:
Since the ammeter has 10 divisions between 0 and 0.5 A, its least count or Vale of 1 division = \(\frac { 0.5A}{ 10 }\) = 0.5 A
Therefore, the value of 17 divisions
= 17 (0.05 A) = 0.85 A

Question 4.
Four resistors, each of 10 Ω, are connected to form a square as shown in the given figure. Find the equivalent resistance between the opposite corner A and C and between two points on any one side (AB).

Answer:
(i) Resistance in the path
ADC = 10 Ω + 10 Ω = 20 Ω
Resistance in the path
ABC = 10 Ω + 10 Ω = 20 Ω
Now, \(\frac{1}{R_{\mathrm{AC}}}=\frac{1}{20}+\frac{1}{20}=\frac{1}{10}\)
∴Equivalent resistance,
R_AC = 10 Ω

(ii) Resistance in the path ADCB
= 10 Ω + 10 Ω + 10 Ω = 30 Ω.
The resistance of 30 Ω is in parallel with AB (= 10 Ω).
Now, \(\frac{1}{R_{\mathrm{AB}}}=\frac{1}{30}+\frac{1}{10}\)
= \(\frac{1+3}{30}=\frac{4}{30}\)
∴ Equivalent resistance,
R_AB = \(\frac { 30 }{ 4 }\) = 7.5 Ω

Question 5.
In the circuit shown in figure given below and find the current recorded by the ammeter (A).

Answer:
Since resistance, AC = 30 Ω and resistances CB = 30 Ω. resistance in the path ACB = 30 Ω + 30 Ω = 60 Ω.
Further, 60 Ω (in the path ACB) and 20 Ω (in the path AB) are in parallel.
Hence, effective resistance (R) between A and B is given by
\(\frac{1}{R}=\frac{1}{60}+\frac{1}{20}=\frac{1+3}{60}=\frac{4}{60}=\frac{1}{15}\)
∴ R = 15 Ω
∴Current I = \(\frac { 7.5V }{ 15Ω }\) = 0.5 A
This is the current recorded by the ammeter.

Question 6.
Read the following information:
(1) The resistivity of copper is lower than that of aluminium which in turn is lower than that of constantan.
(2) Six wires labelled as A, B, C, D, E and F have been designed as per the following parameters :

Wire	Length	Diameter	Material	Resistance
A	l	2d	Aluminium	RA
B	2l	d/2	Constantan	RB
C	3l	d/2	Constantan	Rc
D	l/2	3d	Copper	RD
E	2l	2d	Aluminium	RE
F	l/2	4 d	Copper	RF

Answer the following questions using the above data :
(1) Which of the wires has maximum resistance and why?
(2) Which of the wires has minimum resistance and why?
(3) Arrange R_A, R_B and R_C in ascending order of their values. Justify your answer.
Answer:
(1) The wire C has maximum resistance because it has maximum length, least thickness and highest resistivity.

(2) The wire F has the minimum resistance, since it has least length, maximum thickness and least resistivity. (Using, R = ρ\(\frac { l }{ A }\))

(3) R_C > R_E > R_A
(Using relation, R = ρ\(\frac { l }{ A }\) and comparision)

Question 7.
n resistors, each of resistance R are first connected in series and then in parallel. What is the ratio of the total effective resistance of the circuit in the series to the parallel combination?
Answer:
In the series combination, R_s = nR
In the parallel combination, R_p = \(\frac { R }{ n }\)
∴ \(\frac{R_{\mathrm{s}}}{R_{\mathrm{p}}}=\frac{n R}{R / n}\) = n²

Question 8.
Find out the readings of ammeter and voltmeter in the circuit given below:

Answer:
Equivalent resistance R = (2 + 3 + 3 + 4)Ω
= 12 Ω
Total current drawn = Ammeter reading
= \(\frac { V }{ R }\)
= \(\frac { 6 }{ 12 }\) = 0.5 A
Voltmeter reading = IR₁ = 0.5 x 2 = 1 V

Question 9.
In the given circuit diagram, calculate :
(1) the value of current through each resistor
(2) the total current in the circuit
(3) the total effective resistance of the circuit.

Answer:

Question 10.
Show four different ways in which three resistors of Y ohm, each may be connected in a circuit. In which case is the equivalent resistance of the combination :
(1) Maximum?
(2) Minimum?
Answer:

(1) Circuit (a) has maximum resistance

(2) Circuit (b) has minimum resistance

Question 11.
Find the current drawn from the battery by the network of four resistors shown in the figure.

Answer:
Two combinations of 3V two parallel resistors of 10 Ω each are connected in series.

Question 12.
Find the equivalent resistance across the ends A and B of this circuit.
Answer:

\(\frac{1}{R_{1,2}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}\) = 1
∴ R_1,2 = 1 Ω
Similarly R_3,4 = 1 Ω
R_5,6 = 1 Ω
R_7,8 = 1 Ω
Now R_1,2 and R_3,4 are connected in series
R_1,2,3,4 = R_1,2 + R_3,4 = 1 + 1 = 2Ω
Also, R_5,6 and R_7,8 are connected in series
R_5,6,7,8 = R_5,6 + R_7,8 = 1 + 1 = 2Ω
Now, equivalent resistance
\(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_{1,2,3,4}}+\frac{1}{R_{5,6,7,8}}\)
= \(\frac{1}{2}+\frac{1}{2}=\frac{2}{2}\) = 1
∴ R_eq = 1 Ω

Question 13.
Five resistors are connected in a circuit as shown. Find the ammeter reading when the circuit is closed.

Answer:
Let the resistance between the points B and D be x.
∴ x = \(\frac{3 \Omega \times 6 \Omega}{3 \Omega+6 \Omega}\)
Now, the resistance of the entire circuit is
R = 0.5 Ω + x + 0.5 Ω
= 0.5 Ω + 2.0 Ω + 0.5 Ω
= 3.0 Ω
∴ Ammeter reading = \(\frac { 3 V }{ 3 Ω }\) = 1.0 A.

Question 14.
Draw a diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination.
Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4 Ω resistors? Give the reason.
Answer:
If R_p is the effective resistance of two resistors of 4 Ω each in parallel, then
\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)
∴ R_p = 2 Ω

Since, the combined parallel resistance is also 2 Ω, the potential difference across 2 Ω resistor will be the same as that across the parallel combination of two 4 Ω resistors.

Question 15.
A current of 1A flows in a series circuit containing an electric lamp and a conductor of 5 Ω connected to a 10 V battery. Calculate the resistance of the electric lamp.

Now if a resistance 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place Give the reason.
Answer:

As, I = \(\frac{E}{R_1+R_2}\)
1 = \(\frac { 10 }{ x+5 }\)
or
∴ x = 5 Ω is the resistance of the lamp.
When a resistor of resistance 10 Ω is connected in parallel with x + 5 = 10 Ω, the total resistance R_p of the arrangement is given by,
\(\frac { 10 }{ x+5 }\)
= \(\frac { 1 }{ 5 }\)
∴ R_p = 5 Ω
∴ The current in circuit = \(\frac{E}{R_{\mathrm{p}}}\)
= \(\frac { 10 }{ 5 }\)
= 2 A
This current (2 A) divides eΩually between 10 Ω and (resistance of lamp + 5 Ω).

∴ The current through the lamp will be again 1 A, i.e., it remains the same.
The potential difference across the lamp will be 1 × x = 1 x 5 = 5V.
Therefore, the potential difference across the lamp will remain the same.

Memory Map:

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 6 त्रिभुज Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 6 त्रिभुज

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
दी गई आकृति में रेखाखण्ड XY, त्रिभुज ABC की भुजा AC के समान्तर है तथा त्रिभुज को दो समान भागों में बाँटती है। अनुपात \(\frac{AX}{AB}\) ज्ञात कीजिए।

हल :
दिया है
XY || AC
ΔBXY और ΔBAC में
∠BXY = ∠BAC (संगत कोण)
∠BYX = ∠BCA (संगत कोण)
∴ ΔBXY ~ ΔBAC (AA समरूपता गुणधर्म से )

⇒ \(\frac{AX}{AB}\) = \(\frac{2-\sqrt{2}}{2}\)
अतः \(\frac{AX}{AB}\) = \(\frac{2-\sqrt{2}}{2}\)
∴ AY : AB = (2 – \(\sqrt{2}\)) : 2

प्रश्न 2.
BE और CF एक समकोण ΔABC की माध्यिकाएँ है तथा इस Δ का कोण A समकोण है। सिद्ध कीजिए : 4 (BE² + CF²) = 5BC².
हल :
दिया है : BE तथा CF समकोण ΔABC की माध्यिकाएँ है तथा ∠A = 90° हैं।

सिद्ध करना है: 4(BE² + CF²) = 5BC²
उपपत्ति: समकोण ΔABC में,
BC² = AB² + AC² ……….(1)
समकोण ΔABE में,
BE² = AE² + AB²
BE² = (\(\frac{AC}{2}\))² + AB²
[∵ BE, ΔABC की माध्यिका है
∴ AE = CE = \(\frac{1}{2}\)AC]
BE² = \(\frac{A C^2}{4}\) + 4AB² ……………(2)
अब समकोण ΔCAF में,
CF² = CA² + AF²
CF² = AC² + (\(\frac{AB}{2}\))²
[∵ BF, ΔABC की माध्यिका है
∴ AF = BF = \(\frac{1}{2}\)AB]
CF² = AC² + \(\frac{A B^2}{4}\)
4CF² = 4AC² + AB² …………(3)
समी. (2) और (3) को जोड़ने पर
4 (BE² + CF²) = AC² + 4AB² + 4AC² + AB²
4 (BE² + CF²) = 5 (AC² + AB²)
4 (BE² + CF²) = 5BC²
[समी. (1) के प्रयोग से] इति सिद्धम् ।

प्रश्न 3.
सिद्ध कीजिए कि दो समरूप त्रिभुजों के परिमापों का अनुपात किन्हीं दो संगत भुजाओं के अनुपात के समान होता है।
हल :
दिया है: ΔABC ~ ΔDEF

सिद्ध करना है :

प्रश्न 4.
दी गई आकृति में DE || BC तथा AD : DB = 7 : 5 है, तो क्षेत्रफल (ΔDEF) / क्षेत्रफल (ΔCFB) का मान ज्ञात कीजिए।

हल :
दिया है
DE || BC तथा AD : DB = 7 : 5
ΔADE तथा ΔABC में,
∠ADE = ∠ABC (संगत कोण)
∠AED = ∠ACB (संगत कोण)
∴ ΔADE ~ ΔABC (AA समरूपता गुणधर्म से)
\(\frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}\) ……………(1)
[समरूप त्रिभुज की संगत भुजाएँ समानुपाती होती हैं।]
परन्तु AD : DB = 7 : 5

ΔDFE तथा ΔCFB में
∠DEF = ∠CBF (एकान्तर कोण)
∠DFE = ∠CFB(शीर्षाभिमुख कोण)
∴ ΔDFE ~ ΔCFB
(AA समरूपता गुणधर्म से)

प्रश्न 5.
सिद्ध कीजिए कि एक समकोण त्रिभुज के कर्ण पर खींचे गये समबाहु त्रिभुज का क्षेत्रफल अन्य दो भुजाओं पर खींचे गए समबाहु त्रिभुजों के क्षेत्रफलों के योग के बराबर होता है।
हल :
दिया है : समकोण त्रिभुज ABC की भुजाओं AB, BC तथा AC पर समबाहु त्रिभुज खींचे गए हैं।
सिद्ध करना है : ar (ΔABN) + ar (ΔBCL) = ar(ΔACM)

उपपत्ति : ∵ ΔABN तथा ΔACM के प्रत्येक कोण 60° का होगा ।
अतः ΔABN ~ ΔACM (AA समरूपता गुणधर्म से)

प्रश्न 6.
दिया है कि ΔABC ~ ΔPQR, यदि \(\frac{AB}{PQ}=\frac{1}{3 }\) है, तो ar(ΔABC) / ar(ΔPQR) का मान ज्ञात कीजिए।
हल :
दिया है, ΔABC ~ ΔPQR

प्रश्न 7.
निम्न आकृति में, ∠D = ∠E तथा \(\frac{AB}{DB}=\frac{AE}{EC}\) है, तो सिद्ध कीजिए कि ΔBAC एक समद्विबाहु त्रिभुज है।

हल :
ΔADE
∠D = ∠E (दिया है)
⇒ AD = AE
(∵ बराबर कोणों की सम्मुख भुजाएँ बराबर होती हैं।)
दिया है, \(\frac{AD}{DB}=\frac{AE}{EC}\)
⇒ \(\frac{AD}{DB}=\frac{AD}{EC}\) [समी. (i) से]
⇒ DB = EC …………….(ii)
समीकरण (i) व (ii) को जोड़ने पर,
AD + DB = AE + EC
⇒ AB = AC
अत: ΔABC एक समद्विबाहु त्रिभुज हैं।

प्रश्न 8.
ABC और BDE दो समबाहु त्रिभुज इस प्रकार है कि D, BC का मध्य बिन्दु है । त्रिभुजों ABC और BDE के क्षेत्रफलों में अनुपात ज्ञात कीजिए।
हल :
हम जानते हैं कि समबाहु त्रिभुज समरूप होते हैं,
अर्थात् ΔABC ~ ΔBDE
∵ D, BC का मध्य बिन्दु है,
∴ BC = 2BD ……….(i)

अत: ΔABC का क्षे. ΔBDE का क्षे. = 4 : 1

प्रश्न 9.
निम्न चित्र में, MN || BC और AM : MB = 1 : 2, तो ΔAMN का क्षेत्रफल / ΔABC का क्षेत्रफल का मान ज्ञात कीजिए।

हल :
दिया है, AM = 1K और MB = 2K
∴ AB = AM + MB = 1K + 2K = 3K
ΔAMN और ΔABC में,
∠A = ∠A (उभयनिष्ठ)
∠AMN = ∠ABC (संगत कोण)
∴ ∠AMN ~ ∠ABC (AA समरूपता से)
समरूप त्रिभुजों के क्षेत्रफल के प्रमेय से,

प्रश्न 10.
निम्न चित्र में, यदि AD ⊥ BC, तो सिद्ध कीजिए कि AB² + CD² = BD² + AC²

हल :
समकोण ΔADC मे, पाइथागोरस प्रमेय से,
AC² = AD² + CD²
⇒ AD² = AC² – CD² ……….(i)
पुन: समकोण ΔADB में, पाइथागोरस प्रमेय से
AB² = AD² + BD²
⇒ AD² = AB² – BD² ……….(ii)
समीकरण (i) और (ii) से,
AB² – BD² = AC² – CD²
⇒ AB² + CD² = AC² + BD² इति सिद्धम्

प्रश्न 11.
निम्न चित्र में, यदि ΔABC ~ ΔDEF तथा उनकी भुजाओं की लम्बाइयाँ (सेमी में) उन पर अंकित है, तो प्रत्येक त्रिभुज की भुजाओं की लम्बाई ज्ञात कीजिए।

हल :
दिया है, ΔABC ~ ΔDEF

⇒ 4x – 2 = 18
⇒ x = \(\frac{20}{4}\) = 5
ΔABC में,
AB = 2x – 1 = 2 × 5 – 1 = 9 इकाई
BC = 2x + 2 = 2 × 5 + 2 = 12 इकाई
CD = 3x = 3 × 5 = 15 इकाई
ΔDEF में,
DE = 18 इकाई
EF = 3x + 9 = 3 × 5 + 9 = 24 इकाई
DF = 6x = 6 × 5 = 30 इकाई

प्रश्न 12.
चित्र में, यदि PQ || BC तथा PR || CD है, तो सिद्ध कीजिए कि \(\frac{QB}{AQ}=\frac{DR}{AR}\)

हल :
ΔABC में, PQ || BC
∴ आधारभूत आनुपातिकता प्रमेय से,
\(\frac{AQ}{QB}=\frac{AP}{PC}\) ………….(i)
पुन: ΔADC में,
PR || CD
∴ आधारभूत आनुपातिकता प्रमेय से,
\(\frac{AR}{RD}=\frac{AP}{PC}\) ………….(ii)
समीकरण (i) व (ii) से,
\(\frac{AQ}{QB}=\frac{AR}{RD}\)
अतः \(\frac{QB}{AQ}=\frac{DR}{AR}\) (इति सिद्धम्)

प्रश्न 13.
दो समरूप त्रिभुजों के क्षेत्रफलों का अनुपात 16 : 81 है तो इनकी भुजाओं का अनुपात ज्ञात कीजिए।
हल :
समरूप त्रिभुजों के क्षेत्रफल के प्रमेय से,
समरूप त्रिभुज के क्षेत्रफलों का अनुपात = संगत भुजाओं के वर्गों का अनुपात
⇒ 16 : 81 = संगत भुजाओं के वर्गों का अनुपात
⇒ संगत भुजाओं का अनुपात = \(\sqrt{\frac{16}{81}}\) = \(\frac{4}{4}\) = 4 : 9

प्रश्न 14.
निम्न आकृति में, ABC एक त्रिभुज है। यदि \(\frac{AD}{AB}=\frac{AE}{AC}\), तो सिद्ध कीजिए DE || BC है ।

हल :

आधारभूत आनुपातिकता प्रमेय के विलोम से DE || BC (इति सिद्धम)

प्रश्न 15.
यदि दो समरूप त्रिभुजों की संगत माध्यिकाओं का अनुपात 9 : 16 है, तो इनके क्षेत्रफलों का अनुपात ज्ञात कीजिए।
हल :
∵ दो समरूप त्रिभुजों के क्षेत्रफलों का अनुपात इनकी संगत भुजाओं के वर्ग के अनुपात के बराबर होता है ।
∴ क्षेत्रफलों का अनुपात = (9)² : (16)² = 81 : 256

प्रश्न 16.
दो समरूप त्रिभुजों के परिमाप क्रमश: 25 सेमी तथा 15 सेमी है। यदि पहले त्रिभुज की एक भुजा की लंबाई 9 सेमी है, तो दूसरे त्रिभुज की संगत भुजा की लंबाई ज्ञात कीजिए।
हल :
माना दूसरे त्रिभुज की संगत भुजा की लंबाई x सेमी है।
∵ समरूप त्रिभुज की संगत भुजा की लंबाई का अनुपात उनके परिमाप के अनुपात के बराबर होता है।
∴ \(\frac{25}{15}=\frac{9}{x}\)
⇒ x = \(\frac{15 \times 9}{25}\) = 5.4 सेमी
अत: दूसरे त्रिभुज की संगत भुजा = 5.4 सेमी

प्रश्न 17.
चित्र में ΔABC में, DE || BC है तथा AD = 2.4 सेमी, AB = 3 सेमी और AC = 8 सेमी है, तो AE की लम्बाई क्या है?

हल :
चित्र में,
DE || BC
तथा
AD = 2.4 सेमी
AB = 3.2 सेमी
AC = 8 सेमी

प्रश्न 18.
निम्न आकृति में, ΔABC तथा ΔXYZ दर्शाए गए हैं। यदि ABC = 3.8 सेमी, AC = 3\(\sqrt{3}\) सेमी, BC = 6 सेमी, XY = 6\(\sqrt{3}\) सेमी, XY = 7.6 सेमी, YZ = 12 सेमी तथा ∠A = 65°, ∠B = 70° हों, तो ∠Y का मान ज्ञात कीजिए।
हल :
ΔABC में,
∠A + ∠B + ∠D = 180°
65° + 70° + ∠C = 180°
∠C = 180° – 135° = 45°
अब ΔABC और ΔXYZ में

अर्थात् ΔABC ~ ΔXYZ
∴ ∠A = ∠X = 65°, ∠B = ∠Z = 70°
तथा ∠C = ∠Y = 45°

प्रश्न 19.
निम्न चित्र में, समबाहु त्रिभुज ABC में, AD ⊥ BC, BE ⊥ AC तथा CF ⊥ AB है। सिद्ध कीजिए:
4(AD² + BE² + CF²) = 9AB²

हल :
ΔADB और ΔADC में,
और AB = AC (दिया है)
∠B = ∠C (प्रत्येक कोण 60° है।)
और ∠ADB = ∠ADC (प्रत्येक कोण 90° है।)
∴ ΔADB ≅ ΔADC
⇒ BD = DC
⇒ BD = DC = \(\frac{1}{2}\)BC
∵ ΔADB एक समकोण त्रिभुज है,
∴ AB² = AD² + BD²
⇒ AB² = AD² + (\(\frac{BC}{2}\))²
⇒ AB² = AD² + \(\frac{B C^2}{4}\)
⇒ 4AB² = 4AD² + AB²
⇒ 3AB² = 4AD² …………(i)
इसी प्रकार से, ABEC एक समकोण त्रिभुज है,
∴ BC² = BE² + EC²
⇒ AB² = BE² + (\(\frac{AC}{2}\))²
[∵ BC = AB और EC = AE = \(\frac{AC}{2}\)]
⇒ AB² = BE² + (\(\frac{AC}{2}\))²
⇒ 4AB² = 4BE² + AC²
⇒ 4AB² = 4BE² + AB² [∵ AB = AC]
⇒ 3AB² = 4BE² …………..(ii)
अब समकोण ΔBFC में,
BC² = BF² + CF²
⇒ AB² = (\(\frac{AB}{2}\))² + CF²
⇒ AB² = \(\frac{A B^2}{4}\) + CF²
⇒ 4AB² = AB² + 4CF²
⇒ 3AB² = 4CF² ……..(iii)
समीकरण (i), (ii) और (iii) को जोड़ने पर
9AB = 4(AD² + BE² + CF²) इति सिद्धम

प्रश्न 20.
दी गई आकृति में, ABC एक समद्विबाहु त्रिभुज है, जिसका कोण C है तथा AC = 4 सेमी है। AB की लम्बाई ज्ञात कीजिए।

हल :
दिया है, ΔABC एक समद्विबाहु त्रिभुज है,
∠C = 90° तथा AC = 4 सेमी
∴ ΔABC एक समद्विबाहु त्रिभुज है। अतः
BC = AC
ΔABC में, पाइथागोरस प्रमेय से,
AB² = AC² + BC²
AB² = AC² + AC² (∵BC = AC)
= 2 × AC²
= 2 × 4²
= 32
⇒ AB = \(\sqrt{32}\)
AB = 4\(\sqrt{2}\) सेमी

प्रश्न 21.
कर्ण BC पर एक ही तरफ दो समकोण त्रिभुज ABC, DBC बनाए गए है। यदि AC तथा BD एक- दूसरे बिन्दु पर प्रतिच्छेद करते हैं, तो सिद्ध कीजिए कि :
AP × PC = BP × DP
हल :
दिया है, ΔABC तथा ΔDBC समकोण त्रिभुज हैं।
ΔABC व ΔDBC में,
∠BAC = ∠BDC = 90°
∠BPA = ∠CPD (शीर्षाभिमुख होगा )

अत: AA समरूपता में,
ΔABC ~ ΔDBC
\(\frac{AP}{BP}=\frac{DP}{PC}\)
[∵ समरूप त्रिभुजों की संगत भुजाओं का अनुपात समान होता है।]
⇒ AP × PC = DP × BP इति सिद्धम्

प्रश्न 22.
दी गई आकृति में, DE || BC, AD = 1 सेमी तथा BD = 2 सेमी है। ΔABC तथा ΔADE के क्षेत्रफलों में क्या अनुपात है?

हल :
दिया है, DE || BC, AD = 1 सेमी तथा BD = 2 सेमी
चूँकि, DE || BC
इसीलिए, ∠ADE = ∠DBC ( संगत कोण)

अतः ar(ΔABC) : ar(ΔADE) = 9 : 1

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क)

1. दो आकृतियाँ, जिनके आकार समान हों, परन्तु उनकी आमाप समान न हो, ……………… कहलाती हैं।
2. दो आकृतियाँ जिनके आकार तथा आमाप दोनों समान हों, …………. आकृतियाँ कहलाती है।
3. दो त्रिभुज समरूप होते हैं यदि उनके संगत कोण ……………….. हों ।
4. त्रिभुजों की समरूपता को प्रकट करने के लिए उनक शीर्षों की संगतताओं को सही …………….. में लिखना आवश्यक है।
5. यदि एक रेखा किसी त्रिभुज की दो भुजाओं को एक ही अनुपात में विभाजित करें, तो वह तीसरी भुजा के होती है।

हल :

1. समरूप,
2. सर्वांगसम
3. समान,
4. क्रम,
5. समांतर ।

निम्न में सत्य / असत्य ज्ञात कीजिए :

प्रश्न (ख)

1. दो समरूप त्रिभुजों के क्षेत्रफल त्रिभुजों की संगत भुजाओं के वर्गों के अनुपात में होते हैं।
2. एक समद्विबाहु त्रिभुज में कर्ण का वर्ग शेष दो भुजाओं के वर्गों के योग के बराबर होता है।
3. यदि किसी त्रिभुज की एक भुजा का वर्ग अन्य दो भुजाओं के वर्गों के योग के बराबर हो, तो पहली भुजा का सम्मुख कोण अधिक कोण होता है।
4. पाइथागोरस प्रमेय के बौधायन प्रमेय के नाम से भी जाना जाता है।
5. सभी वृत्त सर्वागसम होते हैं।

हल :

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

(ग) बहुविकल्पात्मक प्रश्न :

प्रश्न 1.
यदि ΔABC ~ ΔDEF इस प्रकार है कि AB = 1.2 सेमी तथा DE = 1.4 सेमी है, तो त्रिभुजों ABC तथा DEF के क्षेत्रफलों में अनुपात है:
(A) 49 : 36
(B) 6 : 7
(C) 7 : 6
(D) 36 : 49
हल :
ΔABC ~ ΔDEF

अत: सही विकल्प (D) है।

प्रश्न 2.
चित्र में, ABC एक समद्विबाहु समकोण त्रिभुज है जिसका समकोण C है। अतः
(A) AB² = 2AC²
(B) BC² = 2AB²
(C) AC² = 2AB²
(D) AB² = 4AC²

हल :
समकोण समद्विबाहु ΔABC में,
AB² = AC² + BC²
AB² = AC² + (AC) ² [∵ BC = AC]
AB² = AC² + AC²
AB² = 2AC²
अत: सही विकल्प (A) हैं।

प्रश्न 3.
चित्र में, DE || BC है। यदि \(\frac{AB}{DB}\) = \(\frac{3}{2}\) तथा AE = 2.7 सेमी है, तो EC बराबर है :

(A) 2.0 सेमी
(B) 1.8 सेमी
(C) 4.0 सेमी
(D) 2.7 सेमी
हल :
ΔABC में DE || BC
∴ आधारभूत समानुपातिक प्रमेय से,

अत: सही विकल्प (B) है।

प्रश्न 3.
यदि ΔABC ~ ΔDEF हो एवं AB = 10 सेमी, DE = 8 सेमी हो, तो ΔABC का क्षेत्रफल : ΔDEF का क्षेत्रफल होगा :
(A) 25 : 16
(B) 16 : 25
(C) 4 : 5
(D) 5 : 4
हल :
ΔABC ~ ΔDEF

अत: सही विकल्प (A) है।

प्रश्न 4.
ΔABC में यदि D, BC का मध्य- बिन्दु इस प्रकार है कि \(\frac{AB}{AC}=\frac{BD}{DC}\) हो एवं ∠B = 70°, ∠C = 70° हो, तो ∠BAD की माप होगी :
(A) 60°
(B) 30°
(C) 20°
(D) 50°
हल :
ΔABC में,
∠A + ∠B + ∠C = 180°
⇒ ∠A + 70° + 70° = 180°

⇒ ∠A = 180° – 140°
∴ ∠A = 40°
अतः \(\frac{AB}{AC}=\frac{BD}{DC}\) और ∠B = ∠C
⇒ ΔABD ~ ΔACD
(SAS समरूपता कसौटी से )
∴ ∠BAD और ∠DAC आपस में बराबर होंगे।
∠BAD + ∠DAC = 40°
⇒ 2∠BAD = 40°
∴ ∠BAD = \(\frac{40°}{2}\) = 20°
अत: सही विकल्प (C) है।

प्रश्न 5.
दी गई आकृति में, XY || QR, \(\frac{PQ}{XQ}=\frac{7}{3}\) तथा PR = 6.3 सेमी है, तो YR बराबर है :
(A) 2.7 सेमी
(B) 18.9 सेमी
(C) 2.1 सेमी
(D) 0.9 सेमी

हल :
दिया है
त्रिभुज PQR में XY || QR, \(\frac{PQ}{XQ}\) = \(\frac{7}{3}\) तथा PR = 6.3 सेमी।
आधारभूत आनुपातिक प्रमेय से,

अत: विकल्प (A) सही है।

प्रश्न 6.
एक 12 मीटर लम्बी ऊर्ध्वाधर छड़ की जमीन पर छाया की लम्बाई 8 मीटर लम्बी है। यदि इसी समय एक मीनार की छाया की लम्बाई 40 मीटर हो, तो मीनार की ऊँचाई होगी :
(A) 60 मीटर
(B) 60 सेमी
(C) 40 सेमी
(D) 80 सेमी
हल :
माना ऊर्ध्वाधर AB छड़ है और BC उसकी छाया है। PQ मीनार है और CQ उसकी छाया है।

ΔABC और ΔPQC में,
∠ACB = ∠PCQ (उभयनिष्ठ)
∠ABC = ∠PQC (प्रत्येक 90°)
कोण-कोण समरूपता गुणधर्म से,
ΔABC ~ ΔPOC
∴ \(\frac{AB}{PQ}=\frac{BC}{QC}\) (∵ समरूप त्रिभुजों की संगत भुजाएँ समानुपाती होती हैं)
⇒ \(\frac{12}{PQ}=\frac{8}{40}\)
∴ PQ = \(\frac{40 \times 12}{8}\)
= 5 × 12 = 60 मीटर
अतः सही विकल्प (A) है।

प्रश्न 7.
आकृति में, DE || BC है यदि AD = x, DB = x – 2, AE = x + 2 तथा EC = x – 1 हो, तो x का मान होगा :

(A) 3
(B) 2
(C) 4
(D) 1
हल :
ΔABC में,
DE || BC
\(\frac{AD}{BD}=\frac{AE}{EC}\)
⇒ \(\frac{x}{x-2}=\frac{x+2}{x-1}\)
⇒ x(x – 1) = (x – 2) (x + 2)
⇒ x² – x = x² – 4
∴ x = 4
अत: सही विकल्प (C) है।

प्रश्न 8.
आकृति में ΔACB ~ ΔAPQ है यदि BC = 8 सेमी, BA = 6.5 सेमी, PQ = 4 सेमी और AP = 2.8 सेमी हो, तो 40 का मान होगा :

(A) 5.6 सेमी
(B) 6.8 सेमी
(C) 10.5 सेमी
(D) 3.25 सेमी
हल :
ΔACB ~ ΔAPQ

अत: सही विकल्प (D) है।

प्रश्न 9.
यदि ΔABC में AB = 5 सेमी, BC = 12 सेमी और AC = 13 सेमी है, तो ∠B का मान होगा:
(A) 120°
(B) 60°
(C) 90°
(D) 45°
हल :
ΔABC में,
(AB)² + (BC)² = (5)² + (12)²
= 25 + 144 = 169
= (AC)²
∴ AB² + BC² = AC²
पाइथागोरस प्रमेय के विलोम से,
ΔABC समकोण त्रिभुज है जिसमें
∠B = 90°
अत: सही विकल्प (C) है।

प्रश्न 10.
दो समरूप ΔABC और ΔPQR हैं जिनके क्षेत्रफल क्रमशः 100 सेमी² तथा 144 सेमी² है और ΔABC की ऊँचाई 6 सेमी है तो ΔPQR की ऊँचाई होगी :
(A) 12 सेमी
(B) 6.3 सेमी
(C) 7.2 सेमी
(D) 4.8 सेमी
हल :
हम जानते हैं कि दो समरूप त्रिभुजों के क्षेत्रफलों का अनुपात उनकी संगत ऊँचाइयों के वर्गों के अनुपात के समान होता है। माना ΔPQR की संगत ऊँचाई x सेमी है।

अत: सही विकल्प (C) है।

प्रश्न 11.
यदि ΔABC और ΔDEF में \(\frac{AB}{DE}=\frac{BC}{FD}\) हो, तो वे समरूप होंगे, यदि :
(A) ∠B = ∠E
(B) ∠A = ∠D
(C) ∠B = ∠D
(D) ∠A = ∠F
हल :
सही विकल्प (C) है।

प्रश्न 12.
यदि ΔDEF और ΔPQR में ∠D = ∠Q और ∠R = ∠E हो, तो निम्नलिखित में से कौन-सा सही नहीं है?
(A) \(\frac{EF}{PR}=\frac{DF}{PQ}\)
(B) \(\frac{DE}{PQ}=\frac{EF}{RP}\)
(C) \(\frac{DE}{QR}=\frac{DF}{PQ}\)
(D) \(\frac{EF}{RP}=\frac{DE}{QR}\)
हल :
सही विकल्प (B) है।

प्रश्न 13.
दो समरूप त्रिभुजों के क्षेत्रफलों का अनुपात 16 : 81 है, तो त्रिभुजों की भुजाओं का अनुपात होगा :
(A) 4 : 9
(B) 16 : 81
(C) 9 : 4
(D) 2 : 3
हल :
हम जानते हैं कि दो समरूप त्रिभुजों के क्षेत्रफलों का अनुपात उनकी संगत भुजाओं के वर्गों के अनुपात के समान होता है ।
अतः दोनों भुजाओं का अनुपात = 4 : 9
अत: सही विकल्प (A) है।