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Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 12 Introduction to Probability Theory Ex 12.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.5

Choose the correct or most suitable answer from the given four alternatives

Question 1.
Four persons are selected at random from a group of 3 men, 2 women and 4 children. The probability that exactly two of them are children is
(1) \(\frac{3}{4}\)
(2) \(\frac{10}{23}\)
(3) \(\frac{1}{2}\)
(4) \(\frac{10}{21}\)
Answer:
(4) \(\frac{10}{21}\)

Explaination:
Number of Men = 3
Number of Women = 2
Number of Children = 4
Number of ways of choosing 2 children out of 4 children = 4C₂
= \(\frac{4 \times 3}{1 \times 2}\) = 6
Number of ways of choosing 4 persons from total 9 persons = 9C₄

= 9 × 2 × 7
= 126
Number of ways of choosing 2 persons (other than 2 children) from the remaining 5 persons (excluding children) = 5C₂

Probability that exactly two of them are children

Question 2.
A number is selected from the set (1, 2, 3, ….., 20}. The probability that the selected number is divisible by 3 or 4 is
(1) \(\frac{2}{5}\)
(2) \(\frac{1}{5}\)
(3) \(\frac{2}{5}\)
(4) \(\frac{2}{5}\)
Answer:
(3) \(\frac{2}{5}\)

Explaination:
Sample space S is
S = {1, 2, 3, ………. , 20}
n(S) = 20
Let A be the event of selecting a number divisible by 3
A = {3, 6, 9, 12, 15, 18}
n(A) = 6
Let B be the event of selecting a number divisible by 4
B = {4, 8, 12, 16, 20}
n(B) = 5
A ∩ B = {12}
n(A ∩ B) = 1
P (Number divisible by 3 or 4) = P (A ∪ B)
= P(A) + P(B) – P(A ∩ B)

Question 3.
A, B and C try to hit a target simultaneously but independently. Their respective probabilities of hitting the target are \(\frac{3}{4}\), \(\frac{1}{2}\), \(\frac{5}{8}\). The probability that the target is hit by A or B but not by C is
(1) \(\frac{21}{64}\)
(2) \(\frac{7}{32}\)
(3) \(\frac{9}{64}\)
(4) \(\frac{7}{8}\)
Answer:
(1) \(\frac{21}{64}\)

Explaination:
Given Probability of hitting the target by A is P(A) = \(\frac{3}{4}\)
Probability of hitting the target by B is P(B) = \(\frac{1}{2}\)
Probability of hitting the target by C is P(C) = \(\frac{5}{8}\)
Given A, B , C are Independent.
∴ P(A ∩ B ∩C) = P(A) P(B) . P(C)
Probability of the target hit by A or B but not by C is
P(A ∩ B ∩ C̅) = P(A ∪ B) . P(C̅)
= [P(A) + P(B) – P (A ∩ B)] [1 – P(C)]
= [P(A) + P(B) – P(A) P(B)] [1 – P(C)]

Question 4.
If A and B are any two events, then the probability that exactly one of them occur is
(1 P(A ∪ B̅) + P(A̅ ∪ B)
(2) P(A ∩ B̅) + P(A̅ ∩ B)
(3) P(A) + P(B) – P(A ∩ B)
(4) P(A) + P(B) + 2P(A ∩ B)
Answer:
(2) P(A ∩ B̅) + P(A̅ ∩ B)

Explaination:
Let A and B be an two events
The probability that exactly one of them occur is
= P(A ∩ B̅) + P(A̅ ∩ B)

Question 5.
Let A and B be two events such that\(\)
P\((\overline{\mathbf{A} \cup \mathbf{B}})\) = \(\frac{1}{6}\), P(A ∩ B) = \(\frac{1}{4}\) and P(A̅) = \(\frac{1}{4}\)
Then the events A and B are
(1) Equally likely but not independent
(2) Independent but not equally likely
(3) Independent and equally likely
(4) Mutually inclusive and dependent
Answer:
(2) Independent but not equally likely

Explaination:


If P(A ∩ B) = p(A) . p(B), then A and B are independent.
we have (A ∩ B) = \(\frac{1}{4}\) ………… (1)
P(A) . P(B) = \(\frac{3}{4} \times \frac{1}{3}=\frac{1}{4}\) ………… (2)
From equations (1) and (2) we get
P(A ∩ B) = p(A) . p(B)
∴ A and B are independent
Since P(A) ≠ P(B), not equally likely.
∴ A and B are independent but not equally likely.

Question 6.
Two items are chosen from a lot containing twelve items of which four are defective, then the probability that at least one of the item is defective
(1) \(\frac{19}{33}\)
(2) \(\frac{17}{33}\)
(3) \(\frac{23}{33}\)
(4) \(\frac{13}{33}\)
Answer:
(1) \(\frac{19}{33}\)

Explaination:
Total number of items = 12
Number of ways of choosing two items from 12 items is = 12C₂
Number of defective items = 4
Number of non defective items = 8
Probability of getting atleast one defective items

Question 7.
A man has 3 fifty rupee notes, 4 hundred rupees notes and 6 five hundred rupees flotes in his pocket. If 2 notes are taken at random, what are the odds in favour of both notes being of hundred rupee denomination?
(1) 1 : 12
(2) 12 : 1
(3) 13 : 1
(4) 1 : 3
Answer:
(1) 1 : 12

Explaination:
Let S be the sample space and A be the event of taking 2 hundred rupee notes.
n(S) = 13C₂ = \(\frac{13 \times 12}{1 \times 2}\) = 13 × 6
n(S) = 78
n(A) = 4C₂ = \(\frac{4 \times 3}{1 \times 2}\) = 2 × 3
n(A) = 6
n(A̅) = n(S) – n(A)
= 78 – 6 = 72
∴ Odds in favour of A is 6 : 72
That is 1 : 12

Question 8.
A letter is taken at random from the letters of the word ‘ASSISTAN’T’ and another letter is taken at random from the letters of the word ‘STATISTICS’. The probability that the selected letters are the same is
(1) \(\frac{7}{45}\)
(2) \(\frac{17}{90}\)
(3) \(\frac{29}{90}\)
(4) \(\frac{19}{90}\)
Answer:
(4) \(\frac{19}{90}\)

Explaination:
Given words ‘ASSISTANT’ ‘STATISTICS
Sample space S = {(A, S, S, I, S, T, A, N, T)
(S,T, A, T, I, S, T, I, C, S)}
n(S) = {(A, S), (A, T), (A, A), (A, T), (A, I), (A, S), (A, T), (A, I), (A, C), (A, S) }
n(S) = 9 × 10 = 90
Let A be the event of se}ecting equal letters.
A = {(A, A), (A, A), (S, S), (S, S), (S, S), (S, S), (S, S), (S, S), (S, S), (S, S), (S, S), (1, 1), (1, 1), (T, T), (T,T) , (T, T), (T, T), (T, T), (T, T) }
n(A) = 19
Probability of getting equal letters = \(\frac{19}{90}\)

Question 9.
A matrix is chosen at random from a set of all matrices of order 2, with elements 0 or 1 only. The probability that the determinant of the matrix chosen is non zero will be
(1) \(\frac{3}{16}\)
(2) \(\frac{3}{8}\)
(3) \(\frac{1}{4}\)
(4) \(\frac{5}{8}\)
Answer:
(2) \(\frac{3}{8}\)

Explaination:
Sample space S = Set of all 2 × 2 matrices with elements 0 or 1 only
Number of elements in S is
n(S) = 2⁴ = 16
Let A be the event of getting 2 × 2 matrices with elements 0 or 1 only whose determinant is non zero.

Question 10.
A bag contains 5 white and 3 black balls. Five balls are drawn successively without replacement. The probability that they are alternately of different colours is
(1) \(\frac{3}{14}\)
(2) \(\frac{5}{14}\)
(3) \(\frac{1}{14}\)
(4) \(\frac{9}{14}\)
Answer:
(3) \(\frac{1}{14}\)

Explaination:
Number of White balls (W) = 5
Number of Black balls (B) = 3
Five balls are drawn successively without replacement.
Probability that they are alternately of different
colours = P (W B W B W) + P (BW BW B)
= P(W) . P(B) . P(W) . P(B) . P(W) + P(B) . P(W) . P(B) . P(W) . P(B)

Question 11.
If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?
(1) P(A/B) = \(\frac{\mathbf{P}(\mathbf{A})}{\mathbf{P}(\mathbf{B})}\)
(2) P(A/B) < P(A) (3) P(A/B) ≥ P(A) (4) P(A/B) >P(A)
Answer:
(3) P(A/B) ≥ P(A)

Explaination:
Given A and B are two events such that A ⊆ B
and P(B) ≠ 0 then
P(A/B) ≥ P(A)

Question 12.
A bag contains 6 green, 2 white , and 7 black balls. If two balls are drawn simultaneously, then the probability that both are different colours is
(1) \(\frac{68}{105}\)
(2) \(\frac{71}{105}\)
(3) \(\frac{64}{105}\)
(4) \(\frac{73}{105}\)
Answer:
(1) \(\frac{68}{105}\)

Explaination:
Number of green balls (G) = 6
Number of white balls (W) = 2
Number of black balls (B) = 7
Two balls are drawn simultaneously
P (Balls are of different colours)
= P[(GW or WG) or (WB or BW) or (BG or GB)]
= P(GW) + P(WG) + P(WB) + P(BW) + P(BG) + P(GB)
= P(G) P(W) + P(W) P(G) + P(W) P(B) + P(B) P(W) + P(B) P(G) + P(G) P(B)
= 2 [P(G) P(W) + P(W) P(B) + P(B) P(G)]

Question 13.
If X and Y be two events such that
P(X/Y) = \(\frac{1}{2}\), P(Y/X) = \(\frac{1}{3}\) and P( X ∩ Y) = \(\frac{1}{6}\),then
(1) \(\frac{1}{3}\)
(2) \(\frac{2}{5}\)
(3) \(\frac{1}{6}\)
(4) \(\frac{2}{3}\)
Answer:
(4) \(\frac{2}{3}\)

Explaination:

Question 14.
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. The probability that the second ball drawn is red will be
(1) \(\frac{5}{12}\)
(2) \(\frac{1}{2}\)
(3) \(\frac{5}{12}\)
(4) \(\frac{5}{12}\)
Answer:
(2) \(\frac{1}{2}\)

Explaination:
Number of Red balls n (R) = 5
Number of Black halls n (B) = 5
Number of elements in the sample space n (S) = 5 + 5 = 10

Case (i)
P (drawing a red ball first) = \(\frac{5}{10}\)
Let P(E₁) = \(\frac{1}{2}\)
Now two red balls are added.
P (drawing a red ball after adding) = \(\frac{7}{12}\)
P(A/E₁) = \(\frac{7}{12}\)

Case (ii)
P (drawing a black ball first) = \(\frac{5}{10}\)
Let P(E₂) = \(\frac{1}{2}\)
Now two black halls are added.
P (drawing a black ball after adding) = \(\frac{5}{12}\)
P(A/E₂) = \(\frac{5}{12}\)
P(A) = P(A/E₁) . P(E₁) + P(A/E₂) . P(E₂)

Question 15.
A number x is chosen at random from the first 100 natural numbers. Let A be the event of numbers which satisfies \(\frac{(x-10)(x-50)}{x-30}\) ≥ 0 then P(A) is
(1) 0.20
(2) 0.51
(3) 0.71
(4) 0.70
Answer:
(3) 0.71

Explaination:
Given x is choosen from the first 100 natural numbers.
n(S) = 100

Satifies when x takes the values 31 to 100 and also at x = 10
∴ A = { 10, 31, 32, 33, ……………, 100}
n(A) = 71

Question 16.
If two events A and B are independent such that P(A) = 0.35 and P(A ∪ B) = 0.6, then P(B) is
(1) \(\frac{5}{13}\)
(2) \(\frac{1}{13}\)
(3) \(\frac{4}{13}\)
(4) \(\frac{7}{13}\)
Answer:
(1) \(\frac{5}{13}\)

Explaination:
Given A and B are independent events
P(A ∩ B) = P(A) P(B)
Also given P(A) = 0.35 and P(A ∪ B) = 0.6
P(A ∪ B) = P(A) – P(B) – P(A ∩ B)
P(A ∪ B) = P(A) + P(B) – P(A) . P(B)
0.6 = 0.35 + P(B) – 0.35 P(B)
0.6 = 0.35 + (1 – 0.35) P (B)
0.6 = 0.35 + 0.65 P (B)
0.65 P (B) = 0.6 – 0.35
P(B) = \(\frac{0.25}{0.65}\)
P(B) = \(\frac{25}{65}\) = \(\frac{5}{13}\)

Question 17.
If two events A and B are such that P(A̅) = \(\frac{3}{10}\) and P(A ∩ B̅) = \(\frac{1}{2}\) then P(A ∩ B) is
(1) \(\frac{1}{2}\)
(2) \(\frac{1}{3}\)
(3) \(\frac{1}{4}\)
(4) \(\frac{1}{5}\)
Answer:
(4) \(\frac{1}{5}\)

Explaination:

Question 18.
If A and B are two events such that P(B) = and P(B/A) = 0.6,then P(A̅ ∩ B) is
(1) 0.96
(2) 0.24
(3) 0.56
(4) 0.66
Answer:
(3) 0.56

Explaination:
Given A and B are two events.
P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6

P (A ∩ B) = 0.6 × 0.4 = 0.24
P(A̅ ∩ B) = P(B) – P(A ∩ B)
= 0.8 – 0.24
P(A̅ ∩ B) = 0.56

Question 19.
There are three events A, B and C of which one and only one can happen. If the odds are 7 to 4 against A and 5 to 3 against B, then odds against C is
(1) 23 : 65
(2) 65 : 23
(3) 23 : 88
(4) 88 : 23
Answer:
(2) 65 : 23

Explaination:

Question 20.
If a and b are chosen randomly from the set {1, 2, 3, 4} with replacement, then the probability of the real roots of the equation x² + ax + b = 0
(1) \(\frac{3}{16}\)
(2) \(\frac{3}{16}\)
(3) \(\frac{3}{16}\)
(4) \(\frac{3}{16}\)
Answer:
(3) \(\frac{3}{16}\)

Explaination:
x² + ax + b = 0 ⇒ x = \(\frac{-a \pm \sqrt{a^{2}-4 b}}{2}\)
Given that the roots are real ⇒ a² – 4b ≥ 0 or a² > 4b
When a = 1, b = 1 or 2 or 3 or 4 a² – 4b < 0
When a = 2, b = 1 a² – 4b = 0
When a = 3, b = 1 or 2 for which a² – 4b ≥ 0
When a = 4, b = 1 or 2, 3 or 4 for which a² – 4b ≥ 0
So, Selecting from the 4 number 4² = 16 ways.
(i.e.,) n(s) = 16
n(A) = (2 or 3 or 4) = 3
n(B) = (1 or 2 or 3 or 4) = 4
P(A) + P(B) = \(\frac{3}{16}+\frac{4}{16}=\frac{7}{16}\)

Question 21.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\) and P(B/A) = \(\frac{2}{3}\). Then P(B) is
(1) \(\frac{1}{2}\)
(2) \(\frac{1}{2}\)
(3) \(\frac{1}{2}\)
(4) \(\frac{1}{2}\)
Answer:
(2) \(\frac{1}{2}\)

Explaination:
Given A and B are two events.
Given P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\)

Question 22.
In a certain college, 4 % of the boys and 1 % of the girls are taller than 1.8 meters. Further 60 % of the students are girls. If a student is selected at random and is taller than 1.8 meters, then the probability that the student is a girl is
(1) \(\frac{2}{11}\)
(2) \(\frac{3}{11}\)
(3) \(\frac{5}{11}\)
(4) \(\frac{7}{11}\)
Answer:
(2) \(\frac{3}{11}\)

Explaination:
Let A₁, A_2, and B be the event of selecting a boy, selecting a girl, and selecting a tall student respectively.

Question 23.
Ten coins are tossed. The probability of getting at least 8 heads is
(1) \(\frac{7}{64}\)
(2) \(\frac{7}{32}\)
(3) \(\frac{7}{16}\)
(4) \(\frac{7}{128}\)
Answer:
(4) \(\frac{7}{128}\)

Explaination:
Favourable events for atleast 8 heads
n(A) = 10C₈ + 10C₉ + 10C₁₀
= 10C₂ + 10C₁ + 10C₀
= \(\frac{10 \times 9}{1 \times 2}\) + 10 + 1
= 5 × 9 + 11 = 45 + 11
n(A) = 56
Ten coins are tossed
∴ n(S) = 2¹⁰ = 1024

Question 24.
The probability of two events A and B are 0.3 and 0.6 respectively. The probability that both A and B occur simultaneously is 0.18. The probability that neither A nor B occurs is
(1) 0.1
(2) 0.72
(3) 0.42
(4) 0.28
Answer:
(4) 0.28

Explaination:
P(A) = 0.3, P(B) = 0.6
P(A ∩ B) = 0.18
So P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.3 + 0.6 – 0.18
= 0.9 – 0.18 = 0.72
P(A’ ∩ B’) = P[(A ∪ B)’] = 1 – P(A ∪B)
= 1 – 0.72 = 0.28

Question 25.
If m is a number such that m ≤ 5, then the probability that quadratic equation 2x² + 2mx + m + 1 = 0 has real roots is
(1) \(\frac{1}{5}\)
(2) \(\frac{2}{5}\)
(3) \(\frac{3}{5}\)
(4) \(\frac{4}{5}\)
Answer:
(3) \(\frac{3}{5}\)

Explaination:
2x² + 2mx + m + 1 = 0


roots are real ⇒ m² – 2m – 2 ≥ 0
Here m ≤ 5 ⇒ n(S) = 5
When m= 1,m² – 2m – 2
When m = 2, m² – 2m- 2
When m = 3, m² – 2m – 2
When m = 4, m² – 2n- 2
When m = 5, m² – 2m – 2
⇒ n{A) = 3 and so P(A) = \(\frac{3}{5}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9

Question 1.
Find by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = 2x², y = 0 and x = 1
Solution:
The region to be revolved is sketched

Since revolution is made about the x axis, the volume of the solid generated is given by

Required volume = \(\frac { 4π }{ 5 }\) cubic units

Question 2.
Find, by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = e^-2x, y = 0, x = 0 and x = 1.
Solution:
Since revolution is made about the x axis, the volume of the solid generated is given

Required volume = \(\frac { π }{ 4 }\) [1 – e^-4] cubic units

Question 3.
Find, by integration, the volume of the solid generated by revolving about the y axis, the region enclosed by x² = 1 + y and y = 3.
Solution:
The region to be revolved is sketched.

Since revolution is made about the y axis, the volume of the solid generated is given by

Required volume = 8π cubic units

Question 4.
The region enclosed between the graphs of y = x and y = x² is denoted by R. Find the volume generated when R is rotated through 360° about x axis.
Solution:
The region to be revolved is sketched.

Find the intersecting point of y = x and y = x²
x² = x
x² – x = 0
x (x – 1) = 0 x = 0, x = 1
If x = 0, y = 0, x = 1, y = 1
∴ Points of intersection are (0, 0), (1, 1)

Required volume = \(\frac { 2π }{ 15 }\) cubic units

Question 5.
Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown to figure.

Solution:
By using integration we have to find the volume of the frustum. So first find the equation of the curve.
Let A(0, 1) and B(2, 2) be two points. Line joining these two points form a straight line. That straight line revolves around x axis.

Volume of the solid revolves around x axis

Volume of the frustum = \(\frac { 14 }{ 3 }\) π m³

Question 6.
A watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major axis 20 cm and minor axis 10 cm about its major axis. Find its volume using integration.
Solution:
A watermelon has an ellipsoid shape.
2a = 20
a = 10
2b = 10
b = 5

∴ Volume of the frustum = \(\frac { 1000π }{ 3 }\) cubic units.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8

Question 1.
Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x axis.
Solution:
3x – 2y + 6 = 0
2y = 3x + 6
y = \(\frac { 1 }{ 2 }\)(3x + 6)

Question 2.
Find the area of the region bounded by 2x – y + 1 = 0, y = -1, y = 3 and y axis.
Solution:
Given straight line is 2x – y + 1 = 0
y = 2x + 1, x = \(\frac { y-1 }{ 2 }\)

= 1 + 1
= 2 sq. units
Area required = 2 sq. units

Question 3.
Find the area of the region bounded by the curve 2 + x – x² + y = 0, x axis, x = -3 and x = 3
Solution:
Given curve is
2 + x – x² + y = 0
y = x² – x – 2
y = (x – 2)(x + 1)

Question 4.
Find the area of the region bounded by the line y = 2x + 5 and the parabola y = x² – 2x.
Solution:
First, we find the point of intersection of
y = 2x + 5 and y = x² – 2x
x² – 2x = 2x + 5
x² – 4x – 5 = 0
(x – 5) (x + 1) = 0
x = 5, x = – 1
when x = 5, y = 15
x = -1. y = 3
(5, 15) (-1, 3) are intersecting points.

54 – 18 = 36
Area required = 36 sq. units

Question 5.
Find the area of the region bounded between the curves y = sin x and y = cos x and the lines x = 0 and x = π
Solution:
First find the intersecting point of two curves
sin x = cos x
tan x = 1
x = \(\frac { π }{ 4 }\)

Area required = 2√2 sq. units

Question 6.
Find the area of the region bounded by y = tan x, y = cot x and the lines x = 0, x = \(\frac { π }{ 2}\), y = 0.
Solution:
First find the intersecting point of y = tan x and y = cot x
tan x = cot x
\(\frac { tan x }{ cot x }\) = 1
tan²x = 1
tan x = 1

= log √2 + log √2
= 2 log √2
= log(√2)²
= log 2 sq. units

Question 7.
Find the area of the region bounded by the parabola y² = x and the line y = x – 2.
Solution:
First find the intersecting point of y² = x and y = x – 2
y = y² – 2
y² – y – 2 = 0
y = 2, y = -1
Intersecting points are (4, 2), (1, -1)

Required Area = \(\frac { 9 }{ 2 }\) sq. units

Question 8.
Father of a family wishes to divide his square field bounded by x = 0, x = 4, y = 4 and y = 0 along the curve y² = 4x and x² = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.
Solution:
Given curve y² = 4x and x² = 4y
Draw these two curves
Also draw the square bounded by the lines
x = 0, x = 4, y = 4 and y = 0
To prove Area A₁ = Area A₂ = Area A₃
Now the point of intersection of the curves y² = 4x and x² = 4y is given by
(\(\frac { y^2 }{ 4 }\))² = 4y
y⁴ = 64y ⇒ y (y³ – 64) = 0
y = 0, y = 4
when y = 0 ⇒ x = 0
y = 4 ⇒ x = 4
Point of intersection are O (0, 0) and B (4, 4)
Now, the area of the region bounded by the curves y² = 4x and x² = 4y is
A₂ = \(\int_{0}^{4}\)(\(\sqrt { 4x }\) – \(\frac { x^2 }{ 4 }\)) dx

Now the area of the region bounded by the curves x² = 4y, x = 4 and x axis is

Similarly the area of the region bounded by the curve y² = 4x, y axis and y = 4 is

Hence we see that
A₁ = A₂ = A₃ = \(\frac { 16 }{ 3 }\) sq. units

Question 9.
The curve y = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.
Solution:
Given curve is y = (x – 2)² + 1
(i.e) (y – 1) = (x – 2)²
Vertex of the parabola is (2, 1)
Minimum point P is (2, 1)
Slope of PQ is 2.

Equation of PQ is y – 1 = 2 (x – 2)
y – 1 = 2x – 4
y = 2x – 3
Intersecting point of y = 2x – 3 and y = (x – 2)² + 1
2x – 3 = (x – 2)² + 1
2x – 4 = (x – 2)²
2(x – 2) = (x – 2)²
x – 2 = 2
x = 4
when x = 4, y = 5

Required Area = \(\frac { 4 }{ 3 }\) sq. units

Question 10.
Find the area of the region common to the circle x² + y² = 16 and the parabola y² = 6x
Solution:
First find the intersecting point of the curves
x² + y² = 16 and y² = 6x
x² + 6x = 16
x² + 6x – 16 = 0
(x + 8) (x – 2) = 0
x = -8, x = 2
x = -8 is impossible
x = 2, y = 2√3
Radius of the circle x² + y² = 16 is 4

Area OABC = 2 (Area of OAB)
= 2 (Area of the curve y² = 6x in [0, 2] + Area of the curve x² + y² = 16 in [2, 4])

Required Area = \(\frac { 4 }{ 3 }\) [4π + √3] sq. units

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 12 Introduction to Probability Theory Ex 12.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.4

Question 1.
A factory has two Machines – I and II. Machines – I produce 60 % of items and Machine – II produces 40 % of the items of the total output. Further 2 % of the items produced by Machine – I are defective whereas 4 % produced by Machine -II are defective. If an item is drawn at random what is the probability that it is defective?
Answer:
Let A₁ be the event that items are produced by machine – I, A₂ be the event that items are produced by machine – II.
Let B be the event of drawing a defective item. We have to find the total probability of event B. That is P(B) clearly A₁ and A₂ are mutually exclusive and exhaustive events.
∴ P(B) = P(A₁) . P(B/A₁) + P(A₂) . P(B/A₂)

P(B) = P(A₁) . P(B/A₁) + P(A₂) . P(B/A₂)
= 0.60 × 0.02 + 0.40 × 0.04
= 0.012 + 0.016
= 0.028

Question 2.
There are two identical urns containing respectively 6 black and 4 red balls, 2 black, and 2 red balls. An urn is chosen at random and a ball is drawn from it.
(i) Find the probability that the ball is black
(ii) if the ball is black, what is the probability that it is from the first urn?
Answer:

(i) Let A₁ be the event of selecting Urn – I and A₂ be the event of selecting Urn – II.
Let B be the event of selecting one black ball.
We have to find the total probability of event B. That is P(B).
Clearly, A₁ and A₂ are mutually exclusive and exhaustive events.
Probability of selecting Urn – I
P(A₁) = \(\frac{1}{2}\)
Conditional Probability of B, given A₁

Probability of selecting Urn – II
P(A₂) = \(\frac{1}{2}\)
Conditional Probability of B, given A₂

(ii) The conditional Probability of A₁ given B is P(A₁/B)
By Bayes’ theorem

Question 3.
A firm manufactures PVC pipes in the three plants viz, X, Y, and Z. The daily production volumes from the three firms X, Y, and Z are respectively 2000 units, 3000 units, and 5000 units. It is known from past experience that 3 % of the output from plant X, 4 % from plant Y, and 2 % from plant Z are defective. A pipe is selected at random from a day’s total production
(i) find the probability that the selected pipe is a defective one?
(ii) if the selected pipe is defective, then what is the probability that it was produced by plant Y?
Answer:
Let A₁ be the daily volume of production by plant X, A₂ be the daily volume of production by plant Y, A₃ be the daily volume of production by plant Z.
Let B be the defective output we have to find P (B).

(i) Find the probability that the selected pipe is a defective one:
Clearly, A₁, A_2, and A₃ are mutually exclusive and exhaustive events.

Probability that the selected pipe is a defective one = \(\frac{7}{250}\)

(ii) If the selected pipe is defective, then what is the probability that it was produced by plant Y?


Probability that the defective pipe produced by plant Y = \(\frac{3}{7}\)

Question 4.
The chances of A, B, and C becoming manager of a certain company are 5 : 3 : 2. The probabilities that the office canteen will be improved if A, B, and C become managers are 0.4, 0.5 and 0.3 respectively. If the office canteen has been improved, what is the probability that B was appointed as the manager?
Answer:
Let A₁, A_2, and A₃ be the events of A, B, and C becoming managers of the company respectively.
Let B be the event that the office canteen will be improved.
We have to find the conditional probability P (A₂/B).
Since A₁, A₂ and A₃ are mutually exclusive and exhaustive events, applying Bayes theorem.

If the office canteen is improved than the probability of that B was appointed as the manager is \(\frac{15}{41}\)

Question 5.
An advertising executive is studying television viewing habits of married men and women during prime time hours. Based on the past viewing records he has determined that during prime time wives are watching television 60 % of the time. It has also been determined that when the wife is watching television, 40 % of the time the husband is also watching. When the wife is not watching the television, 30 % of the time the husband is watching the television. Find the probability that
(i) the husband is watching the television during the prime time of television
(ii) if the husband is watching the television, the wife is also watching the television.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 9 Ratio Analysis Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 9 Ratio Analysis

12th Accountancy Guide Ratio Analysis Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
The mathematical expression that provides a measure of the relationship between two figures is called
(a) Conclusion
(b) Ratio
(c) Model
(d) Decision
Answer:
(b) Ratio

Question 2.
Current ratio indicates
(a) Ability to meet short term obligations
(b) Efficiency of management
(c) Profitability
(d) Long term solvency
Answer:
(a) Ability to meet short term obligations

Question 3.
Current assets excluding inventory and prepaid expenses is called
(a) Reserves
(b) Tangible assets
(c) Funds
(d) Quick assets
Answer:
(d) Quick assets

Question 4.
Debt equity ratio is measure of
(a) Short term solvency
(b) Long term solvency
(c) Profitability
(d) Efficiency
Answer:
(b) Long term solvency

Question 5.
Which of the following is not a tool of financial statement analysis?

Answer:
(a) (i) – 1,(ii) – 4,(iii) – 3,(iv) – 2

Question 6.
To test the liquidity of a concern, which of the following ratios are useful?
(i) Quick ratio
(ii) Net Profit ratio
(iii) Debt – equity ratio
(d) Current ratio
Select the correct answer using the codes given below:
(a) (i) and (ii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and iv)
Answer:
(b) (i) and (iv)

Question 7.
Proportion of share holders’ funds to total assets is called
(a) Proprietary ratio
(b) Capital gearing ratio
(c) Debt equity ratio
(d) Current ratio
Answer:
(a) Proprietary ratio

Question 8.
Which one of the following is not correctly matched?
(a) Liquid ratio – Proportion
(b) Gross profit ratio – Percentage
(c) Fixed assets turnover ratio – Percentage
(d) Debt – equity ratio – Proportion
Answer:
(c) Fixed assets turnover ratio – Percentage

Question 9.
Current liabilities ₹ 40,000; Current assets ₹ 1,00,000; Inventory ₹ 20,000. Quick ratio is
(a) 1:1
(b) 2,5:1
(c) 2:1
(d) 1:2
Hint:
Quick ratio or Liquid ratio = \(\frac{\text { Liquid Assets }}{\text { Current liabilities }}\)
Liquid assets = Current Assets – Inventory
= 1,00,000 – 20,000
= 80,000
= \(\frac{80,000}{40,000}\)
= 2:1
= 110%
Answer:
(c) 2:1

Question 10.
Cost of revenue from operation 3,00,000; Inventory at the beginning of the year 60,000; Inventory at the close of the year’ 40,000. Inventory turnover ratio is.
(a) 2 times
(b) 3 times
(c) 6 times
(d) 8 times
Hint:

Answer:
(c) 6 times

II Very Short Answer Questions

Question 1.
What is meant by accounting ratios?
Answer:
The ratio is a mathematical expression of the relationship between two related or interdependent items. It is the numerical or quantitative relationship between two items. It is calculated by dividing one item by the other related item. When ratios are calculated on the basis of accounting information, these are called ‘accounting ratios’.

Question 2.
What is the quick ratio?
Answer:
The quick ratio gives the proportion of quick assets to current liabilities. It indicates whether the business concern is in a position to pay its current liabilities and when they become due, out of its quick assets.

Question 3.
What is meant by debt-equity ratio?
Answer:
It is calculated to assess the long-term solvency position of a business concern. The debt equity ratio expresses the relationship between long term debt and shareholder’s funds.
Debt equity ratio = \(\frac{\text { Long term debt }}{\text { Shareholders funds }}\)
Capital employed = Shareholder’s funds + Noncurrent liabilities
Greater the return on investment better is than the profitability of a business and vice versa.

Question 4.
What does the return on investment ratio indicate?
Answer:
Return on investment shows the proportion of net profit before interest and tax to capital employed (shareholders’ funds and long term debts). This ratio measures how efficiently the capital employed is used in the business. It is an overall measure of the profitability of a business concern.

Question 5.
Statement any two limitations of ratio analysis.
Answer:
Ratios are only means: Ratios are not ended in themselves but they are only means to achieve a particular purpose. Analysis of related items must be done by the management or experts with the help of ratios. Change in price level: Ratio analysis may not reflect price level changes and current values as they are calculated based on historical data given in the financial statement.

III Short Answer Questions

Question 1.
Explain the objectives of ratio analysis.
Answer:
Following are the objectives of ratio analysis:

• To simplify accounting figures
• To facilitate analysis of financial statements
• To analysis the operational efficiency of a business
• To help in budgeting and forecasting
• To facilitate intra firm and inter-firm comparison of performance

Question 2.
What is the inventory conversion period? How is it calculated?
Answer:
The inventory conversion period is the time taken to sell the inventory. A shorter inventory conversion period indicates more efficiency in the management of inventory. It is computed as follows:

Question 3.
How is operating profit ascertained?
Answer:
Operating profit = Revenue from operations – Operating cost
Cost of revenue from operations = Purchases of stock – in – trade + Change in inventories of stock in trade + Direct expenses.
Operating expenses = Administrative expenses + Selling and distribution expenses.
Operating cost = Cost of revenue from operations + Operating expenses.

Question 4.
State any three advantages of ratio analysis.
Answer:
Following are the advantage of ratio analysis:

• Measuring operational efficiency: Ratio analysis helps to know the operational efficiency of a business by finding the relationship between operating cost and revenues and also by comparison of present ratios with those of the past ratios.
• Intra firm comparison: Comparison of the efficiency of different divisions of an organization is possible by comparing the relevant ratios.
• Inter-firm comparison: Ratio analysis helps the firm to compare its performance with other firms.

Question 5.
Bring out the limitations of ratio analysis:
Answer:

• Consistency in preparation of financial statements: Inter firm comparisons with the help of ratio analysis will be meaningful only if the firms follow uniform accounting procedures consistently.
• Non-availability of standards or norms: Ratios will be meaningful only if they are compared with accepted standards or norms. Only few financial ratios have universally recognized standards. For other ratios, comparison with standards is not possible.
• Change in price level: Ratio analysis may not reflect price level changes and current values as they are calculated based on historical data given in financial statements.

IV Exercises

Liquidity ratios

Question 1.
Calculate the current ratio from the following information.

Solution:
Current ratio = \(\frac{\text { Current Assets }}{\text { Current liabilities }}\)
Current Assets = Current Investments + Inventories + Trade Dr’s + B/R+Cash & Cash equivalents
= 40,000 + 2,00,000 + 1,20,000 + 80,000 + 10,000
= Rs. 4,50,000

Current Liabilities
= Trade Cr’s + B/P + Exps. Payable.
= 80,000 + 50,000 + 20,000
= Rs. 1,50,000
Cur. Ratio = \(\frac{4,50,000}{1,50,000}\)
= 3:1
Answer:
Current ratio : 3:1

Question 2.
Calculate quick ratio: Total current liabilities ₹ 2,40,000; total current assets ₹ 4,50,000; Inventories ₹ 70,000; Prepaid Expenses ₹ 20,000
Solution:
Quick Ratio = \(\frac{\text { Quick assets }}{\text { Current liabilities }}\)
Quick assets = Current Assets – Inventories & Prepaid exps.
= 4,50,000 – (70,000 + 2000)
= Rs.3,60,000
Quick Ratio = \(\frac{3,60,000}{2,40,000}\)
=1:5:1
Answer:
Quick ration: 1:5:1

Question 3.
Following is the balance sheet of Lakshmi Ltd. as of 31st March 2019.

Calculate: (i) Current ratio (ii) Quick ratio
Solution:
Current ratio = \(\frac{\text { Current Assets }}{\text { Current liabilities }}\)
Current Assets = Inventories + Trade Dr’s + Cash & Cash equivalents + Prepaid Exps
= 1,60,000 +3,20,000 + 80,000 + 40,000
= Rs. 6,00,000
Current Liabilities = Short term borrowings + Trade Payable + Expenses payable + Short term provisions.
= 50,000 + 3,10,000 + 15,000 + 25,000
= Rs. 4,00,000

Answer:
(i) Current ratio: 1.5:1;
(ii) Quick ratio: 1:1

Question 4.
From the following information calculate debt equity ratio.

Solution:
Debt Equity Ratio = \(\frac{\text { Long term debt }}{\text { Shareholder’s Funds }}\)
Long term debt = Debenture = Rs. 6,00,000
Shareholder’s Fund = Equity share capital + Reserves & Surplus
= 6,00,000 + 2,00,00 = Rs. 8,00,000
Debt Equity Ratio = \(\frac{6,00,000}{8,00,000}\)
= 0.75:1
Answer:
Debt equity ratio: 0.75:1

Question 5.
From the following Balance Sheet of Sundaram Ltd. Calculate proprietary ratio:

Answer:
Proprietary ratio: 0.75:1

Question 6.
From the following information calculate the capital gearing ratio:

Answer:
Capital gearing ratio: 0.5:1

Question 7.
From the following Balance Sheet of James Ltd. as of 31.03.2019 calculate
(i) Debt- Equity ratio
(ii) Proprietary ratio
(iii) Capital gearing ratio

Solution:
Debt Equity Ratio =  \(\frac{\text { Long Term Debt }}{\text { Shareholder’s Fund }}\)|
Shareholder’s Fund
Long term debt = Debentures = Rs. 3,00,000
Shareholder’s Fund = Eq. share capital +Pref. Shares capital + Reserves & surplus
= 2,50,000 ÷ 2,00,000 + 1,50,000
= Rs. 6,00,000

(3) Capital Gearing ratio

= \(\frac{\text { Funds bearing fixed interest (or) Fixed divided }}{\text { Equity shareholder’s Fund }}\)
Funds bearing fixed interest (or) fixed dividend
= Pref. Share Cap + Debentures
= 2,00,000 + 3,00,000 = Rs. 5,00,000
Equity share holder’s Fund .
= Equity Share cap + Reserves & Surplus
= 2,50,000 + 1,50,000
= Rs. 4,00,000
Capital gearing ratio = \(\frac{5,00,000}{4,00,000}\)
= 1.25:1

Answer:

• Debt-equity ration; 0.5:1;
• Proprietary ration; 0.5:1;
• Capital gearing ratio:1.25:1

Question 8.
From the given information calculate the inventory turnover ratio and inventory conversion period (in months) of Devi Ltd.

Solution:
Inventory Turnover Ratio = \(\frac{\text { cost of revenue from operations }}{\text { Average Inventory }}\)
Cost of revenue from operations = Purchase of stock + change in inventories of finished goods operations + Direct Exps.
= Rs. 6,90,000
AverageInventory = \(\frac{\text { Opening Inventory + Closing inventory }}{2}\)
= \(\frac{1,70,000+1,30,000}{2}\)
= Rs. 1,50,000
Change in inventory = Opening inventory – Closing inventory
= 1,70,000 – 1,30,000
= Rs. 40,000
Cost of revenue from operation
= 6,90,000 + 40,000 + 20,000
= Rs. 7,50,000

Answer:
Inventory turnover ratio 5times; Inventory conversion period 2.4 months

Question 9.
The credit revenue from operations of Velavan Ltd, amounted to ₹ 10,00,000. Its debtors and bills receivables at the end of the accounting period amounted to ₹ 1,10,000 and ₹ 1,40,000 respectively. Calculate trade receivables turnover ratio and also.collection period in months.
Solution:
Trade receivable Turnover ratio = \(\frac{\text { Credit revenue from Operations }}{\text { Average trade receivables }}\)
Average trade receivables = \(\frac{\text { Opening trade receivables + Closing trade receivables }}{2}\)
Trade receivable = Trade Drs + B/R
Inventory Turnovers Ratio = \(\frac{10,00,000}{2,50,000}\)
= 4 times
Average Trade receivable
= 1,10,000 + 1,40,000
= Rs. 2,50,000
Debt collection period = \(\frac{\text { Number of months in a year }}{\text { Trade receivable turnover ratio }}\)
= \(\frac{12}{4}\)
= 3 months
Answer:
Trade receivables turnover ratio: 4 time; Debt collection period: 3 months

Question 10.
From the following figures obtained from Arjun Ltd, calculate the trade payable turnover ratio and credit payment period (in days)

Solution:


Answer:
Trade payable turnover ratio: 10 times; Credit payment period: 36.5 days

Question 11.
From the following information of Geetha Ltd., Calculate fixed assets turnover ratio
(i) Revenue from operations during the year was ₹ 55,00,000.
(ii) Fixed assets at the end of the year ₹ 5,00,000
Solution:

Answer:
Fixed assets turnover ratio: 11 times

Question 12.
Calculate

• Inventory turnover ratio
• Trade receivable turnover ratio
• Trade payables turnover ratio and
• Fixed assets turnover ratio from the following obtained from Aruna Ltd.
  

  Particulars	As of 31st March 2018 ₹	As of 31st March 2019 ₹
  Inventory	3,60,000	4,40,000
  Trade receivables	7,40,000	6,60,000
  Trade Payable	1,90,000	2,30,000
  Fixed assets	6,00,000	8,00,000

   

Additional information:

• Revenue from operations for the year ₹ 35,00,000
• Purchases for the year ₹ 21,00,000
• Cost of revenue from operation ₹ 16,00,000
  Assume that sales and purchases are for credit.

Solution




Answer:

• Inventory turnover ratio; 4 times;
• Trade receivable turnover ratio; 5 times;
• Trade payables turnover ratio: 10 times;
• Fixed assets turnover ratio: 5 times

Question 13.
Calculate gross profit ratio form the following: Revenue from operations ₹ 2,50,000, Cost of revenue from operation ₹ 2,10,000 and Purchases ₹ 1,80,000.
Solution:

Answer:
Gross Profit ratio 16%

Question 14.
Following is the statement of profit and loss of Padma Ltd. for the year ended 31st March, 2018. Calculate the operating cost ratio.


Notes to Accounts

Solution:
Operating cost Ratio = \(\frac{\text { Operating cost }}{\text { Revenue from operation }}\) × 100
Operating cost = Cost of revenue from operations + Operating expenses.
Cost of revenue from operations = Purchase + Change in inventory + Direct Expenses
= 8,60,000 + 40,000 + Nil
= Rs. 9,00,000
Operating Exps = Salaries + Office & Administration Exps + Selling+Distribution Exps
= 1,60,000 + 50,000 + 90,000
= Rs. 3,00,000
Operating cost = 9,00,000 + 3,00,000
= Rs, 12,00,000
Operating cost Ratio = \(\frac {12,00000}{15,00000}\) × 100
= 80%
Answer:
Operating cost ratio 80%

Question 15.
Calculate operating profit ratio under the following cases.
Case 1 : Revenue from operations ₹ 8,00,000 Operating Profit ₹ 2,00,000.
Case 2 : Revenue from operations ₹ 20,00,000 Operating Cost ₹ 14,00,000.
Case 3 : Revenue from operations ₹ 10,00,000 Gross profit 25% on revenue from operations, operating expenses ₹ 1,00,000.
Solution:


Answer:
Operating profit ratio – Case 1: 25%; Case 2:30%; Case 3:15%

Question 16.
From the following details of a business, concern calculates net profit ratio.

Solution
Net Profit Ratio = \(\frac{\text { Net Profit }}{\text { Revenue from operations }}\)× 100
Gross profit = Revenue from operations – Cost of revenue from operation
= 9,60,000-5,50,000 Rs. 4,10,000
Operating Profit = Gross Profit – Operating Exps
Operating Exps = Office & Administrative Exps + Selling & Distribution Exps
= 1,45,000 + 25,000 = Rs. 1,70,000
Operating Profit = 4,10,000 – 1,70,000
Operating Profit = Rs. 2,40,000
Net Profit ratio = \(\frac{2,40,000}{9,60,000}\)× 100 = 25%
Answer :
Net Profit ratio 25%

Question 17.
From the following statement of profit. and loss of Dericston Ltd. Calculate
(i) Gross Profit ratio
(ii) Net Profit ratio.

Solution
Gross Profit Ratio = \(\frac{\text { Gross Profit }}{\text { Revenue from operations }}\)×100
Gross profit = Revenue from operations – Cost of revenue from operation
Cost of revenue from operations = Purchase + Change in inventories
= 18,80,000 – 80,000 = Rs. 18,00,000
Gross Profit = 24,00,000 – 18,00, 000 = Rs. 6,00,000

Answer :

• Gross profit ratio of 25%
• net Profit ratio 10%

Question 18.
From the following trading activities of Jones Ltd. Calculate

• Gross profit ratio
• Net Profit ratio
• Operating cost ratio
• Operating profit ratio

Solution
(1) Gross Profit Ratio = \(\frac{\text { Gross Profit }}{\text { Revenue from operations }}\)× 100
cost of Revenue from operations = Purchase + Change in inventories
= 2,10,000 + 30,000
= Rs. 2,40,000
Gross profit = Revenue from operations – cost of revenue from operations
= 4,00,000 – 2,40,000
= Rs. 1,60,000
Gross Profit ratio = \(\frac{1,60,000}{4,00,000}\)× 100 = 40%


Answer :

• Gross profit ratio 40%
• Net Profit ratio 20%
• Operating cost ratio 75%
• Operating profit ratio 25%

Question 19.
Following is the extract of the balance sheet of Abdul Ltd., as of 31st March 2019

Net Profit before interest and tax for the year was ₹ 60,000. Calculate the return on capital employed for the year.
Solution:

Answer :
Return on capital employed: 15%

12th Accountancy Guide Ratio Analysis Additional Important Questions and Answers

Other Important question & Answers
Question 1.
All solvency ratios are express in terms of
(a) Proportion
(b) Times
(c) Percentage
Answer:
(b) Times

Question 2.
All activity ratios, (or) Turnover ratios in terms of
(a) Proportion
(b) Times
(c) Percentage
Answer:
(b) Times

Question 3.
All profitability ratios are expressed in terms of ………………
(a) Proportion
(b) Times
(c) Percentage
Answer:
(c) Percentage

Question 4.
Shareholders funds include
(a) Equity share capital, preference share capital reserves & Surplus
(b) Loans from banks & financial institutions.
(c) Equity share capital, preference share capital, reserves & surplus, and loans from banks & financial institutions.
Answer:
(a) Equity share capital, preference share capital reserves & Surplus

Question 5.
The current ratio is a
(a) Solvency ratio
(b) Profitability ratio
(c) Liquidity ratio
Answer:
(a) Solvency ratio

Question 6.
The ratio is expressed in ……………… way
(a) 2
(b) 4
(c) 3
Answer:
(c) 3

Question 7.
The cost of revenue from the operation is Rs. 4,00,000. Average inventories Rs. 8,00,000 Inventory turnover ratio is
(a) 5 times
(b) 4 times
(c) 7 times
Answer:
(a) 5 times

Question 8.
The operating ratio is equal to
(a) 100-operating profit ratio
(b) 100 +operating profit ratio
(c) Operating profit ratio
Answer:
(a) 100-operating profit ratio

Question 9.
Operating expenses include
(a) Selling & administration expenses
(b) Selling & administration expens
(c) a & b
Answer:
(c) a & b

Question 10.
Equity share capital Rs. 2,00,000 Reserves & Surplus Rs. 30,000 Debenture Rs. 40,000 and shareholders fund will be.
(a) Rs. 200,000
(b) 2,70,000
(c) Rs. 2,30,000
Hint:
Share holder fund = Equity share capital + Reserve and surplus
= 2,00,000 + 30,000
= ₹ 2,30,000
Answer:
(c) Rs. 2,30,000

III Short Answer Questions

Question 1.
Define Ratio Analysis
Answer:
According to Myers, “Ratio analysis is a study of the relationship among various financial factors in a business”.

Question 2.
What do you mean by ratio Analysis?
Answer:
Ratio analysis is a tool which involves analyzing the financial statements by calculating various. It is a tool of financial statement analysis, in which, inferences are drawn based on the computation and analysis of different ratios.

Question 3.
What are the two ways of classifying the ratios?
Answer:
Ratios may be classified in the following two ways:

• Traditional classification
• Functional classification

Question 4.
What do you mean by the traditional classification of ratio? Explain
Answer:
The traditional classification of ratios is done on the basis of the financial statements from which the ratios are calculated. Under the traditional classification, the ratio is classified as:

• Balance sheet ratio: If both items in a ratio are from the balance sheet, it is classified as a balance sheet ratio.
• Income statement ratio: If the two items in a ratio are from the income statement, it is classified as an income statement ratio.
• Inter – Statements ratio: If a ratio is computed with a tone item from the income statement and another item from the balance sheet, it is called an inter-statement ratio.

Question 5.
What is the functional classification of ratios?
Answer:
Under the functional classification, the rations are classified as follows:

• Liquidity ratios
• Long term solvency ratios
• Turnover ratios.
• Profitability ratios.

Question 6.
What do you mean by Liquidity ratios?
Answer:
Liquidity means the capability of being converted into cash with ease. Liquidity ratios help to assess the ability of a business concern to meet its short term financial obligations. Short term assets (current assets) are more liquid as compared to long term assets (fixed assets). Liquidity ratios are also called as short term solvency ratios.

Question 7.
What do you mean by the current ratio?
Answer:
Current ratio gives the proportion of current assets to current liabilities of a business concern. It is computed by dividing current assets by current liabilities. The current ratio indicates the ability of an entity to meet its current liabilities as and when they are due for payment.

Question 8.
What do you mean by Long-term solves ratios? what are its types?
Answer:
Long term solvency means the firm’s ability to meet its liabilities in the long run. Long term solvency ratios help to determine the ability of the business to repay its debts in the long run. The following ratios are normally computed for evaluating long term solvency of the business:

• Debt equity ratio
• proprietary ratio
• Capital gearing ratio

Question 9.
What do you mean by Turnover Ratios? What are its types?
Answer:
Turnover ratios show how efficiently assets or other items have been used to generate revenue from operations. They are also called as activity ratios or efficiency ratios. They how the speed of movement of various items. They are expressed as number of times in relation to the item compared.
The important turnover ratios are:

• Inventory turnover ratio
• Trade receivable turnover ratio
• Trade payable turnover ratio
• Fixed assets turnover ratio.

Question 10.
What do you mean by Trade Receivable Turnover ratio?
Answer:
Trade receivable turnover ratio is the comparison of credit revenue from operations with average trade receivables during an accounting period. It gives the velocity of the collection of cash from trade receivables.

Question 11.
What do you mean by Debt collection Period?
Answer:
Debt collection period is the average time taken to collect the amount due from trade receivables. Lesser the debt collection period, grater is the efficiency of management in the collection of cash from trade receivables. It is calculated as follows.

What do you mean by Trade payable Turnover ratio?
Trade payable turnover ratio is the comparison of net credit purchases with average trade payables during an accounting period. It gives the velocity to payment of cash towards trade payables.

Question 12.
What do you mean by credit payment period?
Answer:
It is the average time taken by the business for payment of accounts payable. Lesser the credit payment period, greater is the efficiency of the management in managing accounts payable as it indicated quicker settlement of trade payables. It is calculated as follows:

Question 13.
What do you mean by Fixed Assets Turnover
Answer:
The fixed assets turnover ratio gives the number of times the fixed assets are turned over during the year in relation to the revenue from operations. This ratio indicates the efficiency of utilization of fixed assets.

Question 14.
What do you mean by profitability ratios? What are its types?
Answer:
Profitability ratios help to assess the profitability of a business concern. These rations also help to analyze the earning capacity of the business in terms of utilization of resources employed in the business. Generally, these rations are expressed as a percentage.
The profitability ratios commonly used are:

1. Gross profit ratio
2. Operating cost ratio
3. Operating profit ratio
4. Net profit ratio
5. Return on investment.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 5 Admission of a Partner Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 5 Admission of a Partner

12th Accountancy Guide Admission of a Partner Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer

Question 1.
Revaluation A/c is a
(a) Real A/c
(b) Nominal A/c
(c) Personal A/c
(d) Impersonal A/c
Answer:
(b) Nominal A/c

Question 2.
On revluation, the increase in the value of assets leads to
(a) Gain
(b) Loss
(c) Expense
(d) None of these
Answer:
(a) Gain

Question 3.
The profit or loss on revaluation of assets and liabilities is transferred to the capital account of
(a) The old partners
(b) The new partner
(c) All the partners
(d) The Sacrificing partners
Answer:
(a) The old partners

Question 4.
If the old profit sharing ratio is more than the new profit sharing ratio of a partner, the difference is called
(a) Capital ratio
(b) Sacrificing ratio
(a) all the partners
(b) the old partners
(c) the new partner
(d) the sacrificing partners
Answer:
(b) Sacrificing ratio

Question 6.
Which of the following statements is not true in relation to the admission of a partner
(a) Generally mutual rights of the partners change
(b) The profits and losses of the previous years are distributed to the old partners
(c) The firm is reconstituted under a new agreement
(d) The existing agreement does not come to an end
Answer:
(d) The existing agreement does not come to an end

Question 7.
Match List-I with List-II and select the correct answer using the codes given below:

Answer:
b

Question 8.
Select the odd one out
(a) Revaluation profit
(b) Accumulated loss
(c) Goodwill brought by a new partner
(d) Investment fluctuation fund
Answer:
(c) Goodwill brought by a new partner

Question 9.
James and Kamalesh are partners sharing profits and losses in the ratio of 2:1. They admit Yogesh into partnership. The new profit sharing ratio between Balaji, Kamalesh, and Yogesh is agreed to 3:1:1. Find the sacrificing ratio.
(a) 1:3
(b) 3:1
(c) 5:3
(d) 3:5
Hint:

Answer:
(c) 5:3

Question 10.
Balaji and Kamalesh are partners sharing profits and losses in the ratio of 2:1. They admit Yogesh into partnership. The new profit sharing ratio between Balaji, Kamalesh and Yogesh is agreed to 3:1:1. Find the sacrificing ratio between Balaji and Kamalesh.
(a) 1:3
(b) 3:1
(c) 2:1
(d) 1:2
Hint:
Sacrifice ratio = old ratio – new ratio
Old partner’s = Balaji, Kamalesh

Answer:
(d) 1:2

II. Very Short Answer Questions

Question 1.
What is meant by the revaluation of assets and liabilities?
Answer:
When a partner is admitted into the partnership the assets and liabilities are revealed as the current value may differ from the book value. Determination of current values of assets and liabilities is called revaluation of assets and liabilities.

Question 2.
How are accumulated profits and losses distributed among the partner at the time of admission of a new partner?
Answer:
The following and the adjustment are necessary at the time of admission of a partner.

•  Distribution of accumulated profits, reserves, and losses
•  Revaluation of assets and liabilities
•  Determination of new profit – sharing ratio and sacrificing ratio
•  Adjustment for goodwill
•  Adjustment of capital on the basis of the new profit sharing ratio (if no agreed).

Question 3.
What is sacrificing ratio?
Answer:
The sacrificing ratio is the proportion of the profit which is sacrificed or foregone by the old partners in favour of the new partner. The purpose of finding the sacrificing ratio is to share the goodwill brought in by the new partner.
Sacrifice Ratio = Old share – New share

Question 4.
Give the journal entry for writing off existing goodwill at the time of admission of a new partner.
Answer:
Old partners capital /current A/c (in old ratio) Dr. To G/W A/c

Question 5.
State whether the following will be debited or credited in the revaluation account.
Answer:

1. Depreciation on assets – Debited
2. Unrecorded liability – Debited
3. Provision for outstanding expenses – Debited
4. Appreciation of assets – Credited

III Short Answer Questions

Question 1.
What are the adjustments required at the time of admission of a partner?
Answer:
The following adjustments are necessary at the time of admission of a partner.

1. Distribution of accumulated profits, reserves, and losses
2. Revaluation of assets and liabilities
3. Determination of new profit sharing ratio and sacrificing ratio
4. Adjustment for goodwill
5. Adjustment of capital on the basis of new profit sharing ratio

Question 6.
What are the journal entries to be passed on revaluation of ssets and liabilities?
Answer:
Following are the journal entries to be passed to record the revaluation of assets and liabilities:

Question 7.
Write a short note on the accounting treatment of goodwill.
Answer:
Accounting treatment for goodwill on the admission of a partner is discussed below:

Question 1.
When a new partner brings cash towards goodwill
Answer:
When the new brings cash towards goodwill in addition to the amount of capital, it is distributed to the existing partners in the sacrificing ratio.
If the new partner does not bring goodwill in cash or in kind, his share of goodwill must be adjusted through the capital accounts of the partners.

Sometimes the new partner may bring only a part of the goodwill in cash or assets. In such a case, for the cash or the assets brought, the respective account is debited and for the amount not brought in cash or kind, the new partner’s capital account is debited.

If goodwill already appears in the books of accounts, at the time of admission if the partners decide, it can be written off by transferring it to the existing partner s capital account / current account in the old profit sharing ratio.

IV Exercises

Distribution of accumulated profits, reserves, and losses

Question 1.
Arul and Anitha are partners sharing profits and losses in the ratio of 4:3. On 31.03.2018, Ajay was admitted as partner. On the date of admissions, the book of the firm showed a general reserve of ₹ 42,000. Pass the journal entry to distribute the general reserve.
Solution:

(General reserve transferred to old partners capital A/c in the old profit sharing ratio 4:3.)
Answer :
Arul: ₹ 24,000 (Cr.); Anitha: ₹ 18,000 (Cr.)

Question 2.
Anjali and Nithya are partners of firms sharing profits and losses in the ratio of 5:3. They admit Pramila on 01.01.2018. On that date, their balance sheet showed an accumulated loss of ₹ 40,000 on the asset side of the balance sheet. Give the journal entry to transfer the accumulated loss on admission.
Solution:

(Accumulated loss transferred to old partner s capital A/cs in the old ratio.)
Answer:
Profit and loss: ₹ 25,000 (Dr.); Nithya: ₹ 15,000(Dr.)

Question 3.
Oviya and Kavya are partners in firm sharing profits and losses in the ratio of 5:3. They admit Agalya into the partnership. Their balance sheet as of 31st March 2019 is as follows:

The pass journal entry to transfer the accumulated profit and reserve on admission.
Solution:

Accumulated profits and reserve transferred to old partners capital A/c in the old ratio
Answer :
Oviya : ₹ 37,500; Kavya : ₹ 22,500

Question 4.
Hari, Madhavan, and Kesavan are partners, sharing profit and losses in the ratio of 5:32. As from 1st April 2017, Vanmathi is admitted into the partnership and the new profit sharing ratio is decided as 4:3:2:1. The folio 5 adjustments are to be made.
(a) Increase the value of premises by ₹ 60,000.
(b) Depreciate stock by ₹ 5,000, furniture by ₹ 2, 000 and machinery by ₹ 2,500.
(c) Provide for an outstanding liability of ₹ 500.
Pass journal entries and prepare revaluation account.
Solution:

Question 5.
Seenu and Siva are partners sharing profits and losses in the ratio of 5:3. In view of kowsalya admission, they decided.
(i) To increase the value of the building by ₹ 40,000.
(ii) To bring into record investment at ₹ 10,000, Which have not so far been brought into
(iii) To decrease the value of machinery by ₹ 14,000 and furniture by ₹ 12,000.
(iv) To write off sundry creditors by ₹ 16,000.
Pass journal entries and prepare revaluation account
Solution :

Answer:
Revolution Profit in Rs. 40.000

Question 6.
Sai and Shankar are partners, sharing profits and losses in the ratio of 5:3.The firm’s balance sheet as on 31st December, 2017, was as follows:


On 31st December 2017, Shanmugam was admitted into the partnership for 1/4 share of profit with ₹ 12,000 as capital subject to the following adjustments.
(a) Furniture is to be revalued at ₹ 5,000 and the building is to be revalued at ₹ 50,000.
(b) Provision for doubtful debts is to be increased to ₹ 5,500
(c) An unrecorded investment of ₹ 6,000 is to be brought into account.
(d) An unrecorded liability of ₹ 2,500 has to be recorded now.
Pass journal entries and prepare Revaluation Account and capital account of partners after admission.
Solution:


Answer:
Revaluation Profit: ₹16,000; Capital accounts; Sai: ₹ 58,000 (Cr.), Shankar: ₹ 46,000 (Cr.); Shanmugam: ₹ 12,000 (Cr.))

Question 7.
Amal and Vimal are partners in firm sharing profits and losses in the ratio of 7:5. Their valance sheet as of 31st March 2019, is as follows:

Nirmal is admitted as a new partner on 01.04.2018 by introducing a capital of 30,000 for 1/3 share in the future profit subject to the following adjustments.
(a) Stock to be depreciated by ₹ 5,000;
(b) Provision for doubtful debts to be created for ₹ 3,000
(c) Land to be appreciated by ₹ 20,000.
Prepare revaluation account and capital account of partners after admission.
Solution:

Answer:
Revaluation Profit: ₹ 12,000; Capital accounts: Amal ₹ 91,000 (Cr.), Vimal ₹ 65,000 (Cr.), Nirmal ₹ 30,000 (Cr.))

Question 8.
Praveena and Dhanya are partners sharing profits in the ration of 7:3 they admit Malini into the firm. The new ratio among Praveena, Dhanya, and Malini are 5:2:3. Calculate the sacrificing ratio.
Solution:
Sacrificing ratio = Old share – New share
OR = 7:3 NR = 5:2:3
Praveena
SR = OR – NR
\(\frac{7}{10}-\frac{5}{10}=\frac{2}{10}\)
Dhanya
= \(\frac{3}{10}-\frac{2}{10}=\frac{1}{10}\)
SR = OR – NR
SR = 2:1
Answer :
Sacrificing ratio = 2:1

Question 9.
Ananth and Suman are partners sharing profits and losses in the ratio of 3:2. They admit Saran for 1/5 share, which he acquires entirely from Ananth. Find out the new profit sharing ratio and sacrificing ratio.
Solution :

Answer:
New profit sharing ratio: 2:2:1 Sacrificing ratio 1:0

Question 10.
Raja and Ravi are partners, sharing profit in the ratio of 3:2. They admit Ram for 1/4 share of the Profit, he takes 1/20 share from Raja and 4/20 from Ravi. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:


Answer :
New Profit sharing ratio: 11:4:5; Sacrificing ratio 1: 4

Question 11.
Vimala and Kamala are partners, sharing profits in the ratio of 4:3. Vinitha enters into the partnership and she acquires 1/14 from Vimala and 1/14 from Kamala. find out the new profit sharing ratio and sacrificing ratio.
Solution:

Answer :
New Profit sharing ratio 7 : 5 : 2; sacrificing ratio 1 : 1

Question 12
Govinda and Gopal are partners is a firm sharing profit^ in the ratio of 5 : 4. They admit Rahim as a partner. Govind surrenders 2/9 of his share in favour of Rahim. Gopal surrenders 1 /9 of his share in favour of R thim. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:


Answer:
New profit sharing ratio 35:32:14; Sacrificing ratio 5:2

Question 13.
Prema and Chandra share profits in the ratio of 5:3. Hema has admitted a partner. Prema surrendered 1 /8 of her share and Chandra surrendered 1 /8 of her share in favour of Hema. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:
Prema : Chandra → 5:3 (OR)
Sacrificing Ratio

Answer:
New profit sharing ratio 35:21:8; Sacrificing ratio 5:3

Question 14.
Karthik and Kannan are equal partners. They admit Kailash with 1/4 share of the profit. Kailash acquired his share from old partners in the ratio of 7:3 Calculate the new profit sharing ratio and sacrificing ratio.
Solution :
Sacrificing Ratio → Share of New Parter 1/4 → Sacrificed → 7 : 3
karthicks SR

Answer:
New profit sharing ratio 13:17:10; Sacrificing ratio 7:3

Question 15.
Selvam and Senthil are partners sharing profit in the ratio of 2:3. Siva is admitted into the firm with 1/5 share of profit. Siva acquires equally from Selvam and Senthil. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:

Answer:
New profit sharing ratio 3:5:2; Sacrificing ratio 1:1

Question 16.
Mala and Anitha are partners, sharing profits and losses in the ratio of 3:2. Mercy is admitted into the partnership with 1/5 share in the profits. Calculate new profit sharing ratio and sacrificing ratio.
Solution:
Since share sacrificed proportion and new profit sharing ratio are not given, it is assumed that the existing partners sacrifice in their old profit sharing ratio that is 3:2
Sacrificing Ratio

Answer:
New profit sharing ratio 12:8:5; Sacrificing ratio 3:2

Question 17.
Ambika, Dharani and Padma are partners in a firm sharing profile in the ratio of 5:3:2. they admit Ramya for 25% profit. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:

Answer:
New profit sharing ratio 15:9:6:10; Sacrificing ratio 5:3:2

Adjustment for goodwill

Question 18.
Aparna and Priya are partners who share profits and losses in the ratio of 3:2. Brindha joins the firm for 1/5 share of profits and brings in cash for her share of the goodwill of Rs. 10,000. Pass necessary journal entry for adjusting goodwill on the assumption that the fluctuating capital method is followed and the partners withdraw the entire amount of their share of goodwill.
Solution:


Answer:
Share of goodwill: Aparna: ₹ 6,000; Priya ₹ 4,000

Question 19.
Deepak, Senthil, and Santhosh are partners sharing profits and losses equally. They admit Jerald into a partnership for 1/3 share in future profits. The goodwill of the firm is valued at ₹ 45,000 and Jerald brought cash for his share of goodwill. The existing partners withdraw half of the amount of their share of goodwill. Pass necessary journal entries for adjusting goodwill on the assumption that the fluctuating capital method is followed.
Solution:
Jerald’s share of G/w = 45,000 x \(\frac{1}{3}\) = Rs. 15,000

Answer:
Share of goodwill: Deepak: ₹ 5,000; Senthil: ₹ 5,000; Santhosh: ₹ 5,000

Question 20.
Malathi and Shobana are partners sharing profits and losses in the ratio of 5:4. They admit Jayasri into a partnership for 1/3 share of profit. Jayasri pays cash ₹ 6,000 towards her share of goodwill. The new ratio is 3:2:1. Pass necessary journal entry for adjusting goodwill on the assumption that the fixed capital method is followed.
Solution:
= OR – NR
OR = 5:4
New Ratio = 3:2:1

Answer:
Share of goodwill: Malathi’s Current account: ₹ 2,000; Shobana’s Current account: ₹ 4,000;

Question 21.
Anu and Arul were partners in a firm sharing profits and losses in the ratio of 4:1. They have decided to admit Mano into the firm for 2/5 share of profits. The goodwill of the firm on the date f admission was valued at ₹ 25,000. Mano is not able to bring in cash for his share of goodwill. Pass necessary journal entry for goodwill on the assumption that the fluctuating Capital method is followed.
Solution:
Mano’s Share of G/w = 25,000 x 2/5 = Rs. 10,000

Answer:
Share of goodwill: Anu: ₹ 8,000; Arul: ₹ 2,000;

Question 22.
Varun and Barath are sharing profits and losses 5:4. They admit Dhamu into partnership. The new profit sharing ratio is agreed at 1:1:1. Dhamu’s share of goodwill is valued at ₹ 15,000 of which he pays ₹ 10,000 in cash. Pass necessary journal entries for adjustment of goodwill on the assumption that the fluctuating capital method is . followed. ‘
Answer:

Answer:
Share of goodwill: Varun : ₹ 10,000; Barath : ₹ 5,000;

Question 23.
Sam and Jose are partners in the firm sharing profits and losses in the ratio of 3:2. On 1st April 2018, they admitted Joel as a partner. On the date of Joel’s admission, goodwill appeared in the books of the firm at ₹ 30,000. By assuming the fluctuating capital method, pass the necessary journal entry if the partners decide
(a) Write off the entire amount of existing goodwill
(b) write off ₹ 20,000 of the existing goodwill.
Solution :
To write off the entire amount of goodwill

Answer:
Share of goodwill: (a) Sam: ₹ 18,000 (Dr); Jose : ₹ 12,000 (Dr) (b) Sam: ₹ 12,000 (Dr); Jose: ₹ 8,000 (Dr.))

Comprehensive problems:

Question 24.
Rajan and Selva are partners sharing profits and losses in the ratio of 3:1. Their balance sheet as on 31st March 2017 is as under.

On 01.04.2017, they admit Ganesan as a new partner on the following arrangements.
(i) Ganesan brings ₹ 10,000 as capital for 1/5 share of profit.
(ii) Stock and furniture are to be reduced by 10%, a reserve of 5% on debtors for doubtful debts is to be created.
(iii) Appreciate buildings by 20%
Prepare revaluation account, partner’s capital account, and the balance sheet of the firm after admission.
Solution :


Answer:
Revaluation profit:: ₹ 2,100; Capital accounts : Rajan : 27075; Selva; ₹ 15,025; Ganesan : ₹ 10,000 Balance sheet total: ₹ 89,600

Question 25.
Sundar and Suresh are partners sharing profit in the ratio of 3 : 2. Their balance sheet as on 1st January 2017 was as follows:

They decided to admit Sugumar into a partnership for 1/4 share in the profits on the following terms:
(a) Sugumar has to bring in ₹ 30,000 as capital. His share of goodwill is valued at Rs. 5,000. He could not bring cash towards goodwill.
(b) That the stock is valued at ₹ 20,000.
(c) That the furniture is depreciated by ₹ 2,000.
(d) That the value of building be depreciated by 20%
Prepare necessary ledger accounts and the balance sheet after admission.
Solution:



Answer:
Answer:
Revaluation loss : ₹ 15,000; Capital accounts: Sundar : ₹ 39,000; Suresh; ₹ 26,000; Sugumar : ₹ 25,000 Balance sheet total ₹ 1,40,000

Question 26.
The following is the balance sheet of James and Justina as on 1.1.2017. They share the profits and losses equally.

On the above date, Balan is admitted as a partner with a 1/5 share in future profits. Following are the terms for his admission:
(i) Balan brings ₹ 25,000 as capital.
(ii) His share of goodwill is ₹ 10,000 and he brings cash for it.
(iii) The assets are to be valued as under:
Building ₹ 80,000; Debtors ₹ 18,000; Stock ₹ 33,000
Prepare necessary ledger accounts and the balance sheet after admission.
Solution:


Answer:
Revaluation profit:: ₹ 11,000; Capital accounts : James : ₹ 58,000; Justina; ₹ 68,000; Balan : ₹ 25,000 Balance sheet total: ₹ 1,86,000

Question 27.
Anbu and Shankar are partners in a business sharing profits and losses in the ratio of 7:5. The balance sheet of the partners on 31.03.2018 is as follows:

Rajesh is admitted for 1/5 share on the following terms:
(i) Goodwill of the firm is valued at ₹ 80,000 and Rajesh brought cash ₹ 6,000 for his share of goodwill.
(ii) Rajesh is brought ₹ 1,50,000 as his capital.
(iii) Motor car is valued at ₹ 2,00,000; stock at ₹ 3,80,000 and debtors at ₹ 3,50,000.
(iv) Anticipated claim on workmen compensation fund is ^ 10,000
(v) Unrecorded investment of ₹ 5,000 has to be brought into account.
Prepare revaluation account, capital account, and balance sheet after Rajesh’s admission.
Solution:


Answer:
Revaluation profit:: ₹ 15,000; Capital accounts: Anbu: ₹ 5,11,417; Shankar; ₹ 3,79,583; Rajesh : ₹ 1,50,000 Balance sheet total: ₹ 11,71,000

12th Accountancy Guide Admission of a Partner Additional Important Questions and Answers

Question 1.
When A and B sharing profit and losses in the ratio of 3:2. They admit C as a partner giving him 1 /3 share of profits. This will be given by A 8t B
(a) Equally
(b) in the ration of their
(c) in the ratio of profits
Answer:
(c) in the ratio of profits

Question 2.
In order to maintain fair dealings at the time of admission, it is necessary to revalue assets 8t liabilities of the firm to their
(a) Cost Price
(b) Cost price less depreciation
(c) True value.
Answer:
(c) True value.

Question 3.
If the new share of the incoming partner is given without mentioning the details of the sacrifice made by the old partners then the presumption is that old partners sacrifice in the
(a) Old profit sharing ration
(b) Gaining ratio
(c) Capital ratio.
Answer:
(a) Old profit sharing ration

Question 4.
On admission of a new partner, the balance of general Reserve A/c should be transferred to the capital account of
(a) all partners in their new profit sharing ratio
(b) Old partners in their new old profit sharing ratio
(c) Old partners in their new profit sharing ratio.
Answer:
(b) Old partners in their new old profit sharing ratio

Question 5.
The old partners share all the accumulated profit & reserves in their
(a) new profit sharing ratio
(b) Old profit sharing ratio
(c) Capital ratio
Answer:
(b) Old profit sharing ratio

Question 6.
Hie reconstitution of the partnership requires a revision of the existing partners
(a) Profit sharing ratio
(b) Capital ratio
(c) Sacrificing ratio|
Answer:
(a) Profit sharing ratio

Question 7.
……………… ratio is computed at the time by the admission of a partner
(a) gaining ratio
(b) Capitalization
(c) Sacrificing ratio
Answer:
(c) Sacrificing ratio

Question 8.
When unrecorded liability is brought in to books of accounts, it results in
(a) profit
(b) loss
(c) Neither profit nor loss
Answer:
(b) loss

IV Additional Problems:

Question 1.
Anandan and Balaraman partners in a firm with a capital of Rs. 70,000 and Rs. 50,000 respectively. They decided to admit Chandran into the firm with a capital of Rs. 40,000. Give journal entry for Capital brought in by Chandran.
Solution :

Question 2.
Rathai and Kothai are partners sharing profits in the ratio of 3:2. They admit Kanmani for 1/5th share of future profits which she acquires 3/20th from Rathai and 1/20th from Kothai. Calculate new Pro iring ratio and. sacrifice s ratio of old partners.
Solution :

Question 3.
P and Q are partners sharing profits in the ratio of 3:2. They admit R for 1/5th Share which acquires equally from P and Q. Calculate new profit sharing ratio and sacrificing ratio of old partners.
Solution:

Question 4.
Sankar and Saleem are partner in a firm sharing profits and losses in the ration of 3:2 as on 31st March 2005. Their Balarr sheet was as under.

On 1st April 2005, they admit Solomon into partnership on the following condition: Solomon has brought Rs. 1,00,000 as capital.
The value of land and building was to be increased by Rs. 20,000.
Stock and furniture were to be depreciated by Rs. 10,000 and Rs. 5,000 respectively. Rs. 15,000 to be written off from sundry creditors as it is no longer liable.
Provision for doubtful debts is to be increased by Rs. 1,000.
Give journal entries, prepare Revaluation Account and the Balance Sheet.
Solution:

Question 5.
Amar and Akbar are partners in a firm sharing profits and losses in the ratio of 2:1  as 31st March 2005. Their Balan Sh» was as under.

On 1st April 2005, they admit Antony into partnership on the following conditions:
Antony has brought in the capital of Rs. 1,50,000 for 1/5th share of the future profits. Stock and machinery were to be depreciated by Rs. 6,000 and Rs. 15,000 respectively.
Investments Rs. 15,000 not recorded in the books brought into accounts.
Provision for doubtful debts is to be created at 5% on debtors.
A liability of Rs. 4,000 for outstanding repairs has been omitted to be recorded in the books. Give journal entries, prepare Revaluation Account, Capital Account, Bank Account, and the Balance Sheet.
Solution:

Question 6.
Sumathi and Sundari are partners of a firm sharing profit and loss in the ratio of 4 :3. Their Balance Sheet shows Rs. 14,000 as Profit and Loss A/c on the liabilities side. Pass entry.
Solution:

Question 7.
Mahalakshmi and Dhanalakshmi are partners sharing profit and loss in the ratio of 3:2. They admit Deepalakshmi on 1st January 2005. On that date, their Balance Sheet showed an amount of Rs. 25,000 as Profit and Loss A/c in the Asset side. Pass entry.
Solution:

Question 8.
Damodaran and Jagadeesan are partners sharing profits in the ratio of 3:2. They decided to admit Vijayan for 1/5th share of future profit. Goodwill of the firm is to be valued at Rs. 50,000.
Give Journal entries, if
There is no goodwill in the books of the firm.
The goodwill appears at Rs. 30,000
The goodwill appears at Rs. 60,000.
Solution:

Question 9.
Sankari and Sudha are partners sharing profit and loss in the ratio of 3:2. Their Balance Sheet as on 31st March 2005 is as under.

They decided to admit Santhi into the partnership with effect from 1st April 2005 on the following terms:
Santhi to bring in Rs. 60,000 as Capital for 1/3rd share of profits.
Goodwill was valued at Rs. 45,000
The land was valued at Rs. 1,50,000
The stock was to be written down by Rs. 8,000
The provision for doubtful debts was to be increased to Rs. 3,000
Creditors include Rs. 5,000 no longer payable and this sum was to be written off.
Investment of Rs. 10,000 be brought into books.
Prepare Revaluation A/c, Capital A/c, and Balance Sheet of the new firm.
Solution :

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.7

Question 1.
Evaluate the following
(i) \(\int_{0}^{∞}\) x⁵ e^-3x dx
Solution:

(ii) \(\int_{0}^{π/2}\) \(\frac{e^{-tanx}}{cos^6 x}\) dx
Solution:

Question 2.
\(\int_{0}^{∞}\) sin^αx² x³ dx = 32, α > 0, find α
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 1 Electrostatics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

12th Physics Guide Electrostatics Book Back Questions and Answers

Part – I:
I. Multiple choice questions:

Question 1.
Two identical point charges of magnitude – q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?

(a) A₁ and A₂
(b) B₁ and B₂
(c) both directions
(d) No stable
Answer:
(b) B₁ and B₂
Solution:
The potential due to an electric dipole along the equatorial line is zero.

Question 2.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) uniformly charged infinite line
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
c) uniformly charged infinite plane
Solution:

Uniform field lines are represented by equidistant parallel lines.

Question 3.
What is the ratio of the \(\left|\frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}\right|\) charges for the following electric field line pattern?

(a) \(\frac{1}{5}\)
(b) \(\frac{25}{11}\)
(c) 5
(d) \(\frac{11}{25}\)
Answer:
(d) \(\frac{11}{25}\)
Solution:
Here q₁ is a negative charge and q₂ is a positive charge
Hint:
Count the number of lines on each charge.

Question 4.
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 × 10⁵ NC^-1. It experiences a torque equal to 8 Nm. The charge on the dipole if the dipole length is lcm is
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 7 mC
Answer:
(b) 8 mC
Solution:
T = PE sin θ
T = (q × 21)E sin30°
q = (q × 10^-2) × q = 8 × 10^-2c

Question 5.
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order:

(a) D < C < B < A
(b) A < B = C < D
(c) C < A – B < D
(d) D > C > B > A
Answer:
(a) D < C < B < A
Hint :
flux depends on charge

Question 6.
The total electric flux for the following closed surface which is kept inside water:

(a) \(\frac{80 \mathrm{q}}{\varepsilon_{\mathrm{o}}}\)
(b) \(\frac{\mathrm{q}}{40 \varepsilon_{0}}\)
(c) \(\frac{\mathrm{q}}{80 \varepsilon_{\mathrm{o}}}\)
(d) \(\frac{\mathrm{q}}{160 \varepsilon_{0}}\)
Answer:
(b) \(\frac{\mathrm{q}}{40 \varepsilon_{0}}\)
Solution:

Question 7.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before
Solution:

Question 8.
Rank the electrostatic potential energies for the given system of charges in increasing order:

(a) 1 = 4 < 2 < 3
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer:
(a) 1 = 4 < 2 < 3
Solution:

Question 9.
An electric field \(\overrightarrow{\mathrm{E}}\)= 10 × Î exists in a certain region of space. Then the potential difference V = V_o – V_A, where V_o is the potential at the origin and V_A is the potential at x = 2 m is:
(a) 10 V
(b) -20 V
(c) +20 V
(d) -10 V
Answer:
(c) +20 V
Solution:
E = \(-\frac{d v}{d x}\)
dv = E.dx
= l0x
= 10 × 2
dv = 20 V

Question 10.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is
(a)

(b)

(c)

(d)
Answer:
(b)
Solution:
In a spherical shell, the electric field inside is zero. But the electric potential is constant.
\(\mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}}\) as distance increases its potential decrease non-linearly.

Question 11.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is
(a) 8.80 × 10^-17 J
(b) -8.80 × 10^-17 J
(c) 4.40 × 10^-17 J
(d) 5.80 × 10^-17 J
Answer:
(a) 8.80 × 10^-17 J
Solution:
W_A→B = (V_A – V_B)q
=(7+4)ne
= 11 × 50 × 1.6 × 10^-19
=8.8 × 10^-17

Question 12.
If the voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains the same, Q doubled
(d) Both Q and C remain the same
Answer:
(c) C remains the same, Q doubled
Solution:

Question 13.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Question 14.
Three capacitors are connected in a triangle as shown in the figure. The equivalent capacitance between points A and C is

(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) \(\frac{1}{4}\)μF
Answer:
(b) 2 μF
Solution:

Question 15.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10^-2 C and 5 × 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
(a) 3 × 10^-2 C
(b) 4 × 10^-2 C
(c) 1 × 10^-2 C
(d) 2 × 10^-2 C
Answer:
(a) 3 × 10^-2 C
Solution:
Total charge Q = q₁ + q₂ = 4 × 10^-2 C
charge on bigger sphere,
q₂ = Q\(\left(\frac{r_{2}}{r_{1}+r_{2}}\right)\)
= 4 × 10^-2 × \(\frac{3}{4}\)
q₂ = 3 × 10^-2 C

II. Short Answer Questions:

Question 1.
What is meant by the quantization of charges?
Answer:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is an integer (0, ±1, ±2, ±3, ±4).
This is called quantisation of electric charge.

Question 2.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on the point charge q₂ exerted by another point charge q₁

where r̂₂ is the unit vector directed from charge q₁ to charge q₂, and ‘K’ is the proportionality constant K = \(\frac{1}{4 \pi \varepsilon_{0}}\); ε₀ is the permittivity of free space.

Question 3.
What are the differences between the Coulomb force and the gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb’s force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 x 10^-11 N m² kg^-2. The value of the constant k in Coulomb law is k = 9 x 10⁹ N m² C².
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on the nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to Coulomb force.

Question 4.
Write a short note on the superposition principle.
Answer:

1. The superposition principle explains the interaction of multiple charges.
2. The total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
3. The force on q1 exerted by the charge q<₂ is \(\overrightarrow{\mathrm{F}}_{12}=\mathrm{K} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{21}^{2}} \hat{\mathrm{r}}_{21}\)
4. The force on q1 exerted by the charge q<₃ is \(\overrightarrow{\mathrm{F}}_{13}=\mathrm{K} \frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{31}^{2}} \hat{\mathrm{r}}_{31}\)
5. \(\overrightarrow{\mathrm{F}}_{1}^{\text {tot }}=\mathrm{K}\left\{\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{21}^{2}} \hat{\mathrm{r}}_{21}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{31}^{2}} \hat{\mathrm{r}}_{31}+\ldots \frac{\mathrm{q}_{1} \mathrm{q}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}^{2}} \hat{\mathrm{r}}_{\mathrm{n}_{i}}\right\}\)
   

Question 5.
Define ‘electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
\(\vec { E } \) = \(\frac { \vec { F } }{ { q }_{ 0 } } \)
The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Question 6.
What is mean by ‘electric field lines?
Answer:
Electric lines of force is an imaginary straight or curved path in which unit positive charge tends to move in the presence of an electric field.

Question 7.
The electric field lines never intersect. Justify.
Answer:
As a consequence, if some charge is placed at the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

Question 8.
Define ‘electric dipole’. Give the expression for the magnitude of its electric dipole moment and the direction.
Answer:

1. Two equal and opposite charges separated by a very small distance constitute an electric dipole.
   Ex:- Carbondioxide, Water.
2. The magnitude of the dipole moment is given by the product of the magnitude of the one of the charges and the distance between them.
   \(|\overrightarrow{\mathrm{P}}|=2 \mathrm{qa}\)
   \(\overrightarrow{\mathrm{P}}=\mathrm{qr}_{+}+(-\mathrm{q}) \overrightarrow{\mathrm{r}}_{-}\)
   where r̂₊ is the position of vector of +1 from the origin and r̂_– is the position vector of -q from the origin
   \(\overrightarrow{\mathrm{P}}=\mathrm{qai}-\mathrm{qa}(-\hat{i})=2 \mathrm{qai}\)
3. It is a vector quantity having direction along the dipole axis -q to +q.
4. Unit: coulomb metre (Cm).

Question 9.
Write the general definition of electro dipole moment for a collection of point charge.
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from -q to + q. The SI unit of dipole moment is coulomb meter (Cm).
\(\vec { P } \) = qa\(\hat{i} \) -qa(\(\hat{-i} \)) = 2 qa\(\hat{i} \)

Question 10.
Define ‘electrostatic potential’.
Answer:
The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external electric field \(\overrightarrow{\mathrm{E}}\).

Question 11.
What is an equipotential surface?
Answer:
An equipotential surface is a surface on which all the points are at the same potential.

Question 12.
What are the properties of an equipotential surface?
Answer:

1. The work done to move a charge ‘q’ between two points lie on equipotential surface is zero.
2. The electric field is normal to an equipotential surface.

Question 13.
Give the relation between electric field and electric potential.
Answer:
Consider a positive charge q kept fixed at the origin. To move a unit positive charge by a small distance dx in the electric field E, the work done is given by dW = -E dx. The minus sign implies that work is done against the electric field. This work done is equal to an electric potential difference. Therefore,
dW = dV.
(or) dV = -Edx
Hence E = \(\frac { dV }{ dx }\)
The electric field is the negative gradient of the electric potential.

Question 14.
Define ‘electrostatic potential energy’.
Answer:
It is defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 15.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 16.
What is meant by electrostatic energy density?
Answer:

1. The energy stored per unit volume of space is defined electrostatic energy density.
   \(\mathrm{U}_{\mathrm{E}}=\frac{\mathrm{U}}{\text { Volume }}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}\)
2. U – electrostatic potential energy
3. E – electric field
4. £0 – permittivity of free space

Question 17.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 18.
What is Polarisation?
Answer:
Polarisation is defined as the total dipole moment per unit volume of the dielectric.
\(\begin{array}{l}
\vec{p}=\chi_{\mathrm{e}} \overrightarrow{\mathrm{E}}_{\text {ext }} \\
\chi_{\mathrm{e}}=\text { electric susceptibility }
\end{array}\)

Question 19.
What is dielectric strength?
Answer:
The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.

Question 20.
Define ‘capacitance’. Give its unit.
Answer:

• It is defined as the ratio of the magnitude of the charge on either of the conductor plates to the potential difference existing between the conductors.
• Unit: farad (or) C/V

Question 21.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

III. Long Answer Questions:

Question 1.
Discuss the basic properties of electric charges.
Answer:
Basic properties of charges
(i) Electric charge:
Most objects in the universe are made up of atoms, which in turn are made up of protons, neutrons and electrons. These particles have mass, an inherent property of particles. Similarly, the electric charge is another intrinsic and fundamental property of particles. The SI unit of charge is the coulomb.

(ii) Conservation of charges:
Benjamin Franklin argued that when one object is rubbed with another object, charges get transferred from one to the other. Before rubbing, both objects are electrically neutral and rubbing simply transfers the charges from one object to the other. (For example, when a glass rod is rubbed against silk cloth, some negative charges are transferred from glass to silk. As a result, the glass rod is positively charged and silk cloth becomes negatively charged).

From these observations, he concluded that charges are neither created or nor destroyed but can only be transferred from one object to other. This is called conservation of total charges and is one of the fundamental conservation laws in physics. It is stated more generally in the following way. The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

(iii) Quantisation of charges:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ± ….). This is called quantisation of electric charge. Robert Millikan in his famous experiment found that the value of e = 1.6 x 10^-19C. The charge of an electron is -1.6 x 10^-19 C and the charge of the proton is +1.6 x 10^-19C. When a glass rod is rubbed with silk cloth, the number of charges transferred is usually very large, typically of the order of 10¹⁰. So the charge quantisation is not appreciable at the macroscopic level. Hence the charges are treated to be continuous (not discrete). But at the microscopic level, quantisation of charge plays a vital role.

Question 2.
Explain in detail Coulomb’s law and its various aspects.
Answer:
1. Coulomb’s law states that “force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of distance between them.
\(\overrightarrow{\mathrm{F}}=\mathrm{k} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)

2. The force on the charge q₂ exerted by the charge q&lt₁ always lies along the line joining the two charges.
r̂&lt₁2 is a unit vector pointing from q₁ to q₂.

3. In SI units, \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\)= 9 x l0⁹Nm²C^-2
where £0 = 8.85 × 10^-12 C² N^-1m^-2 .

4. The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of lm is
|F| = 9 x l0⁹N
This is a huge quantity, almost equivalent to weight of one million tons.

5. Coulombs law in vacuum takes the form
\(\overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)
In a Medium,
\(\overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)
£ > £0, the force between two point charges in a medium other than vacuum is always less than that in vacuum.

6. Relative permitivitv £r = \(\frac{\varepsilon}{\varepsilon_{o}}\)
for vacuum and air £r = 1
other media, £r > 1

Question 3.
Define ’electric field’ and discuss its various aspects.
Answer:
The electric field is defined as the force experienced by unit positive charge kept at that point.
\(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{q}_{\mathrm{o}}}\)

Important aspects of the Electric field:

1. If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q.
2. If the electric field at a point P is E, then the force experienced by the test charge q0 placed at the point P is
   \(\overrightarrow{\mathrm{F}}=\mathrm{q}_{\mathrm{o}} \overrightarrow{\mathrm{E}}\)
3. Electric field is independent of the test charge q₀ and depends only on the source charge q.
   
4. Since the electric field is a vector quantity, at every point in space, this field has unique direction.
5. The test charge is made sufficiently small such that it will not modify the electric field of the source charge.
6. For continuous and finite size charge distributions, integration techniques must be used.
7. There are two kinds of electric field: uniform or constant electric field and non-uniform electric field.
8. Uniform electric field will have the same direction and constant magnitude at all points in space. Non-uniform electric field will have different directions or different magnitudes or both at different points in space.

Question 4.
Calculate the electric field due to a dipole on its axial line and the equatorial plane.
Answer:
Case (i):

Electric field due to an electric dipole at points on the axial line.

1. Consider an electric dipole placed on the x-axis A point C is located at a distance of r from the midpoint of the dipole along the axial line.
   

2. The electric field at a point C due to +q is
\(\overrightarrow{\mathrm{E}}_{+}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a})^{2}}\) (along BC)

3. Since the electric dipole moment vector p is from -q to +q and is directed along BC, the above equation is rewritten as
\(\overrightarrow{\mathrm{E}}_{+}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a})^{2}} \hat{\mathrm{p}}\) ………..(1)

4. The electric field at a point C due to -q is
\(\overrightarrow{\mathrm{E}}_{-}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}+\mathrm{a})^{2}} \hat{\mathrm{p}}\) ……………(2)

5. Since +q is located closer to the point C than -q, \(\overrightarrow{\mathrm{E}}+\) is stronger than \(\overrightarrow{\mathrm{E}}_{-}\).
Therefore, the length of the \(\overrightarrow{\mathrm{E}}+\) vector is drawn larger than that of \(\overrightarrow{\mathrm{E}}_{-}\) vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

6. The direction of is shown in Figure

Case (ii)
1. Electric field due to an electric dipole at a point on the equatorial plane

2. Consider a point C at a distance r from the midpoint 0 of the dipole on the equatorial plane.

3. Since point C is equidistant from +q and -q and are the same.

4. The direction of \(\overrightarrow{\mathrm{E}}+\) is along BC and the direction of \(\overrightarrow{\mathrm{E}}_{-}\) is along CA.

\(\overrightarrow{\mathrm{E}}+\) and \(\overrightarrow{\mathrm{E}}_{-}\) are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.

5. The perpendicular components \(|\overrightarrow{\mathrm{E}}|\) sinθ and \(\left|\overrightarrow{\mathrm{E}}_{-}\right|\) sinθ are oppositely directed and cancel each other.
\(\mathrm{E}_{\text {tot }}=-\left|\mathrm{E}_{+}\right| \cos \theta \hat{\mathrm{p}}-\left|\overrightarrow{\mathrm{E}}_{-}\right| \cos \theta \hat{\mathrm{p}} \cdots \ldots .(6)\)

6. The magnitudes \(\overrightarrow{\mathrm{E}}+\) and \(\overrightarrow{\mathrm{E}}_{-}\) are the same and are given by

7. By substituting equation (7) into equation (6) we get
\(\overrightarrow{\mathrm{E}}_{\mathrm{tot}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{2 \mathrm{q} \cos \theta}{\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)} \hat{\mathrm{p}}\)
since \(\cos \theta=\frac{a}{\sqrt{r^{2}+a^{2}}}\)

8. At very large distances (r»a), the equation (8) becomes
\(\overrightarrow{\mathrm{E}}_{\mathrm{tot}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\overrightarrow{\mathrm{p}}}{\mathrm{r}^{3}}(\mathrm{r}>>\mathrm{a})\)

Question 5.
Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
1. Consider an electric dipole of dipole moment \(\overrightarrow{\mathrm{p}}\) placed in a uniform electric field . The charge+q will experience a force q\(\overrightarrow{\mathrm{E}}\) in the direction of the field and charge-q\(\overrightarrow{\mathrm{E}}\) will experience a force -qE in a direction opposite to the field. Since the external field \(\overrightarrow{\mathrm{E}}\) is uniform, the total force acting on the dipole is zero.

2. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole.

3. The total torque on the dipole about the point 0, is given by
\(\vec{\tau}=\overrightarrow{\mathrm{0A}} \times(-\mathrm{q} \overrightarrow{\mathrm{E}})+\overrightarrow{\mathrm{0B}} \times \mathrm{q} \overrightarrow{\mathrm{E}}\)
Torque on dipole.

4. Total torque is perpendicular to the plane of the paper and is directed into it. The magnitude of the total torque
\(\vec{\tau}=|\overrightarrow{\mathrm{0A}}|(-\mathrm{q} \overrightarrow{\mathrm{E}})|\sin \theta+| \overrightarrow{\mathrm{0B}}|\mathrm{q} \overrightarrow{\mathrm{E}}| \sin \theta\)

\(\tau\) = qE.2a sin θ

5. Where θ is the angle made by \(\overrightarrow{\mathrm{p}}\) with \(\overrightarrow{\mathrm{E}}\). Since p = 2aq, the torque is written in terms of the vector product as
\(\vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}\)

6. The magnitude of thectrque is \(\tau=p E \sin \theta\) and is maximum when θ = 90^o.
This torque tends to rotate the dipole and align it with the electric field \(\overrightarrow{\mathrm{E}}\). once \(\overrightarrow{\mathrm{p}}\) is aligned with, the total torque on the dipole becomes zero.

7. If the electric field is nonuniform, then the force experienced by +q is different from that experienced by -q. In addition to the torque, there will be a net force acting on the dipole.

Question 6.
Derive an expression for electrostatic potential due to a point charge.
Answer:
1. Consider a positive charge ‘q’ kept fixed at the origin. Let P be a point at distance r from the charge ‘q’.

2. The electric potential at the point P is
\(\mathrm{V}=\int_{\infty}^{\mathrm{r}}(-\overrightarrow{\mathrm{E}}) \cdot \mathrm{d} \overrightarrow{\mathrm{r}}=-\int_{\infty}^{\mathrm{r}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dr}}\)

3. Electric field due to positive point charge is
\(\begin{array}{l}
\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \\
\mathrm{V}=\frac{-1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}
\end{array}\)

4. The infinitesimal displacement vector,
d\(\overrightarrow{\mathrm{r}}\) = drr̂ and using r̂.r̂ = 1, we have
\(\mathrm{V}=\frac{-1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \cdot \mathrm{dr} \hat{\mathrm{r}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \mathrm{dr}\)

5. After the integration,
\(\mathrm{V}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{q}\left\{-\frac{1}{\mathrm{r}}\right\}_{\infty}^{\mathrm{r}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)

6. Hence the electric potential due to a point charge q at a distance r is
\(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)

Question 7.
Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let θ be the angle between the line OP and dipole axis AB.

Potential due to electric dipole:
1. Let r₁ be the distance of point P from +q and r₂ be the distance of point P from -q.

2. Potential at P due to charge,
\(+q=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{1}}\)
Potential at P due to charge
\(-q=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{2}}\)
Total potential at the point P,
\(\mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right)\) (1)

3. By the cosine law for triangle BOP,
r₁² =r² + a² – 2ra cos θ

4. Similarly applying the cosine law for triangle A0P,
r₂² = r² + a²  – 2racos(180-θ)
since cos(180-θ) = – cos θ
r₂² = r² + a² +2racos θ

Using Binomial theorem, we get

\(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\overrightarrow{\mathrm{p}} \cdot \hat{\mathrm{r}}}{\mathrm{r}^{2}}\)
Special Cases:

Question 8.
0btain an expression for potential energy due to a collection of three-point charges which are separated by finite
distances.
Answer:

1. The electric potential at a point at a distance r from point charge q&lt₁ is given by
   \(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}_{1}}{\mathrm{r}}\)
2. The potential V is the work done to bring a unit positive charge front infinity to the point. If the charge q, is brought from infinity to that point at a distance r from q₁.
3. The work done is stored as the electrostatic potential energy W = q₂V
   \(\mathrm{U}=\mathrm{q}_{2} \mathrm{~V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{q}_{1} \mathrm{q}_{2}\) ………….(1)
4. The electrostatic potential energy depends only on the distance between the two point charges.

Three charges are arranged in the following configuration.

1. Bringing a charge q₁ from infinity to point A requires no work because there are no other charges already present in the vicinity of charge q₁.

2. To bring the second charge q₂ to point B, work must be done against the electric field at B created by the charge q₁. So the work done on the charge q₂ is Wq₂V_1B Here V_1B is electrostatic potential due to the charge q₁ at point B.
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}}\) (2)

3. Similarly to bring the charge q₂ to point C, work has to be done against the total electric field due to both the charges q₁ and q₂. So the work done to bring the charge q₃ is the electrostatic potential due to charge q₁ at point C and V₂ is the electrostatic potential due to charge q₂ at point C. The electrostatic potential is
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{13}}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{\mathrm{r}_{23}}\right)\) ………….(3)

4. Adding equations (2) and (3), the total electrostatic potential energy for the system of three charges q₁, q₂ and q₃ is
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{13}}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{\mathrm{r}_{23}}\right)\) ……………. (4)

Question 9.
Derive an expression for the electrostatic potential energy of the_dipole in a uniform electric field.
Answer:
1. Consider a dipole placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\). A dipole experiences a torque when kept in an uniform electric field \(\overrightarrow{\mathrm{E}}\). This torque rotates the dipole to align it with the direction of the electric field.

2. To rotate the dipole an equal and opposite external torque must be applied on the dipole.

3. The work done by the external torque to rotate the dipole from angle θ<super>1 to θ at constant angular velocity is

4. The work done is equal to the potentiaL energy difference between the angular positions θ’ and θ.
U(θ) – U(θ’)= ΔU = – pE cos θ+pE cos θ’

5. If the initial angle is θ’ = 90° and is taken as reference point, then U(θ’)= pE cos 90°=0.

6. The potential energy stored in the system of dipole kept in the uniform electric field is given by
U= – pE cos θ = – \(\overrightarrow{\mathrm{p}}\).\(\overrightarrow{\mathrm{E}}\) ……….(3)

7. In addition to p and E, the potential energy also depends on the orientation 0 of the electric dipole with respect to the external electric field.

8. The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0)

Question 10.
Obtain Gauss law from Coulomb’s law.
Answer:
1. Consider a positive point charge Q is surrounded by an imaginary sphere of radius r.

2. The total electric flux for the entire area is given by,
\(\Phi_{\mathrm{E}}=\oint \mathrm{E} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\oint \mathrm{EdA} \cos \theta \ldots \ldots\) (1)

3. The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore the direction of the area element d\(\overrightarrow{\mathrm{A}}\) is along the electric field \(\overrightarrow{\mathrm{E}}\) and θ = 0°.
\(\Phi_{\mathrm{E}}=\oint \mathrm{E.dA}\) (since cos 0° = 1)
E is a uniform on the surface of the sphere

4. This equation is Gauss Law.

Question 11.
Obtain the expression for electric field due to an infinitely long charged wire.
Answer:

Electric field due to infinite long charged wire:

1. Consider an infinitely long straight wire having uniform linear charge density λ.

2. Let P be a point located at a perpendicular distance r for the wire.

3. The charged wire possesses a cylindrical symmetry.

4. Let us choose a cylindrical Gaussian surface of radius r and length L,

5. The total electric flux in this closed surface is calculated as follows.

for the curved surface. \(\overrightarrow{\mathrm{E}}\) is parallel to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{E}}\) .dA = EdA.
For the top and bottom surfaces,
\(\overrightarrow{\mathrm{E}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{E}}\)
\(\begin{aligned}
&\Phi_{\mathrm{E}}=\int_{\text {Curved }} \mathrm{EdA}=\frac{\mathrm{Q}_{\text {encl }}}{\varepsilon_{0}} \ldots \ldots . \text { (2) }\\
&\text { surface }
\end{aligned}\)

Since the electric field is constant, E is taken out of the integration and
Q_encl is given by Q_encl=λL.
E \(\int_{\begin{array}{l}
\text { Curved } \\
\text { surface }
\end{array}} \mathrm{d} \mathrm{A}=\frac{\lambda \mathrm{L}}{\varepsilon_{\mathrm{o}}}\) ……..(3)
Here,
Φ_E=∫_{curved surface} dA= total area of the curved surface = 2πrL
E.2πrL=\(=\frac{\lambda \mathrm{L}}{\varepsilon_{\mathrm{o}}}\)

6. The electric field due to the infinite charged wired depends on \(\frac{1}{\mathrm{r}}\)rather than \(\frac{1}{r^{2}}\) for a point charge.

7. Equation (5) indicates that the electric field is always along the perpendicular direction (r̂) to wire. If λ > 0 then \(\overrightarrow{\mathrm{E}}\) points perpendicular outward (r̂) from the wire and if λ < 0, then points perpendicular inward (-r̂).

Question 12.
Obtain the expression for electric field due to a charged infinite plane sheet.
Answer:

1. Consider an infinite plane sheet of charges with uniform surface charge density σ.
2. Let P be a point at a distance of r from the sheet.

Electric field due to changed infinite Planar sheet:

1. A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces are chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.
2. Applying Gauss law for this cylindrical surface.
   
3. n̂ is the outward unit vector normal to the plane. The electric field depends on the surface charge density and is independent of the distance r.
4. If σ > 0, the electric field at any point P is outward perpendicular to the plane.
5. If σ < 0, the electric field points inward perpendicularly to the plane.

Question 13.
Obtain the expression for electric field due to a uniformly charged spherical shell.
Answer:
Consider a uniformly charged spherical shell of radius R and total charge Q.
Case (a) At a point outside the shell (r > R)
1. Let us choose a point P outside the shell at a distance r from the center.

2. The charge is uniformly distributed on the surface of the sphere.

3. The electric field must point radially outward if Q > 0 and point radially inward if Q < 0.

4. A spherical Gaussian surface of radius r is chosen and the total charge enclosed by this Gaussian surface is Q.

5. By Gauss law.
\(\oint_{\text {Gaussian }} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\) ………..(1)

6. The magnitude of is also the same at all points due to the spherical symmetry of the charge distribution.
\(\mathrm{E} \oint_{\text {Gaussian }} \mathrm{d} \mathrm{A}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\) ………….(2)

7. The total area of the Gaussian surface,
∫dA = 4πr²
E.4πr²=\(\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\)
For points outside the sphere, a large spherical Gaussian surface is drawn concentric with the sphere.

8. For points outside the sphere, a spherical gaussian surface smaller than the sphere is drawn

Case (c):
The electric field due to a charged spherical cell

• The electric field is radially outward if Q > 0 and radially inward if Q < 0.

Case (b):
At a point on the surface of the spherical shell (r=R)

• The electrical field at points on the spherical shell (r=R) is given by
  
  Electric fields versus distance Gauss spherical shell of radius R

Case (c):
At point inside the spherical shell (r < R)

1. Consider a point P inside the shell at a distance r from the center.
2. A Gaussian sphere of radius r is constructed.
3. Applying Gauss law,
   \(\begin{aligned}
   &\oint_{\text {Gaussian }} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\\
   &\mathrm{E} .4 \pi \mathrm{r}^{2}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}
   \end{aligned}\) …………..(5)
4. Since Gaussian surface encloses no charge, Q = 0.
   E = 0 ……………(6)
   Electric field versus distance for a spherical shell of radius R
5. The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.
6. A graph can be plotted between the electric field and radial distance.

Question 14.
Discuss the various properties of conductors in electrostatic equilibrium,
i) The electric field is zero everywhere, inside the conductor. This is true regardless of whether the conductor is solid or hollow.
Answer:

1. When we apply a uniform electric field, the free electrons accelerated.
2. It causes the left side to be negatively charged and the right side to be positively charged.
3. Due to this realignment of free electrons, there will be an internal electric field created inside the conductor.
4. It increases until it nullifies the external electric field and is said to be in electrostatic equilibrium in the order of
   10^-16 s.
   

ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors.

1. Consider an arbitrarily shaped conductor and a Gaussian surface is drawn inside the conductor.
2. It is very close to the surface of the conductor.
3. Since the electric field is zero inside the conductor, the net electric flux is zero.
4. So, there is no net charge inside the conductor.
5. No net charge inside the conductor

iii) The electric field outside the conductor is perpendicular to the surface of the conductor and has a magnitude of σ/ε. Where σ is the surface charge density at that point.

1. If the electric field has components parallel to the surface of the conductor, then free electrons of the surface of the conductor ‘would experience acceleration.
   
   (a) Electric field is along the surface
   (b) Electric field is perpendicular to the surface of the conductor
2. At electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor.
3. Consider a small cylindrical gaussian surface. 0ne half of this cylinder is embedded inside the conductor.
   
4. Since the electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux.
5. Inside the conductor, the electric field is zero.
6. The bottom flat part of the Gaussian surface has no electric flux.
7. The top flat part of the Gaussian surface contributes to the electric flux.
8. The electric field is parallel to the area vector and the total charge inside the surface is σA.
9. Applying Gauss law, EA = \(\frac{\sigma \mathrm{A}}{\varepsilon_{\mathrm{o}}}\)
10. In Vector form, \(\overrightarrow{\mathrm{E}}\)=\(\frac{\sigma}{\varepsilon_{\mathrm{o}}} \hat{\mathrm{n}}\)

iv) The electrostatic potential has the same value on the surface and inside of the conductor.

1. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface.
2. Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor.
3. At electrostatic equilibrium, the conductor is always at equipotential.

Question 15.
Explain the process of electrostatic induction.
Answer:
Definition: Charging without actual contact is called electrostatic induction.

Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. A negatively charged rod is brought near the conductor without touching. (Fig. a)

1. The negative charge of the rod repels the electrons in the conductor to the opposite side. As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side.
2. Before introducing the charged rod, the free electrons were distributed uniformly on the surface of the conductor and the net charge is zero. 0nce the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

Now the conducting sphere is connected to the ground through a conducting wire, This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere.
(Fig. b)

3. When the grounding wire is removed from the conductor, the positive charges remain near the charged rod.
(Fig. c)

4. Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor. By this process, the neutral conducting sphere becomes positively charged. (Fig. d)

Question 16.
Explain the dielectrics in detail and how an electric field is induced inside a dielectric.
Answer:

1. When an external electric field is applied to a conductor, the charges are aligned in such a way that an internal electric field is created which cancels the external electric field.
2. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced.
3. The magnitude of the internal electric field is smaller than that of the external electric field.
4. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with a magnitude less than that of the external electric field.
5. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor).
   a)

1. The uniform electric field between the plates acts as an external electric field ext which polarizes the dielectric placed between plates.
2. The positive charges are induced on one side surface and negative charges are induced on the other side of the surface.
3. But inside the dielectric, the net charge is zero even in a small volume.
4. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities
   +σ_b and -σ_b.
5. These charges are called bound charges. They are not free to move like free electrons in conductors.

Question 17.
0btain the expression for capacitance for a parallel plate capacitor.
Answer:

1. Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d.
2. The electric field between two infinite parallel plates is uniform and is given by
   \(\mathrm{E}=\frac{\sigma}{\varepsilon_{\mathrm{o}}}\)
   \(\sigma=\frac{Q}{A}\)
   σ = surface charge density on the plates
   
   
3. Capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates.
4. If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.
5. If the distance between the two plates is reduced, the potential difference between the plate decreases with the E constant.
6. The voltage difference between the terminals of the battery increases, leads to an additional flow of charge to the plates from the battery, till the voltage on the capacitor equals the battery’s terminal voltage.
7. If the distance is increased, the capacitor voltage increases and becomes greater than the battery voltage.
8. The charges flow from capacitor plates to the battery till both voltages become equal.

Question 18.
0btain the expression for energy stored in the parallel plate capacitor.
Answer:

1. Capacitor not only stores the charge but also stores energy.
2. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate.
3. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
4. To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
   dW = V dQ ………….(1)
   
5. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.
   \(\mathrm{U}_{\mathrm{E}}=\frac{1}{2}\left(\frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}\right)(\mathrm{Ed})^{2}=\frac{1}{2} \varepsilon_{\mathrm{o}}(\mathrm{Ad}) \mathrm{E}^{2}\) …………(4) where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined \(\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}\) ……….(5)
6. Energy is stored in the electric field existing between the plates of the capacitor. 0nce the capacitor is allowed to discharge, the energy is retrieved.
7. The energy density depends only on the electric field and not on the size of the plates of the capacitor.

Question 19.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Answer:

1. Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by distance d.
2. The capacitor is charged by a battery of voltage V₀ and the charge stored is Q₀
3. The capacitance of the capacitor without the dielectric is
   \(\mathrm{C}_{\mathrm{0}}=\frac{\mathrm{Q}_{\mathrm{0}}}{\mathrm{V}}\)

1. The modified electric field is,
   \(\mathrm{E}=\frac{\mathrm{E}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}} \ldots \ldots \ldots .(1)\)
2. Here E₀ is the electric field inside the capacitors when there is no dielectric and
   Er is the dielectric constant. Since εr >1, the electric E < E₀.
3. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q₀ will remain constant once the battery is disconnected.
4. Hence the new potential difference is,
   \(\mathrm{V}=\mathrm{Ed}=\frac{\mathrm{E}_{0}}{\varepsilon_{\mathrm{r}}} \mathrm{d}=\frac{\mathrm{V}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}}\) …………(2)
5. Thus new capacitance in the presence of a dielectric is,
   \(\mathrm{C}=\frac{\mathrm{Q}_{\mathrm{o}}}{\mathrm{V}}=\varepsilon_{\mathrm{r}} \frac{\mathrm{Q}_{\mathrm{o}}}{\mathrm{V}_{\mathrm{o}}}=\varepsilon_{\mathrm{r}} \mathrm{C}_{\mathrm{o}}\) ……………..(3)
   Since ε_r > 1, we have C > C₀ . Thus insertion of the dielectric constant ε_r increases the capacitance.
   \(\mathrm{C}=\frac{\varepsilon_{\mathrm{r}} \varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}=\frac{\varepsilon \mathrm{A}}{\mathrm{d}}\) ……………(4)
   where ε = ε_r ε₀ is the permititivity of the dielectric medium.
6. The energy stored in the capacitor before the insertion of a dielectric is given by
   \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}\) ………(5)
7. After the dielectric is inserted, Q₀ remains constant but the capacitance is increased.
8. The stored energy is decreased.
9. \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\varepsilon_{\mathrm{r}} \mathrm{C}_{0}}=\frac{\mathrm{U}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}}\) ………………..(6)
   Since ε_r > 1, we get U > U₀.

ii) When the battery remains connected to the capacitor:

1. The battery of voltage V₀ remains connected to the capacitor when the dielectric is inserted into the capacitor.
2. The potential difference V₀ across the plates remains constant.
3. The charge stored in the capacitor is increased by a factor ε_r.
   
4. The energy stored in the capacitor before the insertion of a dielectric is given by \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \mathrm{C}_{\mathrm{o}} \mathrm{V}_{\mathrm{o}}^{2}\) ………..(10)
5. After the dielectric is inserted, the capacitance is increased and the stored energy is also increased
6. \(\mathrm{U}=\frac{1}{2} \mathrm{CV}_{\mathrm{o}}^{2}=\frac{1}{2} \varepsilon_{\mathrm{r}} \mathrm{C}_{\mathrm{o}} \mathrm{V}_{\mathrm{o}}^{2}=\varepsilon_{\mathrm{r}} \mathrm{U}_{\mathrm{o}}\) …………(11)
   Since ε_r>1 we have U > U₀.
7. Since the voltage between capacitor V, is constant, the electric field between the plates also remains constant.
8. The energy density is given by
   \(\mathrm{U}=\frac{1}{2} \varepsilon \mathrm{E}_{0}^{2}\) (12)
   where ε is the permittivity of the given dielectric material.

Question 20.
Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.
Answer:
i) Capacitor in series

1. Consider three capacitors of capacitance C₁, C_2, and C₃
2. connected in series with a battery of voltage V.
   
3. By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different so that the voltage across each capacitor is also different and are denoted as V₁, V₂ and V₃ respectively.
4. The total voltage across each capacitor must be equal to the voltage of the battery.
   V = V₁ + V₂ + V₃
5. Since, Q = CV, we have \(\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}+\frac{\mathrm{Q}}{\mathrm{C}_{2}}+\frac{\mathrm{Q}}{\mathrm{C}_{3}}\)
   \(\mathrm{V}=\mathrm{Q}\left(\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\right)\)
   \(\frac{\mathrm{Q}}{\mathrm{C}_{\mathrm{s}}}=\mathrm{Q}\left(\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\right), \frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\)
6. The inverse of the equivalent capacitance C_s of three capacitors connected in series is equal to the sum of the inverses of each capacitance.
7. The equivalent capacitance C_s is always less than the smallest individual capacitance in the series.

ii) Capacitance in parallel:

1. Consider three capacitors of capacitance C₁, C_2, and C₃ connected in parallel with a battery of voltage V.
2. The potential difference across each capacitor is the same.
   Total charge Q = Q₁ + Q₂ + Q₃
   Q₁, Q2, Q3 – Charge stored in C₁, C₂, C₃
   Now, since Q = CV, we have Q = C₁V + C₂V + C₃V
   
3. If these three capacitors are considered to form a single capacitance Cp which stores the total charge Q.
   Q = C_pV.
   C_pV= C₁V + C₂V + C₃V
   C_p = C₁ + C₂ + C₃
4. The equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitances.
5. The equivalent capacitance C_p in a parallel connection is always greater than the largest individual capacitance.

Question 21.
Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.
Answer:
1. Consider two conducting spheres A and B of radii r₁ and r₂ respectively connected to each other by a thin conducting wire.

2. If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is the same in both the spheres.

3. They are now uniformly charged and attain electrostatic equilibrium.

4. Let q₁ be the charge residing on the surface of sphere A and is the charge residing on the surface of sphere B such that, Q = q₁ + q₂.

5. The charges are distributed only on the surface and there is no net charge inside the conductor.

6. The electrostatic potential at the surface of the sphere A is given by
\(V_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{1}}\)

7. The electrostatic potential at the surface of the sphere B is given by 1 q₂
\(V_{B}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{r_{2}}\)
V_A = V_B

8. The charge density on the surface of sphere A is σ₁, and the charge density on the surface of sphere B is σ₂. This implies that q₁ = 4πr₁² σ₁ and

q₂ = 4πr₂² σ₂ ………..(3)
σ₁r₁ = σ₂r₂ ………..(4)
σr = constant ………..(5)

9. The surface charge density σ is inversely proportional to the radius of the sphere.
Corona discharge:

Consider a charged conductor of irregular shape.

10. The smaller the radius of curvature, the larger is the charge density. The end of the conductor which has larger curvature (smaller radius) has a large charge accumulation.

11. As a result, the electric field near this edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.

12. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Lightning conductor:
Uses to protect tall buildings from lightning strikes.

Principle:
Action at points or corona discharge.

Construction:

1. This device consists of a long thick copper rod passing from the top of the building to the ground.
2. The upper end of the rod has a sharp spike or a sharp needle.
3. The lower end of the rod is connected to the copper plate which is buried deep into the ground.

Working:

1. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike.
2. Since the induced charge density on a thin sharp spike is large, it results in a corona discharge.
3. This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud.
4. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the earth.
5. The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

Question 22.
Explain in detail the construction and working of a Van-de-Graaff generator.
Answer:
Principle:
Electrostatic induction and action at points.

Construction:

1. Hollow metallic sphere A is mounted on an insulating pillar.
2. Pulley B is mounted at the center of the sphere.
3. Pulley C is mounted near the bottom.
4. A silk belt moves over the pulleys.
5. Pulley C is driven continuously by an electric motor.
6. D and E are the comb-shaped conductors mounted near the pulleys.
   

Working:

1. 1. 10⁴ V is given to comb D.
   2. Near the comb, D air gets ionized due to the action of points.
   3. Negative charges move towards the needle and
   4. Positive charges stick to the belt, moves up, and reach the comb E. Due to the electrostatic induction.
   5. Comb E gets a negative charge and the sphere gets a positive charge.
   6. Comb E ionize the air.
   7. Hence descending belt will be. left uncharged.
   8. The machine transfers the positive charge to the sphere.
   9. Leakage of charges can be reduced by enclosing it in a gas-filled steel chamber at very high pressure.
   10. It produces a large potential 10⁷v.

Uses:
To accelerate positive ions (protons, deuteron) for the purpose of nuclear disintegration.

IV.Exercises:

Question 1.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Answer:

Question 2.
The total number of electrons in the human body is typically in the order of 10²⁸. Suppose, due to some reason, you and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of lm. Compare this with your weight. Assume the mass of each person is 60kg and use point charge approximation.
Answer:

Question 3.
Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Answer:


\(\mathrm{F}_{\text {resultent }}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Qq}}{\mathrm{R}^{2}}(1+\sqrt{2}) \hat{\mathrm{i}}\)

Question 4.
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon.
(a) Calculate the value of q required to balance the gravitational attraction between Earth and the Moon
(b) Suppose the distance between the Moon and Earth is halved, would the charge q change?
(Take m_E = 5.9 × 10²⁴ kg, M_m = 7.9 × 10²² (kg)
Answer:

Question 5.
Draw the free body diagram for the following charges as shown in figure (a), (b) and (c).
a)
Answer:

b)
Answer:

c)
Answer:

Question 6.
Consider an electron travelling with a speed y0 and entering into a uniform electric field which is perpendicular to as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity, and position as functions of time.

Answer:

Question 7.
A closed triangular box is kept in an electric field of magnitude E = 2 × 10^-3 NC^-1 as shown in the figure.

Calculate the electric flux through the
(a) vertical rectangular surface
Answer:

(b) slanted surface
Answer:

c) entire surface.
Answer:

Question 8.
The electrostatic potential is given as a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B. and D. Plot the electric field as a function of x for figure (b)

Answer:
Figure a:
\(\overrightarrow{\mathrm{E}}=-\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \hat{\mathrm{i}}\)
From 0 to 0.2 m,

Figure b:

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.

To create the spark, an electric field of magnitude 3 × 10⁶ Vm^-1 is required.
Question (a)
What potential difference must be applied to produce the spark?
Answer:
V = E × d
= 3 × 10⁶ × 0.6 × 10^-3
= 1800 V

Question (b)
If the gap is increased, does the potential difference increase, decrease, or remains the same?
Answer:
If the distance between the plate increased, then its capacitance will decrease which gives rise to an increase in potential

Question (c)
Find the potential difference if the gap is 1 mm.
Answer:
V = E × d
= 3 × 10⁶ × 1 × 10^-3
= 3000 V

Question 10.
A point charge of +10 μC is placed at a distance of 20 cm from another identical point charge of +10 μC. A point charge of -2 μC is moved from point a to b as shown in the figure. Calculate the change in potential energy of the system? Interpret your result.

Answer:

Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.

Question a.

Answer:

Question b.

Answer:

Question c.

Answer:

Question d.

Answer:

Question e.

Answer:

Question 12.
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure. (Take m_p =1.6 × 10^-27 kg, m_e =9.1 × 10^-31 kg and g= 10ms^-2)

Question (a)
Calculate the time of flight for both electron and proton
Answer:
E = \(\begin{array}{l}
\mathrm{V} \\
\hline \mathrm{d}
\end{array}[latex]
= [latex]\frac{5}{10^{-3}}\)
= 5 × 10³ Vm^-1

Question (b)
Suppose if a neutron is allowed to fall, what is the time of flight?
Answer:

Question (c)
Among the three, which one will reach the bottom first?
Answer:
Hence electron will reach the bottom first.

Question 13.
During a thunderstorm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 10⁶ Vm^-1), lightning will occur.

Question (a)
If the bottom part of the cloud is 1000m above the ground, determine the electric potential difference that exists between the cloud and ground.
Answer:
\(\mathrm{E}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}\)
V = E.x
= 3 ×10⁶ ×10³
= 3 × 10⁹ V

Question (b)
In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:

Question 14.
For the given capacitor configuration
a) Find the charges on each capacitor
b) potential difference across them
c) energy stored in each capacitor.

Answer:

Question 15.
Capacitors P and Q have identical cross-sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant Er as shown in the figure. Calculate the capacitance of capacitors P and Q.

Answer:
1. The arrangement can be supposed to be a parallel combination of two capacitors each with plate area A/2 and separation. d. Total capacitance C_p = C_air + C_dielectric

2. The arrangement can be supposed to be a series combination of two capacitors, each with plate area A and separation d/2.

Part – II:

12th Physics Guide Electrostatics Additional Important Questions and Answers

I. Matching Type Questions:

Question 1.
Match Column I and Column II.

Answer:
(A) → (2)
(B) → (1)
(C) → (4)
(D) → (3)

Question 2.
Match Column I and Column II.

Answer:
(A) → (2)
(B) → (3)
(C) → (1)
(D) → (4)

Question 3.
Match the entries of Column I and Column II.

Answer:
(A) → (2)
(B) → (4)
(C) → (3)
(D) → (1)

Question 4.
When a dielectric slab is inserted between the plates of one of the two identical capacitors shown in the figure then match the following:

Answer:
(A) → (1)
(B) → (1)
(C) → (2)
(D) → (2)

II. Assertion – Reason Type Questions:

a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.
c) Assertion is correct, reason is incorrect
d) Assertion is incorrect, reason is correct.

Question 1.
Assertion:
Four-point charges q_1, q₂, q_3, and q₄ are as shown in figure. The flux over the shown Gaussian surface depends only on charges q₁ and q₂.
Reason:
Electric field at all points on Gaussian surface depends only on charges q₁ and q₂,

Answer:
(d) Assertion is incorrect, reason is correct.

Question 2.
Assertion:
On disturbing an electric dipole in stable equilibrium in an electric field, it returns back to its stable equilibrium orientation.
Reason :
A restoring torque acts on the dipole on being disturbed from its stable equilibrium.
Answer:
(a) Assertion is correct, reason is correct; reason is a correct explanation for the assertion.

Question 3.
Assertion:
Work done in moving a charge between any two points in an electric field is independent of the path followed by the charge, between these points.
Reason :
Electrostatic force is a non-conservative force.
Answer:
(c) Assertion is correct, reason is incorrect

Question 4.
Assertion:
Dielectric polarisation means the formation of positive and negative charges inside the dielectric.
Reason:
Free electrons are formed in this process.
Answer:
(c) Assertion is correct, reason is incorrect

III. Statement Type Questions:

Question 1.
Which of the following about potential difference between any two points is true?
I. It depends only on the initial and final position.
II. It is the work done per unit positive charge in moving from one point to other.
III. It is more for a positive charge of two units as compared to a positive charge of one
Answer:
I and II

Question 2.
Select the correct statements from the following.
I. The electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface.
II. Total flux linked with a closed body, not enclosing any charge will be zero.
III. Total electric flux, if a dipole is enclosed by a surface is zero.
Answer:
I, II, and III

Question 3.
An electric dipole of moment \(\overrightarrow{\mathrm{P}}\) is placed in a uniform electric field \(\overrightarrow{\mathrm{E}}\). Then
I. the torque on the dipole is \(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{E}}\).
II. the potential energy of the system is \(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{E}}\).
III. the resultant force on the dipole is zero.
Which of the above statements is/are correct.
Answer:
I and III

Question 4.
Select the incorrect statements from the following.
I. Polar molecules have permanent electric dipole moment.
II. CO₂ molecule is a polar molecule
III. H₂O is a non-polar molecules.
Answer:
II and III

IV. Choose the incorrect statement:

Question 1.
Which of the following statements is incorrect?
I. The charge q on a body is always given by q=ne, where n is any integer positive or negative.
II. By convention, the charge on an electron is taken to be negative charge.
III. The fact that electric charge is always an integral multiple of e is termed as quantisation of charge.
IV. The quantisation of charge was experimently demonstrated by Newton in 1912.
(a) only I
(b) only II
(c) only lV
(d) only III
Answer:
(c) only IV

Question 2.
The energy stored in a parallel plate capacitor is given by \(\mathrm{V}_{\mathrm{E}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\). Now which of the following statements is not true?
I. The work done in charging a capacitor is stored in the form of electrostatic potential energy \(\mathrm{V}_{\mathrm{E}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\)
II. The net charge on the capacitor is Q.
III. The magnitude of the net charge on one plate of a capacitor is Q.
(a) I only
(b) II only
(c) I and II
d) I, II and III
Answer:
(b) II only

Question 3.
Which of the following is not true ?
I. For a point charge, electrostatic potential varies as 1/r .
II. For a dipole, the potential depends on the magnitude of position vector, and dipole moment vector.
III. The electric potential varies as is at large distance.
IV. For a point charge, the electrostatic field varies as \( \frac{1}{\mathrm{r}^{2}}\).
(a) I only
(b) II only
(c) III only
(d) I, II and III
Answer:
(c) III only

V. Diagram – Type Question:

Question 1.
The figure shows a charge +q at point P held in equilibrium in air with the help of four + q charges situated at the vertices of a square. The net electrostatic force on q is given by

(a) Gauss’s law
(b) Coulomb’s law
(c) Principle of superposition
(d) net electric flux out the position of +q.
Answer:
(c) Principle of superposition
Solution:
The weight mg of the charge hole in air is in equilibrium with the net electrostatic force exerted by the four charges situated at the corners. The net electrostatic force is given by the charges at the corners. This is the principle of superposition.

Question 2.
Which of the following graphs shows the correct variation of force when the distance r between two charges varies?


Answer:

Solution:
From Coulombs’ law \(\mathrm{F}=\frac{\mathrm{Kq}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) i.e., \(\mathrm{F} \propto \frac{1}{\mathrm{r}^{2}}\) which is correctly shown by graph (d).

Question 3.
Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that

(a) E_A > E_B > E_C
(b)E_A = E_B = E_C
(c) E_A > E_C > E_B
(d) E_A > E_C > E_B
Answer:
(c) ) E_A > E_C > E_B

VI. Choose the odd man out:

Question 1.
(a) lightning arrestor
(b) Van-de-Graf generator
(c) Photocopying
(d) AC generator
Answer:
(d) AC generator

Question 2.
(a) Gold
(b) Silver
(c) electric potential
(c) Aluminium
(d) ebonite
Answer:
(d) ebonite

Question 3.
(a) Farad
(b) \(\frac{J}{C} m^{-1}\)
(c) Vm^-1
(d) NC^-1
Answer:
(a) Farad

Question 4.
(a) electric field
(b) electric force
(c) electric potential
(d) electric dipole moment
Answer:
(c) electric potential

VII. Choose The Incorrect Pair:
Question 1.

Answer:
C) Electric flux – Vector product

Question 2.

Answer:
A) Coulomb’s law – Force is directly proportional to square of distance

VIII. Choose the correct pair:

Question 1.

Answer:
D) Electric potential due to a uniform charged solid non-conducting sphere – Varies inversely of radius

Question 2.

Answer:
A) Inside a conductor placed in an external electric field – Electric field = 0

IX. Fill In the blanks:

Question 1.
The charge acquired by 5 ×  10¹⁰ electrons _______.
Answer:
q = ne
q = 5 × 10¹⁰ × l.6 × 10^-19 = 8 × 10^-9 C

Question 2.
The unit of permittivity is…………………
Answer:
C²N^-1m^-2

Question 3.
At sharp points, the charge density is ________.
Answer:
maximum

Question 4.
The charges in an electrostatic field are analogous to ………………… in a gravitational field.
Answer:
mass

X. Choose the best answer:

Question 1.
If electric field in a region is radially outward with magnitude E = Ar, the charge contained in a sphere of radius r centred at the origin is
(a) \(\frac{1}{4 \pi \varepsilon_{o}} A r^{3}\)
(b) \(4 \pi \varepsilon_{\mathrm{o}} \mathrm{Ar}^{3}\)
(c) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{A}{r^{3}}\)
(d) \(\frac{4 \pi \varepsilon_{0}}{\mathrm{r}^{3}}\)
Answer:
b) \(4 \pi \varepsilon_{\mathrm{o}} \mathrm{Ar}^{3}\)
Solution:

Question 2.
The electric field intensity just sufficient to balance the earth’s gravitational attraction on an electron will be: (given
(a) -5.6 × 10^-11N/C
(b) -4.8 × 10^-15N/C
(c) -1.6 × 10^-19N/C
(d) -3.2 × 10^-19N/C
Answer:
(a) -5.6 × 10^-11N/C
Solution:

Question 3.
The insulation property of air breaks down when the electric field is 3 × 10⁶ vm^-1 The maximum charge that can be given to a sphere of diameter 5m is approximately
(a) 2 × 10^-2 C
(b) 2 × 10^-3 C
(c) 2 × 10^-4 C
(d) 2 × 10^-5 C
Answer:
(b) 2 × 10^-3 C
Solution:

Question 4.
ABC is an equilateral triangle. Charges +q are placed at each corner as shown in fig. The electric intensity at centre 0 will be

(a) \(\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}}\)
(b) \(\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}}\)
(c) \(\frac{1}{4 \pi \epsilon_{o}} \frac{3 q}{r^{2}}\)
(d) zero
Answer:
(d) zero
Solution:
(Unit positive charge at 0 will be repelled equally by three charges at the three corners of the triangle. By symmetry, the resultant at 0 would be zero.)

Question 5.
A hollow insulated conduction sphere is given a positive charge of 10μC. What will be the electric field at the centre of the sphere if its radius is 2m?
(a) Zero
(b) 5μCm^-2
(c) 20μCm^-2
(d) 8μCm^-2
Answer:
(a) zero
Solution:
(Charge resides on the outer surface of a conducting hollow sphere of radius R. We consider a spherical surface of radius r< R.)

Question 6.
Five balls marked 1, 2, 3, 4, and 5 are suspended by separate threads. The pairs (1, 2) (2, 4) and (4, 1) show mutual attraction and the pairs (2, 3) and (4, 5) show repulsion. The nature of ball marked as 1 is
a) positive
b) negative
c) neutral
d) can’t determine
Answer:
c) neutral

Question 7.
The resultant capacitance of four plates, each is having an area A, arranged, as shown above, will be (plate separation is d)

(a) \(\frac{A \varepsilon_{0}}{d}\)
(b) \(\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{2 \mathrm{~d}}\)
(c) \(\frac{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}\)
(d) \(\frac{3 A \varepsilon_{o}}{d}\)
Answer:
(c) \(\frac{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}\)

Question 8.
At infinity, the electrostatic potential is
(a) Infinity
(b) maximum
(c) minimum
(d) zero
Answer:
(d) zero

Question 9.
The electric field at a point on the equatorial line of a dipole and direction of the dipole moment

(a) will be parallel
(b) will be in the opposite direction
(c) will be perpendicular
(d) are not related
Answer:
(b) will be in the opposite direction
Solution:
(The direction of the electric field at equatorial point A or B will be in opposite direction, like that of direction of dipole moment.)

Question 10.
Debye is the unit of
(a) electric flux
(b) electric dipole moment
(c) electric potential
(d) electric field intensity
Answer:
(b) electric dipole moment

Question 11.
In the given diagram a point charge +q is placed at the origin 0. Work done in taking another point charge -Q from point A to point B is:

(a) \(\frac{q Q}{4 \pi \epsilon_{0} a^{2}}\left(\frac{a}{\sqrt{2}}\right)\)
(b) zero
(c) \(\left[\frac{-\mathrm{q} \mathrm{Q}}{4 \pi \epsilon_{0}} \frac{1}{\mathrm{a}^{2}}\right] \sqrt{2} \mathrm{a}\)
(d) \(\left[\frac{\mathrm{qQ}}{4 \pi \epsilon_{0}} \frac{1}{\mathrm{a}^{2}}\right] \sqrt{2} \mathrm{a}\)
Answer:
(b) zero

Question 12.
The total electric flux emanating from a closed surface enclosing an α -particle is (e-electronic charge)
(a) \(\frac{2 e}{\varepsilon_{0}}\)
(b) \(\frac{e}{\varepsilon_{0}}\)
(c) \(\mathrm{e} \varepsilon_{\mathrm{o}}\)
(d) \(\frac{\varepsilon_{0} \mathrm{e}}{4}\)
Answer:
(a) \(\frac{2 e}{\varepsilon_{0}}\)
Solution:
(According to Gauss’s law total electric flux \(\frac{1}{\varepsilon_{0}}\) through a closed surface is time the total charge inside that surface.
Electric flux, Φ_E = \(\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}}\)
Charge on a-particle = 2e)
Φ_E = \(\frac{2 e}{\varepsilon_{o}}\)

Question 13.
A coil of the area of cross-section 0.5 m² with 10 turns is in a plane that is parallel to a uniform electric field of 100 N/C. The flux through the plane is?
(a) 100 V.m
(b) 500 V.m
(c) 20 V.m
(d) zero
Answer:
(b) 500 V.m

Question 14.
0n moving a charge of Q coulomb by χ cm, WJ of work is done, then the potential difference between the point is
(a) \(\frac{\mathrm{W}}{\mathrm{Q}} \mathrm{v}\)
(b) QWV
(c) \(\frac{Q}{W} V\)
(d) \(\frac{Q^{2}}{W} V\)
Answer:
(a) \(\frac{\mathrm{W}}{\mathrm{Q}} \mathrm{v}\)
Solution:
Potential difference between two points in an electric field is.
V_A – V_B = \(\frac{\mathrm{W}}{\mathrm{q}_{\mathrm{o}}}\)

Question 15.
The positive terminal of 12 V battery is connected to the ground. Then the negative terminal will be
(a) – 6 V
(b) +12 V
(c) zero
(d)- 12V
Answer:
(d) – 12V
Solution:
When negative terminal is grounded, the positive terminal of battery is at +12 V. When positive terminal is grounded, the negative terminal will be at -12 V.

Question 16.
The maximum electric field that a dielectric medium can withstand without break-down is called is
(a) permittivity
(b) dielectric constant
(c) electric susceptibility
(d) dielectric strength
Answer:
(d) dielectric strength

Question 17.
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E is
(a) \(\varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
(b) \(\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
(c) \(\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} / \mathrm{Ad}\)
(d) \(\varepsilon_{0} \mathrm{E}^{2} / \mathrm{Ad}\)
Answer:
(a) \(\varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
Solution:
(Energy required to charge the capacitor is W = u = QV)
∵ E = V/d

Question 18.
Energy is stored in a capacitor in the form of
(a) electrostatic energy
(b) magnetic energy
(c) light energy
(d) heat energy
Answer:
(a) electrostatic energy

Question 19.
Dimension and unit of Electric flux is
(a) ML²T^-3 A^-2, Nm² C^-1
(b) ML³ T^-3 A^-1, Nm² C^-1
(c) ML² T^-1 A^-2, Nm² C^-1
(d) ML^-4 T^-3 A^-2, Nm² C^-1
Answer:
b) ML³ T^-3 A^-1, Nm² C^-1

Question 20.
An air-core capacitor is charged by a battery. After disconnecting it from the battery, a dielectric slab is fully inserted in between its plates. Now, which of the following quantities remains constant?
(a) Energy
(b) voltage
(c) Electric field
(d) Charge
Answer:
(d) Charge

Question 21.
In a charged capacitor, the energy is stored in
(a) the negative charges
(b) the positive charges
(c) the field between the plates
(d) both (a) and (b)
Answer:
(c) the field between the plates

Question 22.
The potential gradient at which the dielectric of a condenser just gets punctured is called
(a) dielectric constant
(b) dielectric strength
(c) dielectric resistance
(d) dielectric number
Answer:
(b) dielectric strength

Question 23.
The unit of permittivity is
(a) C²N^-1m^-2
(b) Nm²C^-2
(c) Hm^-1
(d) NC^-2m^-2
Answer:
(a) C²N^-1

Question 24.
In the case of a Van de Graaff generator, the breakdown field of air is
(a) 2 × 10⁸ Vm^-1
(b) 3 × 10⁶ Vm^-1
(c) 2 × 10⁸ Vm^-1
(d) 2 × 10⁴ Vm^-1
Answer:
(b) 3 × 10⁶ Vm^-1

Question 25.
Van de Graaff generator is used to
(a) store electrical energy
(b) build up the high voltage of a few million volts
(c) decelerate charged particle-like electrons
(d) both (a) and (b)
Answer:
(b) build up the high voltage of a few million volts

XI. Two mark questions:

Question 1.
What is meant by triboelectric charging?
Answer:
Charging the objects through rubbing is called triboelectric charging.

Question 2.
When does an object is said to be electrically neutral?
Answer:
If the net charge is zero in an object, It is said to be electrically neutral.

Question 3.
State Gauss’s Law?
Answer:
Definition:
Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux OE through the closed surface is Φ_E = \(\oint { \vec { E } } \) .d\(\vec { A } \) = \(\frac {{ q }_{encl}}{{ ε }_{0}}\)

Question 4.
State coulomb’s law.
Answer:
Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Question 5.
What is meant by dielectric?
Answer:
A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass, and mica are some examples of dielectrics.

Question 6.
What is meant by the superposition principle?
Answer:
The total force on a given charge is the vector sum of the forces exerted on it due to all other charges. The force on q, due to charges q < ₂, q < ₃ ………… q_n

Question 7.
What are polar molecules? Give examples.
Answer:
In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment.
The net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H₂O, N₂O, HCl, NH₃.

Question 8.
Explain the working of a microwave oven?
Answer:

1. It works on the principle of torque acting on an electric dipole.
2. The food we consume has water molecules which are permanent electric dipoles.
3. Oven produces microwaves that are oscillating electromagnetic fields and produce a torque on the water molecules.
   
4. Due to this torque on each water molecule, the molecules rotate very fast and produce thermal energy.
5. Thus, the heat generated is used to cook the food.

Question 9.
What are permittivity and relative permittivity? How are they related?
Answer:

• Permittivity of a medium = permittivity of free space × relative permittivity, ε = ε₀ ε_r
  Its unit is C² N^-1 m^-2
• Relative permitivity = \(\frac{\text { permitivity of a medium }}{\text { permitivity of vacuum }}=\varepsilon_{\mathrm{r}}=\frac{\varepsilon}{\varepsilon_{\mathrm{o}}}\)
• It has no unit.

Question 8.
What is a capacitor?
Answer:
The capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

Question 11.
State Gauss law?
Answer:
Gauss law states that the total flux of the electric field E over any closed surface is equal to l/£0 times of the net charge enclosed by the surface,
\(\phi_{\mathrm{E}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}}\)

Question 12.
What is meant by electrostatic shielding?
Answer:

• It is the process of isolating a certain region of space from external field.
• It is based on the fact that electric field inside a conductor is zero.

Question 13.
What is meant by electrostatic Induction?
Answer:

• The phenomenon of producing induced charges without any contact with another charge is known as electrostatic induction.
• Electrostatic induction is used in electrostatic machines like Van de Graaff generators and capacitors.

Question 14.
Define farad.
Answer:
A conductor has a capacitance of one farad; if a charge of 1 coulomb given to it, rises its potential by 1 volt.

Question 15.
What are dielectrics? Give examples.
Answer:
A dielectric is an insulating material in which all the electrons are tightly bound to the nucleus of the atom. There are no free electrons to carry current.
E.g:- Ebonite, mica, and oil.

Question 16.
Can two equipotential surfaces intersect? Give reason.
Answer:
Since the electric field is normal to the equipotential surface and also the potential difference between any two points on the surface is nullified, the intersection is not possible.

Question 17.
Distinguish between polar and non-polar molecule?
Answer:

Question 18.
What is meant by dielectric breakdown?
Answer:
When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric breakdown.

Question 19.
The electric field outside a conductor is perpendicular to its surface. Justify.
Answer:

• The electric field has components parallel to the surface of the conductor.
• The free electrons in the conductor would experience acceleration.
• The conductor is not in equilibrium. At electrostatic equilibrium, the electric field must be perpendicular to its surface.

Question 20.
What is called ‘fringing field’ is a capacitor? when does it ignore?
Answer:
For finite-sized plates, the electric field is not strictly uniform between the plates. At both edges, the electric field is bent outward. Under the condition (d²<< A), this effect can be ignored.

Question 21.
How does a capacitor is used in a Computer keyboard?
Answer:
When the key is pressed, the separation between the plates decreases leading to an increase in the capacitance. This is turn triggers the electronic circuits in the computer to identify which key is pressed.

Question 22.
What will happen to the charge, voltage, electric field, the capacitance of a dielectric placed capacitor when it is connected with a battery and then disconnected?
Answer:

Question 23.
Calculate the number of electrons in one Coulomb of negative charge.
Solution:
According to the quantisation of charge q = ne
Here q = 1C. So the number of electrons in 1 coulomb of charge is
\(\mathrm{N}=\frac{q}{e}=\frac{1 \mathrm{C}}{1.6 \mathrm{X} 10^{-19}}=6.25 \times 10^{18} \text { electrons }\)

Question 24.
Define electric potential energy of two point charges?
Answer:
It is equal to work done to assemble the charges or work done in bringing each charge or work done in bringing a charge from an infinite distance.

Question 25.
Why is it safer to be inside a car than standing under a tree during lightning?
Answer:

• The metal body of the car provides electrostatic shielding.
• The electric field inside the car is zero.
• During lightning, the electric discharge passes through the body of the car. So during lightning, it is safer to sit inside a car than on open ground or under a tree.

Question 26.
Define the electric potential energy of an electric dipole placed in an electric field.
Answer:
Electric potential energy of an electric dipole in an electrostatic field is the work done in rotating the dipole to the desired position in the field.

Question 27.
What happens if a polar molecule is placed in an electric field?
Answer:

1. When a polar molecule is placed in an electric field, the dipoles orient themselves in the direction of the electric field. Hence a net dipole moment is produced.
2. The alignment of dipole moments in the direction of the applied electric field is called polarisation of electric polarisation.
3. The magnitude of the induced dipole moment p is directly proportional to the electric field E. The dipole moment, p = αE when α is called molecular polarisability.

Question 28.
Define volt.
Answer:

1. The unit of potential difference is volt.
2. The potential difference between the two points is I volt if 1 joule of work is done in moving 1 coulomb of charge from one point to another against the electric force.

Question 29.
What does an electric dipole experience when kept in a uniform electric field and non-uniform electric field?
Answer:
Uniform electric field:  When a dipole is kept in a uniform electric field at an angle θ, the net force is zero. It experiences a torque \(\vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}} \quad \tau=p \mathrm{E} \sin \theta\)

Non-uniform electric field: If the dipole is placed in a non-uniform electric field at an angle θ in addition to torque, it also experiences a force.

Question 30.
What is the Effective Capacitance of capacitors connected in series?
Answer:
When a number of capacitors are connected in series, the reciprocal of the effective capacitance is equal to the sum of reciprocal of the capacitances of the individual capacitors.

Question 31.
What is the effective capacitance of capacitors connected in parallel?
Answer:
The effective capacitance of the capacitors connected in parallel is equal to the sum of the capacitances of the individual capacitors.
C_p = C₁ + C₂ + ……….. + C_n

Question 32.
What are conductors and insulators? Give examples.
Answer:

1. Bodies which allow the charges to pass through are called conductors.
   e.g. Metals, human body, Earth, etc.
2. Bodies that do not allow the charge to pass through are called insulators.
   (e.g) Glass, mica, ebonite, plastic, etc.

Question 33.
What is meant by point charge?
Answer:
Any charge which occupies a space with dimensions less than its distance away from an observation point considered a point charge.

Question 34.
What is a capacitor?
Answer:
A capacitor is a device used to store electric charge and electrical energy.

Question 35.
Why does a balloon after rubbing stick to a wall?
Answer:
It is due to the polarisation of the wall due to the electric field due to a balloon.

Question 36.
How does the lightning conductor prevent the lightning stroke from the damage of the building?
Answer:

1. When a negatively charged cloud passes over the building, a positive charge will be induced on the pointed conductor.
2. The positively charged sharp points will ionize the air in the vicinity.
3. This will partly neutralise the negative charge of the cloud, thereby lowering the potential of the cloud.
4. The negative charges that are attracted to the conductor travel down to the earth. Thereby preventing the lightning stroke from the damage of the building.

Question 37.
Define the physical quantity whose unit is Vm, and state whether it is scalar or Vector.
Answer:

1. Electric flux has the unit Vm.
2. The number of electric field lines passing through a given area.
3. q = EA cosθ
4. It is a scalar quantity.
5. The other unit is Nm²c^-1.

XII. Three Mark Questions:

Question 1.
Give some important inference, over the expression of the electric field due to an electric dipole on its axial line and equational line.
Answer:
Important inferences
i) The magnitude of the electric field at points on the dipole axis is twice the magnitude of the electric field at points on the equatorial plane. The direction of the electric field at points on the dipole axis is directed along the direction of dipole moment vector \(\overrightarrow{\mathrm{p}}\) but at points on the equatorial plane it is directed opposite to the dipole moment vector, that is along \(\overrightarrow{\mathrm{p}}{-}\).

ii) At very large distances, the electric field due to a dipole varies as \(\frac{1}{\mathrm{r}^{3}}\) electric field due to a dipole at very large distances goes to zero faster than the electric field due to a point charge.

iii) The distance 2a approaches zero and q approaches infinity such that the product 2aq = p is finite, then the dipole is called a point dipole.

Question 2.
Mention some important points on an expression of electric potential due to point charge.
Answer:
Important points :
i) If the source charge q is positive, V > 0. If q is negative, then V is negative and equal to
\(\mathrm{V}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)
ii) The potential due to positive charge decreases as the distance increases, but for a negative charge the potential increases as the distance is increased. At infinity (r = 0) electrostatic potential is zero (V=0).
iii) The electric potential at a point P due to a collection of charges q₁, q₂, q₃,……..q_n is equal to sum of the electric potentials due to individual charges.

Question 3.
Mention some important points over the derivation of electric potential due to an electric dipole.
Answer:
Important points:
i) The potential due to an electric dipole falls as \(\frac{1}{r^{2}}\) and the potential due to a single point
charge falls as \(\frac{1}{\mathrm{r}}\). Thus the potential due to the dipole falls faster than that due to a monopole. As the distance increases from electric dipole, the effects of positive and negative charges nullify each other.

ii) The potential due to a point charge is spherically symmetric since it depends only on the distance r. But the potential due to a dipole is not spherically symmetric because the potential depends on the angle between \(\overrightarrow{\mathrm{p}}\) and position vector \(\overrightarrow{\mathrm{r}}\) of the point.

iii) However the dipole potential is axially symmetric. If the position vector \(\overrightarrow{\mathrm{r}}\) is rotated about \(\overrightarrow{\mathrm{p}}\) by keeping θ fixed, then all points on the cone at the same distance r will have the same potential

Question 4.
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let r be the radius of a small drop and R that of the large drop. Then, since the volume remains conserved,
\(\frac { 1 }{ 2 }\) πR² = \(\frac { 4 }{ 3 }\) πR³n
⇒ R³ = r³n
R = r³(n)^1/3
Further, since the total charge remains conserved, we have, using Q = CV
C_large V = n C_small v
Where V is the potential of the large drop.
4πε₀ RV = n (4πε₀r)v
V = \(\frac { nrv }{ R }\) = \(\frac { nrv }{{ r(n) }^{1/3}}\)
V = vn^2/3

Question 5.
Derive an expression for electric flux in a non-uniform electric field and an arbitrarily shaped area.
Answer:
1. Suppose the electric field is not uniform and the area is not flat, then the entire area is divided into n small area segments such that each area element is almost flat and the electric field through each area element is considered to be uniform.

2. The electric flux for the entire area A is approximately written as
\(\begin{array}{l}
\Delta \overrightarrow{\mathrm{A}}_{1}, \Delta \overrightarrow{\mathrm{A}}_{2}, \Delta \overrightarrow{\mathrm{A}}_{3} \ldots \ldots \Delta \overrightarrow{\mathrm{A}}_{\mathrm{n}} \\
\Phi_{\mathrm{E}}=\overrightarrow{\mathrm{E}}_{1} \cdot \Delta \overrightarrow{\mathrm{A}_{1}}+\overrightarrow{\mathrm{E}}_{2} \cdot \Delta \overrightarrow{\mathrm{A}}_{2}+\overrightarrow{\mathrm{E}}_{3} \cdot \Delta \overrightarrow{\mathrm{A}}_{3} \ldots . \overrightarrow{\mathrm{E}}_{\mathrm{n}} \cdot \Delta \overrightarrow{\mathrm{A}_{\mathrm{n}}} \\
=\sum_{\mathrm{i}=1}^{\mathrm{n}} \overrightarrow{\mathrm{E}}_{\mathrm{i}} \cdot \Delta \overrightarrow{\mathrm{A}_{\mathrm{i}}}
\end{array}\)

3. By taking the limit \(\Delta \overrightarrow{\mathrm{A}}_{i} \rightarrow 0\) (for all i) the summation in equation becomes integration. The total electric flux for the entire area is given by \(\Phi_{\mathrm{E}}=\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}\)

Electric flux for non-uniform electric Field:

1. The electric flux for a given surface depends on both the electric field pattern on the surface area and orientation of the surface with respect to the electric field.

Question 6.
The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow. Justify.
Answer:

• This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field.
•  As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside the conductor.
• We can also understand this fact by applying an external uniform electric field on the conductor.

Question 7.
The electrostatic potential has the same value on the surface and inside the conductor. Justify.
Answer:

• We know that the conductor has no parallel electric component on the surface without doing any work.
• This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface.
• Since the electric field inside the conductor is zero inside the conductor, the potential is same as the surface of the conductor.
• Thus at electrostatic equilibrium, the conductor is always at equipotential.

Question 8.
Explain Faraday Cage’s experiment? Faraday cage is an instrument to demonstrate the effect of electrostatic shielding.
Answer:

• It is made up of metal bars configured in the instrument.
•  If an artificial lightning bolt is created outside the person inside is not affected.
• This is because the metal bar provides electrostatic shielding.
• The Electric field inside becomes zero.
• The charges flow through the metal bar to the ground with no effect on the person inside.

Question 9.
What do you understand from the expression of capacitance in a parallel plate capacitor?
Answer:
Capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates.
1. If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.

2. If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant. As a result, the voltage difference between the terminals of the battery increases which in turn leads to an additional flow of charge to the plates from the battery, till the voltage on the capacitor equals to the battery’s terminal voltage. Suppose the distance is increased, the capacitor voltage increases and becomes greater than the battery voltage. Then, the charges flow from capacitor plates to the battery till both voltages become equal.

Question 10.
Mention the applications of capacitors?
Answer:

1. The flash which comes from the digital camera when we take photographs is due to the energy released from the capacitor, called a flash capacitor.
2. During cardiac arrest, a device called heart defibrillator is used to give a sudden surge of a large amount of electrical energy to the patient’s chest to retrieve the normal heart function.
3. Capacitors are used in the ignition system of automobile engines to eliminate sparking.
4. Capacitors are used to reduce power fluctuations in power supplies and to increase the efficiency of power transmission.

XIII – Five mark questions:

Question 1.
Explain the historical background of electric charges?
Answer:

• Two millenniums ago, Greeks noticed that amber after rubbing with animal fur attracted small pieces of leaves and dust.
• The amber possessing this property is said to be charged.
• A glass rod rubbed with a silk cloth, attracts a piece of paper. So glass rod also becomes ‘charged’ when rubbed with a suitable material.
• Consider a charged rubbed rod hanging from a thread.
• Suppose another charged rubber rod is brought near the first rubber rod; the rods repel each other.
• Now if we bring a charged glass rod close to the charged rubber rod, they attract each other.
• At the same time, if a charged glass rod is brought near another charged glass rod both the rods repel each other.

Inferences:
i) The charging of the rubber rod and that of the glass rod are different from one another.
ii) The charged rubber rod repels another charged rubber rod, which implies that Tike charges repel each other’.
iii) The charged amber rod attracts the charged glass rod, implying that the charge in the glass rod is not the same kind of charge present in the rubber. Thus unlike charges attract each other.
Therefore, two kinds of charges exist in the universe. Benjamin Franklin called one type of charge as positive (+) and another type of charge as negative (-). Based on Franklin’s convention, rubber and amber rods are negatively charged while the glass rod is positively charged. If the net charge is zero in the object, it is said to be electrically neutral.

iv) The atom is electrically neutral and is made up of negatively charged electrons, positively charged protons, and neutrons which have zero charges. When an object is rubbed with another object, some amount of charge is transferred from one object to another due to the friction between them, and the object is then said to be ‘electrically charged’. Charging the object through rubbing is called ‘triboelectric charging’.

Question 2.
Derive an expression for electric field due to a system of point charges.
Answer:
Suppose a number of point charges are distributed in space, to find the electric field at some point P due to this collection of point charges, superposition principle is used.

The electric field due to a collection of point charges at an arbitrary point is simply equal to the vector sum of the electric fields created by the individual point charges. This is called the superposition of electric fields.

Question 3.
Discuss the electric flux of a uniform electric field?
Answer:
Electric flux for uniform Electric field:

1. Consider a uniform electric field in a region of space. Let us choose an area A normal to the electric field lines.
Φ_E = EA ……….(1)

2. Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pass through area A. The electric flux for this case is zero.
Φ_E > E = 0 ………..(2)

3. If the area is inclined at an angle e with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux.

4. The electric field component parallel to the surface area will not contribute to the electric flux. For this case, the electric flux  Φ_E = (E cosθ) A …………..(3)

5. Further, e is also the angle between the electric field and the direction normal to the area, Hence in general, for a uniform electric field, the electric flux is defined as
Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = EA cosθ ………………(4)

6. \(\overrightarrow{\mathrm{A}}\) is the area vector \(\overrightarrow{\mathrm{A}}\) = A n̂. Its magnitude is simply the area A and the direction are along the unit vector n̂ perpendicular to the area.

7. Using this definition for flux,
Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\)
equations (2) and (3) can be obtained as special cases.
In Figure (a), θ=0° Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = EA
In Figure (b), θ = 90° Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = 0
Here \(\overrightarrow{\mathrm{A}}\) = An̂

Question 4.
Discuss some salient points about Gauss law?
Answer:
1. The total electric flux through the closed surface depends on the charges enclosed by the surface and the charges present outside the surface will not contribute to the flux and the shape of the closed surface which can be chosen arbitrarily.

2. The total electric flux is independent of the location of the charges inside the closed surface.

3. Chosen imaginary surface is called a Gaussian surface. The shape of the Gaussian surface to be chosen depends on the type of charge configuration and the kind of symmetry existing in that charge configuration.

4. The electric field E is due to charges present inside and outside the Gaussian surface but the charge Qencl denotes the charges which lie only inside the Gaussian surface.

5. The Gaussian surface cannot pass through any discrete charge but it can pass through continuous charge distributions.

6. Gauss law is another form of Coulomb’s law and it is also applicable to the charges in motion. Because of this reason, Gauss law is treated as much more general law than Coulomb’s law.

Question 5.
Derive an expression for electric field due to two parallel charged infinite sheet.
Answer:

1. Consider two infinitely large charged plane sheets with equal and opposite charge densities +σ and -σ which are placed parallel to each other.
2. The electric field between the plate and outside the plates is found using Gauss law.
3. The magnitude of the electric field due to an infinite charged plane sheet is \(\frac{\sigma}{2 \varepsilon}\) and it points perpendicularly outward if σ>0 and points inward if σ<0.
   
4. At the points P₁ and P₂, the electric field due to both plates are equal in magnitude and opposite in direction. As a result, electric field at a point outside the plate is zero. But inside the plate, electric fields are in the same direction
5. The direction of the electric field inside the plates is directed from positively charged plate to the negatively charged plate and is uniform everywhere inside the plate.
   \(\mathrm{E}_{\text {inside }}=\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}\)

Question 6.
Explain the principle of a capacitor?
Answer:

1. A capacitor is a device used to store electric charge and electric energy.
2. It consists of two conducting objects separated by some distance.
3. A simple capacitor consists of two parallel metal plates separated by a small distance.
4. When a capacitor is connected to a battery of potential difference V, the electrons are transferred from the battery to one plate and the other plate to the battery so that one plate becomes negatively charged with a charge of – Q and the other plate becomes positively charged with + Q.
5. The potential difference between the plates is equivalent to the battery’s terminal voltage. If the battery voltage is increased, the amount of charges stored in the plates also increase.
6. In general, the charge stored in the capacitor is proportional to the potential difference between the plates.
   Q α V, So that Q = CV, where C is the proportionality constant called capacitance.
   
7. The capacitance of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between conductors.
8. Unit:- Farad or Coulomb per volt. Farad is a very large unit of capacitance.
9. In Practice, we use micro & picofarad.
10. Total charge stored in the capacitor is zero when we say the capacitor stores charges, it means that amount of charge that can be stores charge, it means that amount of charge that can be stored in any one of the plates.
11. Nowadays it is available in different shapes and types. It is used in various kinds of electronic circuits.

Question 7.
Write the properties of lines of force:
Answer:

1. It starts from positive charge and ends at negative charge .
2. For a positive charge, the electric field lines point radially outward and for a negative charge, electric field lines point radially inward.
3. Tangent drawn at any point gives the direction of the electric field at the point.
4. The electric field lines are denser in a region where the electric field has a larger magnitude and less dense in a region of smaller magnitude.
5. They do not intersect
6. Field lines emanating from the positive charge are proportional to the magnitude of the charge.

XlV. Additional Problems (Two Marks):

Question 1.
An ele1tric dipole of charges 2 × l0^-10C separated by a distance 5 mm, is placed at an angle of 60 to a uniform field of 10 Vm^-1. Find the
i) magnitude and direction of the force acting on each charge and
ii) torque exerted by the field.
Data:
+q = 2 × 10^-10 C
-q = – 2 × 10^-10 C
θ = 60°
E = 10Vm^-1
2d = 5mm = 5 × 10^-3 m
i) Magnitude and directions of the force acts on each charge =?
ii) Torque exerted by the field =?
Solution:

1. Positive charge experience a force in the direction of the field.
2. Negative charge experiences a force opposite to the direction of the field.
   i) To find the force on each charge:
3. Force acts on positive charge: F = qE
   F = 2 × 10^-10 × 10
   F = 2 × l0^-9 × N in the direction of E
   Force acts on negative charge:
   F = qE = 2 × 10^-10 × 10
   F= 2 × l0^-9 N in the direction of E

ii) To find Torque exerted bj the field:

Question 2.
An electric dipole of charges 2 × 10^-6 C, – 2 × 10^-6 C is separated by a distance of 1 cm. Calculate the electric field due to dipole at a points
i) axial line 1 m from its centre and
ii) equatorial line 1 m from its centre.
Answer:


i) Electric field at a point on axial line = 360NC^-1
ii) Electric field at a point on equatorial line = 180NC^-1

Question 3.
Two charges +q and -3q are separated by a distance of lm. At what point in between the charges on its axis is the potential zero?
Answer:
Data :
q₁ = +q; q₂ = -3q; r = 1m
Solution:
Let the potential be zero at point 0 at a d . 1 from +q charge.


The potential is zero at a distance of 0.25m from the charge +q.

Question 4.
A parallel plate capacitor is maintained at some potential difference. A 3 mm thick slab is introduced between the plates. To maintain the plates at the same potential difference, the distance between the plates is increased by 2.4mm. Find the dielectric constant of the slab.
Data:
t = 3 mm; d’- d = 2.4 mm
V is same;
ε_r = ?
Solution:
Capacitance of air-filled parallel plate capacitor:

Since the potential are made the same capacitance in the above two cases is also the same.

The dielectric constant of the slab ε_r = 5

Question 5.
Three capacitors each of capacitance 9 pF are connected in series.
i) What is the total capacitance of the combination?
ii) What is the potential difference across each capacitor if the combination is connected to 120 V supply?
Answer:
Data:
C₁ = C₂ = C₃ = 9pF
V = 120V
C_s = ?
V₁, V₂, V₃ =?
To find effective capacitance:
Three capacitors are in series

∵ \(\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\)
C_s = 3pF
C_s = 3 × 10^-12 F

V₁, V₂, V₃ are the potential differences across the three capacitors as shown in the figure.

i) C_s = 3 × 10^-12 F
ii) V₁, V₂, V₃ = 40 V

XV. Five Mark Problems:

Question 1.
Four-point charges +q, +q, and -q is to be arranged respectively at the four corners of a square PQRS of side r. Find the work needed to assemble this arrangement.
Solution:

Here PQ = QS = SR= RP = r
\(\mathrm{PS}^{\prime}=\mathrm{RQ}=\mathrm{r} \sqrt{2}\)
W = Potential Energy of the system of four charges.
\(=\frac{(-q)(+q)}{4 \pi \varepsilon_{o} r}+\frac{(+q)(+q)}{4 \pi \varepsilon_{o} r}+\frac{(+q)(-q)}{4 \pi \varepsilon_{o} r}+\frac{(-q)(-q)}{4 \pi \varepsilon_{o} r}\)

Question 2.
Two small charged spheres repel each other with a force of 2 × 10^-3 N. The charge on one sphere is twice that of the other. When one of the charges is moved 10 cm away from the other, the force is 5 × 10^-4 N. Calculate the charges and the initial distance between them.
Answer:
Data:
F₁= 2 × 10^-3N when distance = r
F₂ = 5 × 10^-4N
when distance (i.e.) r’= (r+ 0. 1)rn
r = r + 10cm
q₁ = ?, q₂ = ?, r =?
Solution:
According to Coulomb’s law
\(\mathrm{F}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \cdot \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{2}}\)
Let q₁ = q and q₂ = 2q and distance between the charges be r experience a repulsive force F₁.


We know,
q₁ = q = 33.3 × 10^-9 C
q₂ = 2q = 2 × 33.3 × 10^-9 C
q₂ =66.6 × 10^-9 C
Initial distance r = 0.1m
Charges:
q₁ = 33.3 × 10^-9 C
q₂ = 66.6 × 10^-9 C

Question 3.
Two capacitors of unknown capacitance are connected in series and parallel. If net capacitances in two combinations are 6µF and 25µF respectively. Find their capacitances.
Answer:
Given:
Cs = 6µF
Cp = 25µF
\(\frac{1}{C_{S}}=\frac{1}{C_{1}}+\frac{1}{C_{2}} \quad \cdots \cdots\)

Question 4.
Three charges + 1 µC, + 3 µC, and – 5 µC are kept at the vertices of an equilateral triangle of sides 60 cm. Find the electrostatic potential energy of the system of charges.
Answer:


The electrostatic potential energy of the system of charges = – 0.255 J

Question 5.
Calculate the force between electron and proton in a Hydrogen atom.
(e = 1.6 × 10^-9 and r₀ = 0.53Å)
Solution:
The electron and proton attract each other. The force between these two particles is given by

Question 6.
Four point charges are placed at the four corners of a square in two ways (a) and (b) as shown in figure. Will the (i) electric potential and (ii) electric field, at the centre of the square be the same or different in the two configurations, and why?

Answer:
a) Electric potential is a scalar quantity. 0pposite charges cancel each other So, in both the configurations electric potential will be zero and the same.
b) Electric field is a vector quantity. It depends on positions of charge.
In the first diagram Electricfield,
\(E=2 \sqrt{2} \frac{\mathrm{kq}}{(\mathrm{r} / \sqrt{2})^{2}}\)
In the second diagram, E= 0
Because opposite charges are placed at the corner.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 12 Introduction to Probability Theory Ex 12.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.3

Question 1.
Can two events be mutually exclusive and independent simultaneously?
Answer:
When A and B are independent
P(A ∩ B) = P(A) P(B)
But when A and B are mutually
Exclusive P(A ∩ B) = 0

Question 2.
If A and B are two events such that
P ( A ∪ B ) = 0.7 , P (A ∩ B) = 0.2 , and P (B) = 0.5 , then show that A and B are independent.
Answer:
Given A and B are twp events such that
P(A ∪ B) = 0.7, P(A ∩ B) = 0.2 and P(B) = 0.5
To prove A and B are independent it is enough to prove
P(A ∩ B) = P(A) . P(B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.7 = P (A) + 0.5 – 0.2
0.7 = P(A) + 0.3
P(A) = 0.7 – 0.3 = 0.4
P(A) . P(B) = 0.4 × 0.5 = 0.20
= P(A ∩ B)
∴ P(A∩B) = P(A) . P(B)
∴ A and B are independent.

Question 3.
If A and B are two independent events such that P(A ∪ B) = 0.6, P(A) = 0.2, find p(B).
Answer:
Given A and B are independent.
⇒ P(A ∪ B) = P(A).P(B)
Here P(A ∪ B) = 0.6 and P(A) = 0.2
To find P(B):
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) P(A ∪ B) = P(A) + P(B) – P(A) . P(B)
(i.e.,) 0.6 = 0.2 + P(B) (1 – 0.2)
P(B) (0.8) = 0.4
⇒ P(B) = \(\frac{0.4}{0.8}=\frac{4}{8}=\frac{1}{2}\) = 0.5

Question 4.
If P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8, find P(A/B) and P(A ∪ B).
Answer:
Given P(A) = 0.5, P(B) = 0.8
and P(B/A) = 0.8

P(A ∩ B) = P(B/A) P(A)
Substituting in equation (1) we get

P(A/B) = 0.5
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ………. (2)

P(A ∩ B) = P(A/B) . P(B)
= 0.5 × 0.8
P(A ∩ B) = 0.40
(2) ⇒ P(A ∪ B) = 0.5 + 0.8 – 0.40
= 1.3 – 0.40
P(A ∪ B) = 0.90

Question 5.
If for two events A and B, P(A) = \(\frac{3}{4}\) P(B) = \(\frac{2}{5}\) and A ∪ B = S (sample space), find the conditional probability p (A/B).
Answer:

Question 6.
A problem in Mathematics is given to three students whose chances of solving it are \(\frac{1}{3}\), \(\frac{1}{4}\) and \(\frac{1}{5}\).
(i) What is the probability that the problem is solved?
Answer:
(i) What is the probability that the problem is solved?
Let A₁ denote the event of that first student solves the problem.
A₂ denote the event that second student solves the problem.
A₃ denote the event that third student solves the problem.
Given P(A₁) = \(\frac{1}{3}\), P(A₂) = \(\frac{1}{4}\) P(A₃) = \(\frac{1}{5}\)
We note that A₁, A₂, A₃ are independent events.
The problem will be solved if atleast one of them
solves it we have to find P(A₁ ∪ A₂ ∪ A₃)
Probability of at least one solves the problem = 1 – Probability of no one solving it
P(A₁ ∪ A₂ ∪ A₃) = 1 – P(A̅₁ ∪ A̅₂ ∪ A̅₃)
= 1 – P(A̅₁) . P(A₂) . P(A₃)
A₁, A₂, A₃ are independent then A̅₁, A̅₂, A̅₃ are also independent.
= 1 – [1 – p(A₁)] [1 – P(A₂)] [1 – P(A₃)]

(ii) What is the probability that exactly one of them will solve it?
Answer:
p(A̅₁) = 1 – P(A₁) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
p(A̅₁) = 1 – P(A₂) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
p(A̅₁) = 1 – P(A₃) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Probability of Exactly one student solving the problem

[Probability of exactly one student solving the problem = Probability [(A₁ solving the problem and A₂, A₃ non solving the problem) or (A₁, A₃ non solving and A₂ solving) or (A₁, A₂ solving and A₃ non solving)]

Question 7.
The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that the both the oil and filter need changing is 0.15.
(i) If the oil had to be changed, what is the probability that a new oil filter is needed?
(ii) If a new oil filter is needed, what is the probability that the oil has to be changed?
Answer:
Let A be the event of changing oil, B be the event of changing oil filter.
Given P(A) = 0.30, P(B) = 0.40, P(A ∩ B) = 0.15

(i) Probability of new oil filter B needed when the oil A changed is

(ii) Probability of oil A changed when new oil filter B is changed is

Question 8.
One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that
(i) both are white
(ii) both are black
(iii) one white and one black
Answer:
First Bag contains 5 white and 3 black balls Total number of balls in the first bag 8 Second Bag contains 4 white and 6 black halls Total number of balls in the second bag = 10
One ball is drawn from each bag.

(i) Probability of both are white:
P (getting both are white) = P (getting white ball from the first bag) × P (getting the white ball from the second bag)

(ii) Probability of both are black:
P (getting both are black) = P (getting black ball from the first bag) × P (getting the ball from the second bag)

(iii) Probability of one white and one black.
P (getting one white and one black) = P ( getting one white from the first bag or one white from the second bag) + P (getting one black from the first bag or one black from the second bag)

Question 9.
Two-thirds of students in a class are boys and the rest girls. It is known that the probability of a girl getting the first grade is 0.85 and that of boys is 0.70. Find the probability that a student chosen at random will get first-grade marks.
Answer:
Let G be the event of choosing a boy and G be the event of choosing a Girl.
Given P(B) = \(\frac{2}{3}\), P(G) = \(\frac{1}{3}\)
Let B₁ be the event of a boy getting first grade
P(B₁) = 0.70
Let G₁ be the event of a girl getting first grade
P(G₁) = 0.85
Probability of a. student getting a first grade = Probability of a boy getting first grade or Probability
of a Girl getting first grade
= P(B) × P(B₁) + P(G) × P(G₁)
= \(\frac{2}{3}\) × 0.70 + \(\frac{1}{3}\) × 0.85
= \(\frac{1.4+0.85}{3}\)
= \(\frac{2.25}{3}\) = 0.75

Question 10.
Given P(A) = 0.4 and P(A ∪ B) = 0.7 Find P(B) if
(i) A and B are mutually exclusive
(ii) A and B are independent events
(iii) P(A/B) = 0.4
(iv) P(B/A) = 0.5
Answer:
P(A) = 0.4, P(A ∪ B) = 0.7
(i) When A and B are mutually exclusive
P(A ∪ B) = P(A) P(B)
(i.e.,) 0.7 = 0.4 + P(B)
0.7 – 0.4 = P(B)
(i.e.,) P(B) = 0.3

(ii) Given A and B are independent
⇒ P(A ∩ B) = P(A). P(B)
Now, P(A ∪ B) = P(A) + P(B) – P (A ∩ B)
(i.e.,) 0.7 = 0.4 + P(B) – (0.4) (P(B))
(i.e.,) 0.7 – 0.4 = P(B) (1 – 0.4)
0.3 = P (B) 0.6
⇒ P(B) = \(\frac{0 \cdot 3}{0 \cdot 6}=\frac{3}{6}\) = 0.5

(iii) P(A/B) = 0.4
(i.e.,) \(\frac{P(A \cap B)}{P(B)}\) = 0.4
⇒ P(A ∩ B) = 0.4 [P(B)] …………. (i)
But We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
⇒ P(A ∩ B) = 0.4 + P(B) – 0.7
= P(B) – 0.3 …………. (ii)
from (i) and (ii) (equating R.H.S) We get
0.4 [P(B)] = P(B) – 0.3
0.3 = P(B) (1 – 0.4)
0.6 (P(B)) = 0.3 ⇒ P(B) = \(\frac{0.3}{06}=\frac{3}{6}\) = 0.5

(iv) P(B/A) = 0.5
(i.e.,) \(\frac{P(A \cap B)}{P(A)}\) = 0.5
(i.e.,) P(A ∩ B) = 0.5 × P(A)
= 0.5 × 0.4 = 0.2
Now P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.7 = 0.4 + P(B) – 0.2
⇒ 0.7 = P(B) + 0.2
⇒ P(B) = 0.7 – 0.2 = 0.5

Question 11.
A year is selected at random. What is the probability that
(i) it contains 53 Sundays
(ii) it is a leap year which contains 53 Sundays.
Answer:
Probability of year being a leap year = \(\frac{1}{4}\)
Probability of year being non – leap year = \(\frac{3}{4}\)

(i) It contains 53 Sundays:
A non – leap year has 365 days.
365 days = 52 weeks + 1 day.
52 weeks contain 52 Sundays.
In order to get 53 Sundays in a non – leap year the remaining I day must be a Sunday.
Remaining one day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday.
Probability of getting Sunday from the remaining one day = \(\frac{1}{7}\)
A leap year has 366 days.
366 days = 52 weeks + 2 odd days
52 weeks contain 52 Sundays.

In order to get 53 Sundays in a leap year the remaining 2 days must contain a Sunday. Remaining Two days may be

S = { (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday ), ( Friday, Saturday), (Saturday, Sunday) }
n(S) = 7

Let A be the event set of getting a Sunday then
A = { (Sunday, Monday), ( Saturday , Sunday) }
n(A) = 2

P (getting a Sunday from the remaining 2 days)

P(getting 53 Sundays in a year) = P(getting a leap year) × P(getting a Sunday from the remaining 2 days) + P(getting a non-leap year) × P(getting a Sunday from the remaining 1 day)

∴ Probability of getting 53 Sundays in a year = \(\frac{5}{28}\)

(ii) A leap year has 366 days
\(\frac{366}{7}\) = 52 weeks + 2 days
In 52 weeks, we get 52 Sundays.
From the remaining two days we should get one Sunday, the remaining two days can be any one of the following combinations.
Saturday and Sunday, Sunday and Monday, Monday and Tuesday, Tuesday and Wednes¬day, Wednesday and Thursday, Thursday and Friday, Friday and Saturday of the seven combinations two have Sundays.
∴ (Probability of getting a Sunday = \(\frac{2}{7}\)
Selecting a leap year = \(\frac{1}{4}\)
{∴ In every four consecutive years we get one leap year}
∴ Probability of getting 53 Sundays = \(\frac{2}{7} \times \frac{1}{4} \times \frac{1}{14}\)

Question 12.
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?
Answer:
Given
Probability X hitting the targent P (X) = \(\frac{3}{4}\)
Probability Y hitting the targent P (Y) = \(\frac{4}{5}\)
Probability Z hitting the targent P (Z) = \(\frac{2}{3}\)
P(X̅) = 1 – P(X) = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
P(Y̅) = 1 – P(Y) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
p(Z̅) = 1 – P(Z) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Probability hitting the target exactly by 2 hits

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 12 Introduction to Probability Theory Ex 12.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.2

Question 1.
If A and B are mutually exclusive events P(A) = \(\frac{3}{8}\) and P(B) = \(\frac{1}{8}\), then find
(i) P(A̅)
(ii) P(A ∪ B)
(iii) P(A̅ ∩ B)
(iv) P(A̅ ∪ B̅)
Answer:
(i) P(A̅)
P(A̅) = 1 – P(A)
= 1 – \(\frac{3}{8}\)
P(A̅) = \(\frac{8-3}{8}\) = \(\frac{5}{8}\)

(ii) P(A ∪ B)
P(A ∪ B) = P(A) + P(B)
since A and B are mutually exclusive events.

(iii) P(A̅ ∩ B)
P(A̅ ∩ B) = P(B) – P(A ∩ B)
Since A and B are mutually exclusive we have A ∩ B = Φ
∴ P(A ∩ B) = 0
∴ P(A̅ ∩ B) = p(B)
= \(\frac{1}{8}\)

(iv) P(A̅ ∪ B̅)
P(A̅ ∪ B̅) = P\((\overline{A \cup B})\)
= 1 – P(A ∩ B)
Since A and B are mutually exclusive we have A ∩ B = Φ
∴ P(A ∩ B) = 0
P(A̅ ∪ B̅) = 1 – 0 = 1

Question 2.
If A and B are two events associated with a random experiment for which P (A) = 0.35, P (A or B ) = 0.85 , and P (A and B) = 0.15 find (i) P (only B) (ii) P (B) (iii) P (only A)
Answer:
Given P (A) = 0.35 ,
P (A and B) = P(A ∩ B) = 0.15
P(A or B) = P(A ∪ B) = 0.85
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.85 = 0.35 + P (B) – 0.15
0.85 + 0.15 – 0.35 = P(B)
P(B) = 1 – 0.35
P(B) = 0.65

(i) P (only B)
P (only B) = P(A̅ ∩ B)
= P(B) – P(A ∩ B)
= 0.65 – 0.15
P (only B) = 0.50

(ii) P (B̅)
P (B̅) = 1 – P(B)
= 1 – 0.65
P (B̅) = 0.35

(iii) P (only A)
P (only A) = P(A ∩ B̅)
= P(A) – P(A ∩ B)
= 0.35 – 0.15
P (only A) = 0.20

Question 3.
A die is thrown twice. Let A be the event, ‘First die shows 5’ and B be the event , ‘second
die shows 5’. Find P(A ∪ B).
Answer:
A die is thrown twice. Let S be the sample space
s = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3 ,2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2) ,(5, 3 ) ,(5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
n(S) = 36
Let A be the event ‘First die shows 5’
A = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,0)}
n(A) = 6

Let B be the event ‘Second die shows 5’
B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
n(B) = 6

Question 4.
The probability of an event A occurring is 0.5 and B occurring is 0.3. If A and B are mutually exclusive events, then find the probability of
(i) P(A∪B)
(ii) P(A ∩ B̅)
(iii) P(A̅ ∩ B)
Answer:
P(A) = 0.5, P(B) = 0.3
Here A and B are mutually exclusive.
(i) P(A ∪ B) = P(A) + P(B)
= 0.5 + 0.3 = 0.8
(ii) P(A ∩ B) = P(A) + P(B) – P(A ∪ B) = 0.5 + 0.3 – 0.8
P(A ∩ B) = 0
P(A ∩ \(\overline{B}\)) = P(A) – P(A ∩ B) = 0.5 – 0 = 0.5
(iii) P(\(\overline{A}\) ∩ B) = P(B) – P(A ∩ B) = 0.3 – 0 = 0.3

Question 5.
A town has 2 fire engines operating independently. The probability that a fire engine is available when needed is 0.96.
(i) What is the probability that a fire engine is available when needed?
(ii) What is the probability that neither is available when needed?
Answer:
A be the event of availability of a fire B be the event of a fire engine when needed. Availability of a second fire engine when needed.
Given P(A) = 0.96, P(B) = 0.96
Then A̅ is the event of non-availability of the first fire engine and B̅ is the event of non-availability of second fire engine when needed.
P (A̅) = 1 – P(A)
= 1 – 0.96
= 0.04
Also P(B) = 0.04

(i) P(atleast one engine is available) = (1 – probability of no engine available)
= 1 – P(A’ ∩ B’)
= 1 – P (A’) P(B’)
= 1 – (0.04) (0.04)
= 1 – 0.0016
= 0.9984

(ii) P (A’ ∩ B’) = P (A’) P(B’)
= 0.04 × 0.04
= 0.0016

Question 6.
The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that
(i) it will get at least one of the two awards
(ii) it will get only one of the awards.
Answer:
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.48 + 0.36 – 0.2
= 0.64.

(ii) P (Getting only one award)
= P(A) – P(A ∩ B) + P(B) – P(A ∩ B)
= (0.48 – 0.2) + (0.36 – 0.2)
= 0.28 + 0.16
= 0.44.