Srinivas

Solving these TS 10th Class Maths Bits with Answers Chapter 13 Probability Bits for 10th Class will help students to build their problem-solving skills.

Probability Bits for 10th Class

Question 1.
A sample space consists of 80 elementary events that are equally likely. Probability of each of them is ………….
A) 1
B) 0
C) \(\frac{1}{80}\)
D) 80
Answer:
C) \(\frac{1}{80}\)

Question 2.
If I calculate the probability of an events as – 0.5, then
A) The probability of not happening is 0.5
B) The probability of happening is 0.5
C) The event is not going to happen
D) I made a mistake
Answer:
D) I made a mistake

Question 3.
On a multiple choice test, each question has 4 possible choices. If you make a random guess, probability that you are correct is …………..
A) \(\frac{1}{4}\)
B) 1
C) 0
D) 4
Answer:
A) \(\frac{1}{4}\)

Question 4.
A bag contains 6 red marbles, 3 blue marbles and 7 green marbles. If a marble is randomly selected from the bag, the probability that it is blue ……………
A) \(\frac{1}{6}\)
B) \(\frac{1}{3}\)
C) \(\frac{3}{16}\)
D) \(\frac{13}{16}\)
Answer:
C) \(\frac{3}{16}\)

Question 5.
If an individual is selected at random, prob ability that he has a birthday in July in 2012?
A) \(\frac{30}{365}\)
B) \(\frac{31}{365}\)
C) \(\frac{30}{366}\)
D) \(\frac{31}{366}\)
Answer:
D) \(\frac{31}{366}\)

Question 6.
When a card is picked up from a deck of cards, it should be either a red or a black card because these events are :
A) Mutually exclusive
B) Equally likely
C) Complementary
D) All of these
Answer:
C) Complementary

Question 7.
Probability of getting an even or odd number in throwing a dice is …………..
A) \(\frac{1}{2}\)
B) 1
C) 0
D) \(\frac{1}{4}\)
Answer:
B) 1

Question 8.
Probability of getting 7 on a 6 faced die when it is thrown is ……………
A) 1
B) 0
C) \(\frac{1}{6}\)
D) \(\frac{1}{7}\)
Answer:
B) 0

Question 9.
Among the following probability of an event E, P(E) = …………….
A) -0.5
B) 3
C) 0.2
D) 500%
Answer:
C) 0.2

Question 10.
Two unbiased coins are tossed simultaneously. Probability of getting atmost two heads ………….
A) \(\frac{1}{4}\)
B) \(\frac{1}{2}\)
C) 1
D) \(\frac{3}{4}\)
Answer:
D) \(\frac{3}{4}\)

Question 11.
A card is pulled from a deck of 52 cards. The probability of obtaining a club is
A) \(\frac{1}{3}\)
B) \(\frac{13}{26}\)
C) \(\frac{2}{11}\)
D) \(\frac{1}{4}\)
Answer:
D) \(\frac{1}{4}\)

Question 12.
If a coin is tossed, then the probability that a head turns up is
A) \(\frac{1}{2}\)
B) \(\frac{1}{4}\)
C) \(\frac{1}{3}\)
D) \(\frac{1}{6}\)
Answer:
A) \(\frac{1}{2}\)

Question 13.
If a die is rolled then the probability of getting an even number is
A) \(\frac{1}{6}\)
B) \(\frac{1}{3}\)
C) \(\frac{1}{2}\)
D) \(\frac{2}{5}\)
Answer:
C) \(\frac{1}{2}\)

Question 14.
If two dice are thrown simutlaneously, the probability of showing the same numbers on their faces is
A) \(\frac{1}{6}\)
B) \(\frac{1}{4}\)
C) \(\frac{1}{12}\)
D) \(\frac{1}{3}\)
Answer:
A) \(\frac{1}{6}\)

Question 15.
If a card is drawn from a deck of 52 cards the probability that it is a club card is
A) \(\frac{1}{52}\)
B) \(\frac{1}{4}\)
C) \(\frac{1}{13}\)
D) \(\frac{1}{26}\)
Answer:
B) \(\frac{1}{4}\)

Question 16.
A box contains pencils and pens. The probability of picking out a pen at random is 0.65. Then the probability of not picking a pen is
A) 0.45
B) 0.55
C) 0.65
D) 0.35
Answer:
D) 0.35

Question 17.
In a simultaneous toss of two coins, prob-ability of no tails is
A) \(\frac{1}{2}\)
B) \(\frac{1}{3}\)
C) \(\frac{1}{4}\)
D) \(\frac{3}{4}\)
Answer:
D) \(\frac{3}{4}\)

Question 18.
In a simultaneous toss of two coins, the probability of atleast one head is
A) \(\frac{1}{3}\)
B) \(\frac{2}{4}\)
C) \(\frac{3}{4}\)
D) \(\frac{1}{4}\)
Answer:
C) \(\frac{3}{4}\)

Question 19.
In a single throw of two dice, the probability of getting a total of 12 is
A) \(\frac{1}{18}\)
B) \(\frac{1}{36}\)
C) \(\frac{1}{9}\)
D) \(\frac{1}{12}\)
Answer:
B) \(\frac{1}{36}\)

Question 20.
In a single throw of two dice, the probability of getting a total of 11 is
A) \(\frac{1}{9}\)
B) \(\frac{1}{18}\)
C) \(\frac{1}{12}\)
D) \(\frac{35}{36}\)
Answer:
B) \(\frac{1}{18}\)

Question 21.
In a single throw of two dice, the probability of getting a doublet is
A) \(\frac{5}{6}\)
B) \(\frac{3}{11}\)
C) \(\frac{5}{12}\)
D) \(\frac{1}{6}\)
Answer:
D) \(\frac{1}{6}\)

Question 22.
In a single throw of two dice, the probability of getting distinct numbers is
A) \(\frac{5}{6}\)
B) \(\frac{5}{12}\)
C) \(\frac{5}{36}\)
D) \(\frac{4}{36}\)
Answer:
A) \(\frac{5}{6}\)

Question 23.
In a single throw of two dice, the probability of getting even doublet is
A) \(\frac{3}{13}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{15}\)
D) \(\frac{1}{18}\)
Answer:
B) \(\frac{1}{12}\)

Question 24.
When two dice are rolled, probability of getting odd doublet is
A) \(\frac{1}{12}\)
B) \(\frac{1}{18}\)
C) \(\frac{1}{9}\)
D) \(\frac{1}{6}\)
Answer:
A) \(\frac{1}{12}\)

Question 25.
Two dice are rolled, the probability of getting 6 as the product is
A) \(\frac{1}{18}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{9}\)
D) \(\frac{1}{6}\)
Answer:
C) \(\frac{1}{9}\)

Question 26.
A page is opened at a random from a book containing 90 pages. Then the probability of a page number is a perfect square is …………….
A) \(\frac{90}{90}\)
B) \(\frac{2}{90}\)
C) \(\frac{1}{90}\)
D) None
Answer:
C) \(\frac{1}{90}\)

Question 27.
The probability of picking a red king card from a well shuffled deck of playing cards is ……………. (A.P. June. ’15)
A) \(\frac{1}{3}\)
B) \(\frac{1}{26}\)
C) \(\frac{1}{2}\)
D) 1
Answer:
B) \(\frac{1}{26}\)

Question 28.
Getting a prime or composite number is a ………………. event (A.P. Mar. ’15)
A) mutually exclusive
B) equally likely
C) 0
D) none
Answer:
A) mutually exclusive

Question 29.
P(E) = 0.65 then P(\(\overline{\mathrm{E}}\)) = ……………. (T.S. Mar. ’15)
A) 0.25
B) 1
C) 0.35
D) 0
Answer:
C) 0.35

Question 30.
If P(E) = 0.82 then P(\(\overline{\mathrm{E}}\)) = ……………..
A) 0.18
B) 0.28
C) 0.38
D) P(E) = P(\(\overline{\mathrm{E}}\))
Answer:
A) 0.18

Question 31.
Let E, \(\overline{\mathrm{E}}\)E be the complimentary events, in a random experiment, then which of the following is true ? (T.S. Mar. ’16)
A) P(E) + P(\(\overline{\mathrm{E}}\)) = 2
B) P(E) + P(\(\overline{\mathrm{E}}\)) = 3
C) P(\(\overline{\mathrm{E}}\)) + P(E) = 1
D) P(E) + P(\(\overline{\mathrm{E}}\)) = 4
Answer:
C) P(\(\overline{\mathrm{E}}\)) + P(E) = 1

Question 32.
Two fair dice are rolled and the face values are added. The probability of getting an odd number greater than 8 is ……………..
A) \(\frac{2}{9}\)
B) \(\frac{1}{6}\)
C) \(\frac{1}{4}\)
D) \(\frac{1}{9}\)
Answer:
B) \(\frac{1}{6}\)

Question 33.
A jar contains 3 mangoes and x guavas. Two fruits are pulled from the jar without replacement. An expression that represents the probability one fruit is mango and the next fruit is guava is …………..

Answer:
B)

Question 34.
Three different greeting cards and their corresponding covers are randomly strewn about on a table. If Sita puts the greeting cards into the covers at random, the probability of correctly matching of all the greeting cards and covers is …………..
A) \(\frac{5}{6}\)
B) \(\frac{2}{3}\)
C) \(\frac{1}{6}\)
D) \(\frac{1}{9}\)
Answer:
C) \(\frac{1}{6}\)

Question 35.
If two dice are rolled at a time then the probability that the two faces show different numbers is
A) \(\frac{1}{6}\)
B) \(\frac{35}{36}\)
C) \(\frac{5}{6}\)
D) \(\frac{1}{36}\)
Answer:
C) \(\frac{5}{6}\)

Question 36.
The probability of getting a number less than 5 when a die is rolled is
A) \(\frac{4}{5}\)
B) \(\frac{2}{3}\)
C) \(\frac{3}{6}\)
D) \(\frac{1}{6}\)
Answer:
B) \(\frac{2}{3}\)

Question 37.
If a ball is drawn at random from a box containing 11 red balls, 6 white balls and 9 green balls then, the probability that the ball is not green is
A) \(\frac{9}{26}\)
B) \(\frac{17}{26}\)
C) \(\frac{11}{26}\)
D) \(\frac{6}{26}\)
Answer:
B) \(\frac{17}{26}\)

Question 38.
Which of the following are equally likely events ?
A) Getting a Head or Tail in tossing a coin.
B) In a throw of a die, getting prime or composite number.
C) Drawing a number card from 1 – 50, a number divisible by 6 or 8.
D) Picking a heart or black card from a deck of playing cards.
Answer:
A) Getting a Head or Tail in tossing a coin.

Question 39.
In a single throw of two dice, the probability of getting a total of 12 is
A) \(\frac{1}{18}\)
B) \(\frac{1}{36}\)
C) \(\frac{1}{9}\)
D) \(\frac{5}{6}\)
Answer:
B) \(\frac{1}{36}\)

Question 40.
Two dice are rolled, the probability of getting 6 as the product is
A) \(\frac{1}{18}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{9}\)
D) \(\frac{1}{6}\)
Answer:
C) \(\frac{1}{9}\)

Question 41.
The ‘event’ of getting a number less than or equal to 6 is a …………….
A) base event
B) possible event
C) element
D) sure event
Answer:
D) sure event

Question 42.
When a coin is tossed, the probability of getting a head is ………….
A) \(\frac{1}{2}\)
B) 2
C) -1
D) \(\frac{3}{2}\)
Answer:
A) \(\frac{1}{2}\)

Question 43.
From a deck of cards, a card is drawn at random then, the probability of getting a black face card is …………
A) \(\frac{9}{2}\)
B) \(\frac{1}{4}\)
C) \(\frac{3}{2}\)
D) \(\frac{3}{26}\)
Answer:
D) \(\frac{3}{26}\)

Question 44.
From a bag containing 6 red balls, 5 green balls and 3 blue balls, the probability of getting a green ball at random ……………
A) \(\frac{5}{14}\)
B) \(\frac{4}{5}\)
C) \(\frac{5}{4}\)
D) None
Answer:
A) \(\frac{5}{14}\)

Question 45.
There are 50 cards numbered from 1 to 50. A card is drawn at random, then the probability that the number on the card is divisible by 8 is …………….
A) \(\frac{25}{3}\)
B) \(\frac{3}{25}\)
C) \(\frac{19}{4}\)
D) None
Answer:
B) \(\frac{3}{25}\)

Question 46.
The probability of a certain event is ………..
A) 9
B) 7
C) 0
D) 1
Answer:
D) 1

Question 47.
Probability of an event lies between ………. and ……..
A) 0, 1
B) 2, 3
C) 7, 1
D) 4, 9
Answer:
A) 0, 1

Question 48.
P(E) + P(\(\overline{\mathrm{E}}\)) = ……………….
A) 0
B) 2
C) 1
D) None
Answer:
C) 1

Question 49.
In a box, there are 28 marbles of which x are green and the rest are white. If the probability of getting a green marble is \(\frac{2}{7}\), then number of green marbles = ………….
A) 8
B) 9
C) 10
D) 13
Answer:
A) 8

Question 50.
If E is an event whose probability is \(\frac{2}{5}\), then the probability of not E is …………
A) \(\frac{1}{2}\)
B) \(\frac{5}{3}\)
C) \(\frac{3}{5}\)
D) \(\frac{1}{3}\)
Answer:
C) \(\frac{3}{5}\)

Question 51.
If two dice are rolled simultaneously then the ‘sum’ with greatest possibility to happen is …………..
A) 71
B) 7
C) 3
D) None
Answer:
B) 7

Question 52.
The probability of raining in a day is …………..
A) \(\frac{-1}{2}\)
B) \(\frac{1}{2}\)
C) \(\frac{1}{4}\)
D) None
Answer:
B) \(\frac{1}{2}\)

Question 53.
If one side is chosen at random from the sides of a right triangle, then the probability that it is hypotenuse is ……………
A) 2
B) \(\frac{1}{2}\)
C) 3
D) \(\frac{1}{3}\)
Answer:
D) \(\frac{1}{3}\)

Question 54.
When a dice is thrown, the probability of getting neither a prime nor composite number is ……………
A) \(\frac{1}{3}\)
B) \(\frac{1}{2}\)
C) \(\frac{1}{6}\)
D) None
Answer:
C) \(\frac{1}{6}\)

Question 55.
Getting a Tail or Head ………………..
A) equally likely
B) unlikely
C) exclusive
D) None
Answer:
B) unlikely

Question 56.
Getting a prime (or) composite
A) mutually exclusive
B) likely !
C) 0
D) None
Answer:
D) None

Question 57.
Getting a red card (or) black card is ……………..
A) mutually exclusive
B) more likely
C) less likely
D) None
Answer:
A) mutually exclusive

Question 58.
P (Sure event) = ……………….
A) 1
B) 0
C) -1
D) 2
Answer:
A) 1

Question 59.
P (Impossible event) = ……………..
A) 4
B) 3
C) -1
D) 0
Answer:
D) 0

Question 60.
The probability of a face card from red cards is ……………….
A) \(\frac{3}{13}\)
B) \(\frac{13}{3}\)
C) \(\frac{2}{17}\)
D) None
Answer:
A) \(\frac{3}{13}\)

Question 61.
The probability of drawing a black king from the deck is ………………
A) \(\frac{1}{14}\)
B) \(\frac{1}{3}\)
C) \(\frac{1}{2}\)
D) \(\frac{1}{26}\)
Answer:
D) \(\frac{1}{26}\)

Question 62.
The probability of drawing a black card front he black cards is ………………
A) 3
B) 2
C) 0
D) 1
Answer:
D) 1

Question 63.
The probability of getting two tails when two coins are tossed is ……………..
A) \(\frac{1}{4}\)
B) \(\frac{1}{2}\)
C) \(\frac{2}{3}\)
D) None
Answer:
A) \(\frac{1}{4}\)

Question 64.
There are …………. cards in a pack of playing cards.
A) 19
B) 16
C) 52
D) 50
Answer:
C) 52

Question 65.
P(E) = 0.05 then P(\(\overline{\mathrm{E}}\)) = ……………
A) 1.35
B) 0.95
C) 9.5
D) 1.5
Answer:
B) 0.95

Question 66.
P(G) = \(\frac{4}{17}\), P(\(\overline{\mathrm{G}}\)) = …………..
A) \(\frac{13}{17}\)
B) \(\frac{3}{17}\)
C) \(\frac{7}{17}\)
D) \(\frac{1}{17}\)
Answer:
A) \(\frac{13}{17}\)

Question 67.
P(N) + P(\(\overline{\mathrm{N}}\)) = …………….
A) 0
B) 1
C) 3
D) 7
Answer:
B) 1

Question 68.
A baby is born the probability that it is a boy (or) girl is …………….
A) 1
B) \(\frac{-1}{2}\)
C) \(\frac{1}{3}\)
D) \(\frac{1}{2}\)
Answer:
D) \(\frac{1}{2}\)

Question 69.
P(E) + P (not E) = ………………
A) 1
B) 2
C) 3
D) None
Answer:
A) 1

Question 70.
Identify true statement. ( )
A) 0 < P(E) < 1
B) 0 < P(E) < 2
C) p < P(E)
D) None
Answer:
A) 0 < P(E) < 1

Question 71.
There are ……………… face cards.
A) 1
B) 2
C) 4
D) None
Answer:
D) None

Question 72.
Probability can never be ……………
A) 0
B) 1
C) 0.5
D) -2
Answer:
D) -2

Question 73.
A dice is tossed once then the probability of getting an even number or a multiple of 3 is
A) \(\frac{1}{2}\)
B) \(\frac{2}{3}\)
C) \(\frac{1}{4}\)
D) None
Answer:
B) \(\frac{2}{3}\)

Question 74.
The probability that a leap year has 53 Sundays is ……………….
A) \(\frac{2}{7}\)
B) \(\frac{3}{7}\)
C) \(\frac{1}{7}\)
D) \(\frac{21}{17}\)
Answer:
A) \(\frac{2}{7}\)

Question 75.
Two dice are thrown once together What is the probability of getting a doublet ?
A) \(\frac{1}{4}\)
B) \(\frac{1}{2}\)
C) \(\frac{1}{6}\)
D) None
Answer:
C) \(\frac{1}{6}\)

Question 76.
P(E) – 1 + P(\(\overline{\mathrm{E}}\)) = ………………
A) -2
B) 0
C) 9
D) 2
Answer:
B) 0

Question 77.
P(E) = 0.455 then P(\(\overline{\mathrm{E}}\)) = ……………….
A) 0.545
B) 0.145
C) 0.345
D) None
Answer:
A) 0.545

Question 78.
P(A¹) = …………………
A) Φ
B) A
C) 1 – P(A)
D) None
Answer:
C) 1 – P(A)

Question 79.
Karishma and Reshma are playing chess. The probability of winning Karishma is 0.59. Then probability of Reshma winnig the match is …………… (A.P. Mar. ’15)
A) 1
B) 0.46
C) 0.5
D) 0.41
Answer:
D) 0.41

Question 80.
Vineeta said that probability of impossible events is 1. Dhanalakshmi said that probability of sure event is ‘O’ and Sireesha said that probability of any event lies in between 0 and 1. In the above with whom will you agree? (A.P. Mar. ’15)
A) Vineetha
B) Dhanalakshmi
C) Sireesha
D) All the three
Answer:
C) Sireesha

Question 81.
From the figure the probability of getting blue ball is ……………. (A.P. Mar. ’15, ’16)

A) \(\frac{3}{5}\)
B) \(\frac{3}{3}\)
C) \(\frac{5}{5}\)
D) \(\frac{5}{3}\)
Answer:
A) \(\frac{3}{5}\)

These TS 10th Class Maths Chapter Wise Important Questions Chapter 11 Trigonometry given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Previous Exams Questions

Question 1.
If sin A = cos A then find the value of A. (A.P. Mar. ’15)
Solution:
sin A = cos A (given)
then A + A = 90 ⇒ 2A = 90
⇒ A = \(\frac{90}{2}\) = 45°
∴ A = 45°

Question 2.
If 4 sin² θ – 1 = 0 then find ‘θ’ (θ < 90) also, find the value of θ and the value of cos²θ + tan²θ (AP. Mar. ’15)
Solution:
Given, 4 Sin²θ – 1 = 0
4 Sin²θ = 1
Sin²θ = \(\frac{1}{4}\)
Sinθ = ± \(\sqrt{\frac{1}{4}}\) = ±\(\frac{1}{2}\)
Given θ is less than 90°
∴ Sin θ = \(\frac{1}{2}\)
sinθ = sin 30°
∴ θ = 30°
Cosθ = Cos 30° = \(\frac{\sqrt{3}}{2}\)
Tan θ = Tan 30° = \(\frac{1}{\sqrt{3}}\)
Cos² θ + Tan² θ = Cos² 30° + Tan² 30°
= \(\left(\frac{\sqrt{3}}{2}\right)^2\) + \(\left(\frac{1}{\sqrt{3}}\right)^2\)
= \(\frac{3}{4}\) + \(\frac{1}{3}\) = \(\frac{9+4}{12}\) = \(\frac{13}{12}\)

Question 3.
Show that tan²θ – \(\frac{1}{\cos ^2 \theta}\) = 1 (T.S. Mar. ’15)
Solution:
Method – I : Since \(\frac{1}{\cos ^2 \theta}\) = sec²θ

Question 4.
Find the value of (T.S. Mar. ’15)

Solution:

Question 5.
If tan θ = \(\sqrt{3}\) (θ is acute angle) then find the value of 1 + cos θ. (T.S. Mar. ’16)
Solution:
tan θ = \(\sqrt{3}\) = tan 60 (∵ θ is acute)
⇒ θ = 60
⇒ 1 + cos θ = 1 + cos 60 = 1 + \(\frac{1}{2}\) = \(\frac{3}{2}\)
∴ 1 + cos θ = \(\frac{3}{2}\)

Question 6.
Show that \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = sec θ – tan θ. (T.S. Mar. ’16)
Solution:
\(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = \(\sqrt{\frac{(1-\sin \theta)(1-\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}}\)

Hence proved

Question 7.
If tan (A + B) = 1, and cos (A – B) = \(\frac{\sqrt{3}}{2}\) 0° < A + B < 90, A > B then find values of A and B. (T.S. Mar. ’16)
Solution:
tan (A + B) = 1 = tan 45°
∴ A + B = 45° ……………. (1)
cos (A – B) = \(\frac{\sqrt{3}}{2}\) = cos 30°
⇒ A – B = 30° ……………… (2)
Solving the equation (1) and (2) we get

then A + B = 45
37.5 + B = 45 ⇒ B = 45 – 37.5 = 7.5
So, A = 37.5°, B = 7.5°

Question 8.
Find the value of tan² 60 + 4 cos² 45 + 3 sec² 30 + 5 cos² 90 = cosec 30 + sec 60 – cot² 30 (T.S. Mar. ’16)
Solution:
Put the following values in the given problem
tan 60° = \(\sqrt{3}\) , cos 45° = \(\sqrt{2}\), sec 30° = \(\frac{2}{\sqrt{3}}\)
cos 90° = 0, cosec 30° = 2, sec 60° = 2, cot 30° = \(\sqrt{3}\)
We get
tan² 60 + 4 cos² 45 + 3 sec² 30 + 5 cos² 90 = cosec 30 + sec 60 – cot² 30

Additional Questions

Question 1.
In a right angled triangle ABC, with right angle at B in which a = 5 units, b = 13 units and ∠BCA = θ, then find sin θ and tan θ.
Solution:
Given a = 5 units = BC
b = 13 units = CA or AC
∠BCA = θ
By pythagoras theorem
AC² = AB² + BC²
⇒ 13² = AB² + 5²
⇒ AB² = 13² – 5² = 169 – 25 = 144
⇒ AB = \(\sqrt{144}\) = 12 units
Now from the figure
Sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{13}\)
and Tan θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{5}\)

Question 2.
If Cos c = \(\frac{3}{5}\), then find Sin c and Tan c
Solution:
We have cos c = \(\frac{3}{5}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)

⇒ AC = 5, BC = 3
By Pythagoras theorem
AC² = AB² + BC²
⇒ 5² = AB² + 3²
⇒ 25 = AB² + 9 = 16
⇒ AB = \(\sqrt{16}\) = 4
From Sin c = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{4}{5}\), Tan c = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{4}{3}\)

Question 3.
If 12 Tan A = 9, then find Sin A and Cos A.
Solution:
Given 12 Tan A = 9

⇒ Tan A = \(\frac{9}{12}\) = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ BC = 9, AB = 12
By Pythagoras theorem
AC² = AB² + BC²
= 12² + 9²
= 144 + 81
AC² = 225
AC = \(\sqrt{225}\) = 15
Now, sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) = \(\frac{9}{5}\) and cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{15}\)
∴ sin A = \(\frac{9}{15}\) and cos A = \(\frac{12}{15}\)

Question 4.
If 5Cot A = 12, find Cos A and Cosec A.
Solution:
Given Cot A = \(\frac{12}{5}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
⇒ AB = 12, BC = 5
By Pythagoras theorem
AC² = AB² + BC²
= 12² + 5²
= 144 + 25
= AC² = 169
AC = \(\sqrt{169}\) = 13
From the figuref ∆ABC,
Cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{13}\), Cosec A = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{13}{5}\)
∴ Cos A = \(\frac{12}{13}\) and Cosec A = \(\frac{13}{5}\)

Question 5.
Evaluate the following.
i) Sin 60° + Cos 60°
Solution:
Sin 60° + Cos 60° = \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) = \(\frac{\sqrt{3}+1}{2}\)

ii) \(\frac{{Sin} 45^{\prime \prime}}{{Sin} 30^{\prime \prime}+{Cos} 60}\)
Solution:

iii) Tan² 30° + Cot² 45° – Cos 60
Solution:
Tan² 30° + Cot² 45° – Cos 60
= \(\left(\frac{1}{\sqrt{3}}\right)^2\) + (1)² – \(\frac{1}{2}\)
= \(\frac{1}{3}\) + \(\frac{1}{1}\) – \(\frac{1}{2}\)
= \(\frac{2+6-3}{6}\)
= \(\frac{5}{6}\)

iv) 2 Tan² 45° + Sin² 60° – Cos² 30°
Sol:
2 Tan² 45° + Sin² 60° – Cos² 30°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^2\) – \(\left(\frac{\sqrt{3}}{2}\right)^2\)
= 2 × 1 + \(\frac{3}{4}\) – \(\frac{3}{4}\) = 2

v) Cot² 30° + 4 Sin² 45° + 3 Cosec 60°
Solution:
Cot² 30° + 4 Sin² 45° + 3 Cosec 60°
= (\(\sqrt{3}\))² + 4\(\left(\frac{1}{\sqrt{2}}\right)^2\) + 3\(\left(\frac{2}{\sqrt{3}}\right)^2\)
= 3 + 4 × \(\frac{1}{2}\) + 3 × \(\frac{4}{3}\)
= 3 + 2 + 4
= 9

vi) \(\sqrt{2}\) Sin 45° + Cos 90° + Sin 90°
Solution:
\(\sqrt{2}\) .Sin 45° + Cos 90° + Sin 90°
= \(\sqrt{2}\) . \(\frac{1}{\sqrt{2}}\) + 0 + 1
= 1 + 1
= 2

Question 6.
Evaluate Cos 60°, Cos 30° – Sin 60° Sin 30°
What is the value of Cos (60° + 30°) ? What can you conclude ?
Solution:
Take Cos 60° Cos 30° – Sin 60°. Sin 30°
= \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\)
= 0 …………….. (1)
Now take Cos(60° + 30°) = Cos (90°) = 0 ………….. (2)
From equations (1) and (2), I conclude that
Cos(60° + 30°) = Cos 60° . Cos 30° – Sin 60°. Sin 30°
i.e.,Cos(A + B) = Cos A . Cos B – Sin A . Sin B.

Question 7.
Is it right to say Sin (60° – 30°) = Sin 60°. Cos 30° – Cos 60°. Sin 30° ?
Solution:
LHS = Sin(60° – 30°)
= Sin 30° = \(\frac{1}{2}\)
RHS = Sin 60°. Cos 30° – Cos 60°. Sin 30°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{-1}{4}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
Yes, it is right to say
Sin(60° – 30°) = Sin 60°. Cos 30° – Cos 60°. Sin 30°
i.e., Sin (A – B) = Sin A Cos B – Cos A. Sin B.

Question 8.
Is it right to say that Cos (A + B) = Cos A + Cos B ?
Solution:
Take A = 60°, B = 30°
Then Cos(A + B) = Cos (60° + 30°)
= Cos 90° = 0
Cos A + Cos B = Cos 60° + Cos 30°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{1+\sqrt{3}}{2}\)
∴ Cos (A + B) ≠ Cos A + Cos B
It is not right to say that Cos (A + B) ≠ Cos A + Cos B

Question 9.
i) \(\frac{{Sin} 66^{\prime \prime}}{{Cos} 24 “}\)
ii) \(\frac{{Sin} 18^{\prime \prime}}{{Cos} 72^{\prime \prime}}\)
iii) \(\frac{{Tan} 80^{\prime \prime}}{{Cot} 10^{\prime \prime}}\)
iv) \(\frac{{Sec} 69^{\prime \prime}}{{Cosec} 21 “}\)
v) \(\frac{{Tan} 54 “}{{Cot} 36^{\prime \prime}}\)
Solution:

Question 10.
Find the value of
i) Sin 75° – Cos 15°
ii) Sec 23° – Cosec 67°
iii) Sec 70° – Sec 20°
iv) Tan 68° – Tan 22°
Solution:
i) Given Sin 75° – Cos 15°
= Sin 75° – Cos(90 – 75°)
= Sin 75° – Sin 75°
= 0

ii) Given Sec 23° – Cosec 67°
= Sec 23° – Cosec (90 – 23°)
= Sec 23° – Sec 23°
= 0

iii) Sec 70° Sec 20°
= Sin 70° . \(\frac{1}{\cos 20 “}\) = \(\frac{{Sin} 70^{\prime \prime}}{{Cos}\left(90^{\prime \prime}-70^{\prime \prime}\right)}\)
= \(\frac{{Sin} 70 “}{{Cos}(90 “-70 “)}\) = \(\frac{{Sin} 70^{\prime \prime}}{{Sin} 70^{\prime \prime}}\) = 1

iv) Given Tan 68°. Tan 22°
Tan 68° . Tan(90° – 68°)
Tan 68° . Cot 68°
= \(\frac{{Tan} 68^{\prime \prime}}{{Tan} 68^{\prime \prime}}\) = 1

Question 11.
If Cot 2A = Tan (A – 18), when 2A is an acute angle, find the value of A.
Solution:
Given that Cot 2A = Tan (A – 18°)
⇒ Tan (90° – 2A) = Tan (A – 18°) [∵ Cot θ = Tan (90 – θ)]
⇒ 90 – 2A = A – 18°
⇒ 90 + 18° = A + 2A
⇒ 3A = 108°
⇒ A = \(\frac{108 “}{3}\) = 36°
Hence, the value of A is 36°.

Question 12.
If Cos 4A = Sin(A – 20), when 4A is an acute angle, find the value of A.
Solution:
Given that Cos 4A = Sin(A – 20°)
⇒ Sin(90° – 4A) = Sin(A – 20°)
[∴ Cos 0 = Sin(90 – 0)]
⇒ 90 – 4A = A – 20°
⇒ 90 + 20 = A + 4A
⇒ 110° = 5A
⇒ A = \(\frac{110 “}{15}\) = 22°
Hence, the value of A is 22°.

Question 13.
If Sin θ + Cosec θ = 2, find the value of Sin² θ + Cosec² θ
Solution:
Given Sin θ + Cosec θ = 2
Squaring on both sides
(Sin θ + Cosec)² = 2²
⇒ Sin² θ + Cosec² θ + 2 . Sin θ . Cosec θ = 4
⇒ Sin² θ + Cosec² θ + 2 . Sin θ . \(\frac{1}{{Sin} \theta}\) = 4
⇒ Sin² θ + Cosec² θ + 2 = 4
⇒ Sin² θ + Cosec² θ + 4 – 2
⇒ Sin² θ + Cosec² θ = 2

Question 14.
Show that \(\frac{{Tan} A+{Cot} B}{{Tan} B+{Cot} A}\) = Tan A . Cot B.
Solution:

= Cot B . Tan A
= Tan A . Cot B = RHS
LHS = RHS

Question 15.
Show that (Sin θ + Cos θ)² – (Sin θ – Cos θ)² = 4 Sin θ Cos θ
Solution:
LHS = (Sin θ + Cos θ)² – (Sin θ – Cos θ)²
= (Sin² θ + Cos² θ + 2 Sin θ . Cos θ) – (Sin² θ + Cos² θ – 2 Sin θ . Cos θ)
= (2 Sin θ + Cos θ) – (-2 Sin θ . Cos θ)
= 2 Sin θ + Cos θ + 2 Sin θ . Cos θ
= 4 Sin θ . Cos θ
= RHS
∴ LHS = RHS

Question 16.
Find the value of \(\frac{{Tan}^2 60^{\prime \prime}+4 {Sin}^2 45^{\prime \prime}+3 {Sec}^2 30^{\prime \prime}+10 {Cos}^2 90^{\prime \prime}}{{Cosec}^2 60^{\prime \prime}+{Cos} 60^{\prime \prime}-{Cot}^2 60^{\prime \prime}}\)
Solution:
Put the following values in the given problem.

Question 17.
Show that \(\frac{1}{{Sin} \theta}\) – Sin θ = Cot θ. Cos θ
Solution:

Question 18.
Show that \(\frac{{Tan}^2 q}{1+{Sec} q}=\frac{1-{Cos} q}{{Cos} q}\)
Solution:

∴ LHS = RHS

Question 19.
Evaluate : log₄ (1 + tan² 45°)²
Solution:
∴ log₄ (1 + 1)²
= log₄ = 1

Question 20.
Is it true to say that cos (60°+30°) = cos 60° cos 30° + sin 60° sin 30°
Solution:
Here L.H.S = cos (60° + 30°) = cos 90° = 0
R.H.S – cos 60° cos 30° + sin 60° sin 30°
= \(\frac{1}{2}\) × \(\frac{\sqrt{3}}{2}\) + \(\frac{\sqrt{3}}{2}\) × \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) + \(\frac{\sqrt{3}}{4}\)
= \(\frac{\sqrt{3}}{4}\)
= \(\frac{\sqrt{3}}{2}\)
Here L.H.S ≠ R.H.S, so it’s not justify.

Question 21.
In ∆ABC, ∠C = 90° If BC + CA = 17 cm; BC – CA = 7 cm. Find
(i) sin A
(ii) sin B
Solution:
Here, given ∠C = 90°; BC + CA = 7 cm;
BC – CA = 7 cm

BC = \(\frac{24}{2}\) = 12 cm; figure script pno.4
BC = 12 cm
We apply BC = 12 cm. In BC + CA = 17 cm
Then 12 + CA = 17
CA = 17 – 12 = 5 cm
We know from ∆ABC, AB² = BC² + CA²
AB = \(\sqrt{\mathrm{BC}^2+\mathrm{CA}^2}\)
AB = \(\sqrt{12^2+5^2}\) = \(\sqrt{144+5}\)
AB = \(\sqrt{169}\) = 13
We know = sin A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{12}{13}\)
sin B = \(\frac{\mathrm{AC}}{\mathrm{AB}}\) = \(\frac{5}{13}\)

Question 22.
Find the value of cos² 1° + cos² 2° + cos² 3° + ……………. + cos² 90°
Solution:
We know cos² (90° – 89°) + cos² (90° – 88°) + …………… + cos² 89° + cos² 90°
Here total 90° terms is there we know sin²θ + cos²θ = 1
= (sin² 89° + cos² 89) + (sin² 89 + cos² 88) + ………….. 44 terms.
= 44(1) + \(\frac{1}{\sqrt{2}}\) + 1
= 45 + \(\frac{1}{\sqrt{2}}\)

Question 23.
If cosec θ + cot θ = k then prove that cos θ = \(\frac{\mathrm{k}^2-1}{\mathrm{k}^2+1}\).
Solution:
Given cosec θ + cot θ = k ………….. (1)
We know cosec²θ – cot²θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1 k(cosec θ – cot θ) = 1
k(cosec θ – cot θ) = 1
cosec θ – cot θ = …………….. (2)

These TS 10th Class Maths Chapter Wise Important Questions Chapter 13 Probability given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 13 Probability

Previous Exams Questions

Question 1.
5 red and 8 white balls are present in a bag. If a ball is taken randomly from the bag then find the probability of it to be
i) white ball
ii) not to be white ball (A.P. Mar. ’16)
Solution:
Total number of balls present in bag
= 5 (red) + 8 (white) = 13
Probability for taking out a white ball
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total No. of outcomes }}\)
= \(\frac{8}{13}\)
Probability for not to be a white ball
P(\(\overline{\mathrm{E}}\)) = \(\frac{8}{13}\)
we know P(E) + P(\(\overline{\mathrm{E}}\)) = 1
⇒ P(\(\overline{\mathrm{E}}\))
= 1 – P(E) = 1 – \(\frac{8}{13}\)
= \(\frac{5}{13}\)

Question 2.
When die is rolled once unbiased what is the probability of getting a multiple of 3 out of possible out comes ? (T.S. Mar. ’15)
Solution:
P(E) = \(\frac{\text { favourable outcomes }}{\text { Total outcomes }}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Question 3.
There are 12 red, 18 blue and 6 white balls in a box. When balls is drawn at random from the box, what is the probability of not getting a red ball ? (T.S. Mar. ’15)
Solution:
Total Number of balls = 12 + 8 + 6
= 36
Number of red balls = 12
probability of getting red ball
P(\(\overline{\mathrm{R}}\)) = \(\frac{\text { favourable outcomes }}{\text { Total outcomes }}\)
= \(\frac{12}{36}\) = \(\frac{1}{3}\)
∴ Probability of not getting red ball
P(R) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Total Number of balls = 12 + 18 + 6 = 36
Exclude, the red balls, the number of remaining balls = 18 + 6 = 24
∴ Probability of not getting a
Red ball = \(\frac{24}{36}\) = \(\frac{2}{3}\)

Question 4.
There are 100 flash cards labelled from 1 to 100 in a bag. When a card is drawn from the bag at random, what is the probability of getting …….
(i) a card with prime number from possible outcomes ?
(ii) a card without prime number from possible outcomes ? (T.S. Mar. ’15)
Solution:
Number of prime numbers between 1 and 100 = 25
Probability of getting a card with prime numbers = \(\frac{25}{100}\) = \(\frac{1}{4}\) = 0.25
Probability of getting a card without prime number = \(\frac{75}{100}\) = 0.75
1 – 0.25 = 0.75

Question 5.
Find the probability of setting a sum of the numbers on them is 7, when two dice are rolled at a time. (T.S. Mar. ’16)
Solution:
When two dice are rolled at a time the total outcomes are = 6² = 36.
Number of outcomes such that their sum of numbers on face is 7 = 6
∴ Probability of getting sum of numbers on faces to be
7 = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Question 6.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of blue balls in the bag. (T.S. Mar. ’16)
Solution:
Number of red balls present in a bag = 5
Let the No. of blue balls = x (say)
Then the total No. of balls = 5 + x
From those (5 + x) balls in the bag the number of favourable outcomes to take a red ball randomly = 5
So, the probability of taking a red ball = \(\frac{5}{5+x}\)
Now
The number of favourable outcomes to take a blue ball randomly = x
So, the probability of taking a blue ball = \(\frac{x}{5+x}\)
From the given problem
Probability of blue bell = (Probability of red ball) (2)
\(\frac{x}{5+x}\) = \(\left[\frac{5}{5+x}\right]\) 2
\(\frac{x}{5+x}\) = \(\frac{10}{5+x}\) ⇒ x = 10
∴ No. of blue balls in the bag = 10.

Additional Questions

Question 1.
Kishore buys a fruit from a shop. The shopkeeper have one box. The box contain 18 mangoes, 32 apples so shopkeeper takes out one fruit at random what is the probability that the mango taken out from box.
Solution:
Given
Number of mangoes in the box = 18
Number of apples in the box = 32
Total number of fruits in the box = 18 + 32
So total number of outcomes = 50
Let E be the even that the mango taken out of the box = 18
We know that,
P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { Total No. of all possible outcomes }}\)
= \(\frac{18}{50}\) = \(\frac{9}{25}\)

Question 2.
A room contains 30 green chairs and some white chairs if the probability of drawing a white chair is triple that of green chair determine the number of white chairs in the room.
Solution:
Let the number of white chairs = x
Given number of green chairs = 30
Total number of chairs in room = x + 30
Total outcomes in drawing a chair at random = x + 30
Number of outcomes favourable to green chair = 30
∴ P(G) = \(\frac{30}{x+30}\)
So, given in problem P(W) = 3 × \(\frac{30}{x+30}\)
= \(\frac{90}{x+30}\)
We know that P(G) + P(W) = 1
\(\frac{30}{x+30}\) + \(\frac{90}{x+30}\) = 1
\(\frac{120}{x+30}\) = 1
x + 30 = 120
x = 90
∴ No. of white chairs x = 90.

Question 3.
There are 25 cards of same size in a bag on which number 1 to 25 are written one card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 5.
Solution:
Given total number of cards = 25
The number which are divisible by ‘5’ are 5, 10, 15, 20, 25
No. of all possible outcomes n(5) = 25
Number of out comes favourable to
E = n(E) = 5
∴ P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{5}{25}\) = \(\frac{1}{5}\)
We know that P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(\(\overline{\mathrm{E}}\)) = 1 – (P(E))
p(\(\overline{\mathrm{E}}\)) = 1 – \(\frac{1}{5}\)
= \(\frac{5-1}{5}\)
P(\(\overline{\mathrm{E}}\)) = \(\frac{4}{5}\)
∴ probability that the number selected card is not divisible by 5 = \(\frac{4}{5}\)

Question 4.
A jar contains 18 marbles, some are red and other white if a marble is drawn at random from the jar the probability that it is white \(\frac{5}{6}\) . Find the number of white marbles.
Solution:
Total number of marbles in the jar = 18
Let the number of red marbles = k
The number of white marbles = 18 – k
probability of drawing a red marble = \(\frac{\mathrm{k}}{18}\)
From problem = \(\frac{\mathrm{k}}{18}\) = \(\frac{5}{6}\)
⇒ k = \(\frac{90}{6}\)
⇒ k = 15
No. of red marbles = k = 15
No. of white marbles = 18 – 15 = 3.

Question 5.
A game consists of tossing a one rupee coin 2 times and noting its outcome each time. Ravi wins if all the wins give the same result, i.e., two heads or two tails and lose otherwise. Calculate the probability that he will lose the game.
Solution:
We know if a coin is tossed for n times, then the total number of outcomes = 2ⁿ
So, a coin is tossed for 2 times, then the total number of outcomes 2² = 4.
see here
T T
T H
H T
H H
of the above no. of outcomes with different result = 2.
probability of lossing the game
= \(\frac{\text { No. of favourable outcomes of lose }}{\text { No. of total outcomes }}\)
= \(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 6.
A lot consists of 200 ball pens of which 50 are defective and others are good. The shop keeper draws one pen at random and gives to sindhu. what is the probability that,
1) She will buy it ?
2) She will not buy it ?
Solution:
i) Total no. of ball pens = 200
∴ Number of all possible outcomes = 200
Number of defective ball pens = 50
Number of good ball pens = 200 – 50
⇒ No. of favourable outcomes = 150
probability that sindhu will buy it
= \(\frac{\text { No. of favourable outcomes }}{\text { Total No. of all possible outcomes }}\)
= \(\frac{150}{200}\) = \(\frac{3}{4}\)

ii) Probability that sindhu will not buy it
= 1 – (probability that sindhu will buy it)
= 1 – \(\frac{3}{4}\)
= \(\frac{4-3}{4}\)
= \(\frac{1}{4}\)

These TS 10th Class Maths Chapter Wise Important Questions Chapter 12 Applications of Trigonometry given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 12 Applications of Trigonometry

Previous Exams Questions

Question 1.
A person from the top of a building of height 25 m has observed another building top and bottom at an angle of elevation 45° and at an angle of depression 60° respectively. Draw a diagram for this data. (T.S. Mar. ’15)
Solution:

Question 2.
A ladder of 3.9 m length is laid against a wall. The distance between the foot of the wall and the ladder is 1.5 m. Find the height at which the ladder touches the wall. (T.S. Mar. ’15)
Solution:

h² = (3.9)² – (1.5)²
= (3.9 + 1.5) (3.9 – 1.5)
= 5. 4 × 2.4
= (0.6 × 9) × (0.6 × 4)
= (0.6)² × 6²
∴ h = 6 × 0.6 = 3.6 m

Question 3.
An observer flying in an aeroplane at an altitude of 900 m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships. (T.S. Mar. ’15)
Solution:

In ∆ABC
Tan 60 = \(\frac{900}{\mathrm{x}}\)
\(\sqrt{3}\) = \(\frac{900}{\mathrm{x}}\)
⇒ x = \(\frac{900}{\sqrt{3}}\) = 300\(\sqrt{3}\)
In ∆ ABD
Tan 30 = \(\frac{900}{\mathrm{x+d}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{900}{300 \sqrt{3}+d}\)
d = 600\(\sqrt{3}\) m

Question 4.
If the angle of elevation of sun increases from ‘O’ to 90 then the length of shadow of a tower decreases. Is this true ? Justify your answer. (T.S. Mar. ’16)
Solution:
Yes, this statement is true.
We observe this in day to day life.

AD – ground
BC – tower
AB – shadow

Question 5.
A boat has to cross a river. It crosses river by making an angle of 60° with bank, due to the stream of river it travels a distance of 450 m to reach another side of river. Draw a diagram to this data. (T.S. Mar. ’15)
Solution:

AB – width of river
AD, BC are river banks
AC – The distance travelled in river = 450 m
A – initial point, C – terminal point

Question 6.
Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road, the angles of elevation of top of the poles are 60° and 30°. Find the height of poles, (T.S. Mar. ’16)
Solution:
As shown in the figure

AD = width of road = 80 m.
AB, DE are two poles
AB = DE (∵ they have equal heights)
‘C’ is a point on road.
∠ACB = 30°, ∠DCE = 60°
Then in ∆ ACB
tan C = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ tan 30 = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ AC = AB\({\sqrt{3}}\) ……………….. (1)
In ∆ CDE
tan C = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
⇒ tan 60 = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
\({\sqrt{3}}\) = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
⇒ CD = \(\frac{\mathrm{DE}}{\sqrt{3}}\) ……………… (2)
but AC + CD = AD
⇒ AB\({\sqrt{3}}\) + \(\frac{\mathrm{DE}}{\sqrt{3}}\) = 80
But DE = AB
⇒ AB\({\sqrt{3}}\) + \(\frac{\mathrm{AB}}{\sqrt{3}}\) = 80
⇒ \(\frac{3 \mathrm{AB}+\mathrm{AB}}{\sqrt{3}}\) = 80
⇒ 4AB = 80\({\sqrt{3}}\)
⇒ AB = \(\frac{80 \sqrt{3}}{4}\) = 20 \({\sqrt{3}}\) m.
So height of the pole = 20 m.

Additional Questions

Question 1.
The angle of elevation of the top of a Tree from the foot of building is 60° and the angle of elevation of the top of the building from the foot of the tree is 30° what is the ratio of heights of tree and building.
Solution:

Let the height of the tree = x m = AB
Let the height of the building = y m = CD
distance between the tree and building = d = BD
Angle of elevation of the top of the tree = 60°
From the figure tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\) = \(\frac{\mathrm{x}}{\mathrm{d}}\)
\(\sqrt{3}\) = \(\frac{\mathrm{x}}{\mathrm{d}}\)
d = \(\frac{\mathrm{x}}{\sqrt{3}}\) ……………. (1)
∴ tan 30° = \(\frac{\mathrm{CD}}{\mathrm{BD}}\) = \(\frac{\mathrm{y}}{\mathrm{d}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{y}}{\mathrm{d}}\)
d = y\(\sqrt{3}\) ………………. (2)
\(\frac{\mathrm{x}}{\sqrt{3}}\) = y\(\sqrt{3}\)
\(\frac{\mathrm{x}}{\mathrm{y}}\) = \(\sqrt{3}\) × \(\sqrt{3}\)
\(\frac{\mathrm{x}}{\mathrm{y}}\) = \(\frac{3}{1}\)
x : y = 3 : 1
Hence, the ratio of the heights of the tree and building = 3 : 1.

Question 2.
The angle of the top of a pillar from two points are at a distance of 7m and 12m Find the height of the pillar from the base of the pillar and in the same straight line with it are complementary.
Solution:

From the figure,
Let AB be a height of the pillar = h m
Let the two points on the ground be ’c and d’ such that they make a distance 7 m and 12 m
From foot of the pillar
AC = 7m ; AD = 12m
Angles of elevation are ∠ACB = 0 ;
∠ADB = (90 – θ)
In the right angled ∆ABC, we have
tan θ = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)
tan θ = \(\frac{\mathrm{h}}{7}\) …………… (1)
From.the right angled ∆ABC we have
tan (90 – θ) = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)
cot θ = \(\frac{\mathrm{h}}{12}\)
\(\frac{1}{\tan \theta}\) = \(\frac{\mathrm{h}}{12}\)
tan θ = \(\frac{12}{\mathrm{~h}}\) …………….. (2)
From (1) and (2)
\(\frac{\mathrm{h}}{12}\) = \(\frac{12}{\mathrm{~h}}\)
h² = 84
h = \(\sqrt{4 \times 21}\)
h = 2 \(\sqrt{21}\) m
∴ The height of the pillar = h = 2 \(\sqrt{21}\)

Question 3.
A wire of length 25m had been tied with electric pole at an angle of elevation 30° with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60° with the ground how much length of the wire was cut.
Solution:

In the figure
Let PQ be a height of the electric pole = h m
given QS be the actual length of the wire = 25 m
let Q is length of the wire was cut S, R are the first and second points of observations.
Let PS = a + b ; PR = b
Angles of elevations are ∠PSQ = 30°
∠PRQ = 60°
From ∆PSQ
sin 30° = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)
\(\frac{1}{2}\) = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)
2 PQ = 25
PQ = 12.5
From ∆PRQ
tan 60° = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)
\(\sqrt{3}\) = \(\frac{12.5}{\mathrm{b}}\)
PR = b = \(\frac{12.5}{\sqrt{3}}\)
∆PQR, From
cos 60° = \(\frac{\mathrm{PR}}{\mathrm{QR}}\)
\(\frac{1}{2}\) = \(\frac{\frac{12.5}{\sqrt{3}}}{\mathrm{QR}}\)
\(\frac{1}{2}\) = \(\frac{12.5}{\sqrt{3 \mathrm{QR}}}\)
QR = \(\frac{25}{\sqrt{3}}\)
∴ Length of the wire was cut = \(\frac{25}{\sqrt{3}}\)

Question 4.
Two boys are on opposite sides of a tower 200m height. They measure the angle of elevation of the top of the tower as 45°, 60° respectively. Find the distance through which the boys are separated.
Solution:

Given height of Tower = 200 m
We say ‘x’ is the distance between 1^st person and base of the tower and ‘y’ is the distance between 2^nd person and base of the tower.
From ∆ ABD
Tan 45° = \(\frac{\mathrm{BD}}{\mathrm{AD}}\)
1 = \(\frac{200}{\mathrm{x}}\)
x = 200m
From ∆ BDC
Tan 60° = \(\frac{\mathrm{BD}}{\mathrm{DC}}\)

Question 5.
The tower height is l5mts and length of shadow is 15 \(\sqrt{3}\) m what is the angle of elevation of the sun.
Solution:

Let AC be the height of the tower = 15 m
Length of shadow = 15\(\sqrt{3}\) m
Let angle of elevation be θ.
From ∆ ABC Tan θ = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{15}{15 \sqrt{3}}\)
Tan θ = \(\frac{1}{\sqrt{3}}\)
Tan θ = Tan 30°
θ = 30°
∴ the angle of elevation of the sun θ = 30°

These TS 10th Class Maths Chapter Wise Important Questions Chapter 14 Statistics given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 14 Statistics

Previous Year Exam Questions

Question 1.
Find the mean of 5, 6, 9, 10, 6, 12, 3, 6, 11, 10. (A.P. Mar. 15)
Solution:
Mean = \(\frac{\text { Sum of scores }}{\text { No.of scores }}\)
= \(\frac{5+6+9+10+6+12+3+6+11+10}{10}\)
= \(\frac{78}{10}\) = 7.8

Question 2.
Write the formula for the median of a grouped data. Explain symbol with their used meaning. (A.P. Mar. ’15)
Solution:
Median (M) = l + \(\left(\frac{\frac{\mathrm{n}}{2}-c . f}{\mathrm{f}}\right)\) × h
l = lower limit of the median class.
n = sum of the frequency
c.f = cumulative frequency of the class preceding the median class
f = frequency of the median class
h = length of the class

Question 3.
Find the median of 5, 3, 1, -4, 6, 7, 0 (A.P. June ’15)
Solution:
The given observations are 5, 3, 1, -4, 6, 7, 0
Writing the observations in ascending order, we have -4, 0, 1, 3, 5, 6, 7.
There are 7 observations. Hence, the median will be \(\left(\frac{-7+1}{2}\right)^{\mathrm{th}}\) observation
i.e., 4^th observation
The 4^th observation is 3.
Hence, the median is 3.

Question 4.
Find the mode of the following data : (A.P. June ’15)

Sol:

Since, the maximum number of consumers (is 20) have got monthly consumption in the interval 120 – 140, the modal class is 120 – 140.
The lower boundary (l) of the modal class = 120.
The class size (h) = 20
The frequency of modal class (f₁) = 20
The frequency of the class preceding the modal class (f₀) = 16.
The frequency of the class succeeding the modal class (f₂) = 14.
Now, using the formula :
Mode = l + \(\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right]\) × h
= 120 + \(\left[\frac{20-16}{2 \times 20-16-14}\right]\) × 20
= 120 + \(\left[\frac{4}{40-30}\right]\) × 20
= 120 + \(\left[\frac{4}{10}\right]\) × 20 = 120 + 8 = 128

Question 5.
Find the mode of a5, 6, 9, 6, 12, 3, 6, 11, 6, 7 (A.P. Mar. ’16)
Solution:
In the given data 5, 6, 9, 6,12, 3, 6, 11, 6 and 7 the frequency of 6 is maximum.
Hence, mode = 6

Question 6.
Find the mean of first ‘n’ natural numbers. (A.P. Mar. ’16)
Solution:
Mean = \(\frac{\text { Sum of first ‘n’ natural number }}{n}=\frac{\Sigma n}{n}=\frac{n(n+1)}{2(\mathrm{n})}=\frac{n+1}{2}\)
∴ \(\frac{n+1}{2}\) is the average (mean) of first ‘n’ natural number.

Question 7.

How do you find the deviation from the assumed mean for the above data ? (T.S. Mar. ’15)
Solution:
The assumed value in calculation of mean of a grouped data is the mid value of the class interval which has maximum frequency.

Question 8.
In a village, an enumerator has surveyed for 25 households. The size of the family and the number of families is tabulated as follows : (T.S. Mar ’15)

Find the mode of the data.
Solution:

Mode = l + \(\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)\) × h
= 5 + \(\left[\frac{9-7}{18-7-2}\right]\) × 2
= 5 + \(\frac{4}{9}\)
= 5 + 0.44 = 5.44

Question 9.
Daily expenditure of 25 householders is given in the following table: (T.S. Mar. ’15)

Draw a “less than type” cumulative frequency ogive curve for this data.
Solution:

Points (150, 4) (200, 9) (250, 21) (300, 23) (350, 25)

Question 10.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers In the factory by using step deviation method. (T.S. Mar. ’16)
Solution:

Assumed mean (A) = 325
Σf_iu_i = 14; Σf_i = 50
Class interval (h) = 50
Formula for the mean in step-deviation method \(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\right]\)
Now substituting the above values in the formula we get
\(\overline{\mathrm{x}}\) = 325 + [\(\frac{14}{50}\) × 50]
= 325 + 14 = 339
So mean daily wage of workers = 339

Question 11.
The following table gives production yield per hectare of wheat of 100 farmers of village. (T.S. Mar. ’16)

Draw both ogives for the above data. Hence obtain the median production yield.
Solution:
We consider upper limits of class on X-axis and cumulative frequency on Y-axis to draw more than ogive.

So the points (55, 2) (60, 26) (65, 42) (70, 50) (75, 88) and (80, 100) are to be plotted by choosing the
Scale :-
on X – axis 1 unit = 50
on Y – axis 1 unit = 10
We get more than ogive

Part II
To draw less than ogive, we choose lower limits on X- axis and less than cumulative frequency on Y – axis.

Now the points to be plotted on graph (50. 100) (55, 98) (60, 74) (65, 58) (70, 50) and (75, 12)
Scale :-
on X – axis 1 cm = 5 units
on Y – axis 1 cm = 10 units
The above two curves cross at some point. Now we draw a perpendicular line to X-axis from this point.

The coordinate on X-axis (foot of perpendicular) is the median.

Additional Questions

Question 1.
Calculate the mean for the following.

Solution:

Here Σ f_i = 51 ; Σ f_ix_i = 2015
∴ We know \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
\(\overline{\mathrm{x}}\) = \(\frac{2015}{51}\) = 39.50
⇒ \(\overline{\mathrm{x}}\) = 39.50

Question 2.
Calculate the mean for the following.

Solution:

Here, assumed Mean a = 40
Size of the class h = 10
Σ f_i = 64 ; Σ f_iμ_i = 17
we know \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{j}} \mu_{\mathrm{i}}}{\Sigma \mathrm{f}_1}\) × h
\(\overline{\mathrm{x}}\) = 40 + \(\frac{17}{64}\) × 10
\(\overline{\mathrm{x}}\) = 40 + 2.65
\(\overline{\mathrm{x}}\) = 42.65

Question 3.
Find the Median for the following.

Solution:

Here, n = 82 ; \(\frac{\mathrm{n}}{2}\) = \(\frac{82}{2}\) = 41
and median Lines on the 40 – 60
Lower limit l = 40
Cf = 29
f = 26
h = 20
Median = l + \(\frac{\left(\frac{n}{2}-c . f\right)}{f}\) × h
= 40 + \(\frac{(41-29) \times 20}{26}\)
= 40 + \(\frac{12 \times 20}{26}\)
= 40 + \(\frac{240}{26}\)
= 40 + 923 = 49.23

Question 4.
Find the median of the following data.

Solution:

Here, n = 130 ; \(\frac{\mathrm{n}}{2}\) = \(\frac{130}{2}\) = 65
and median Lines on the = 100 – 125
Lower limit l = 100
f = 32
cf = 62
h = 25
Median = l + \(\frac{\left(\frac{n}{2}-c . f\right)}{f}\) × h
= 100 + \(\frac{(65-62) \times 25}{32}\)
= 100 + \(\frac{3 \times 25}{32}\)
= 100 + \(\frac{75}{32}\)
= 100 + 2.3 = 102.3

Question 5.
Find the Mode of the frequency distribution given below.

Solution:

Here, Lower Limit l = 15
f₁ = 14
f₀ = 06
f₂ = 21
We know mode = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right]\) × h
= 15 + \(\left[\frac{14-6}{28-6-21}\right]\) × 5
= 15 + \(\frac{8 \times 5}{28-27}\)
= 15 + 40
= 55

Question 6.
The Median of the data \(\frac{\mathrm{K}}{\mathrm{6}}, \frac{\mathrm{K}}{4}, \mathrm{K}, \frac{\mathrm{K}}{\mathrm{3}}, \frac{\mathrm{K}}{2}\) is 14. Then find the “k” value.
Solution:
\(\frac{\mathrm{K}}{\mathrm{6}}, \frac{\mathrm{K}}{4}, \mathrm{K}, \frac{\mathrm{K}}{\mathrm{3}}, \frac{\mathrm{K}}{2}\)
Given Median \(\overline{\mathrm{x}}\) = 14
Here, x = \(\frac{\mathrm{K}}{\mathrm{3}}\)
∴ \(\frac{\mathrm{K}}{\mathrm{3}}\) = 14
K = 42

Question 7.
If the Mean and Median of a data are 32.5 and 37.62 respectively. Find the mode of the data.
Solution:
Given mean = 32.5
Median = 37.62
We know, mode = 3 (Median) – 2 (Mean)
= 3 (37.62) – 2 (32.5)
= 112.86 – 65
= 47.86

Question 8.
Find the mean when the median is 72.8 mode is 65.
Solution:
Given median = 72.8
Mode = 65
Mean = ?
We know, mode = 3 (Median) – 2 (Mean)
∴ Mean = \(\frac{3(\text { Median })-\text { Mode }}{2}\)
= \(\frac{3(72.8)-65}{2}\)
= \(\frac{218.4-65}{2}\)
= \(\frac{153.4}{2}\) = 76.7

Question 9.
Find the Median of the data given below.
Solution:
15, 20, 2, 17, 18, 76, 5
Ascending order : 2, 5, 15, 17, 18, 20, 76
So, median = 17

Question 10.
Find the Median for the following data
30, 17, 12, 21, 33, 22
Ascending order: 12, 17, 21, 22, 30, 33
Solution:
Here following data have two numbers have in median so, that
Median = \(\frac{21+22}{2}\)
= \(\frac{43}{2}\) = 21.5

Question 11.
Find the mode of the following data.
20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3
Solution:
Mode = 3, 7

Question 12.
If the median of 60 observations, below Is 28.5 find the value of x and y.

Solution:

Median = l + \(\frac{\left(\frac{\mathrm{n}}{2}-\mathrm{cf}\right)}{\mathrm{f}}\) × h
given that Σf = n = 60
⇒ 45 + x + y = 60
⇒ x + y = 15 ……………. (1)
The median is 28.5 which lies between 20 and 30.
∴ Median class = 20 – 30
∴ l = lower boundary of the median class = 20
\(\frac{\mathrm{n}}{2}\) = \(\frac{60}{2}\) = 30
c.f = cumulative frequency = 5 + x and
h = 10
median = l + \(\frac{\left(\frac{\mathrm{n}}{2}-\mathrm{cf}\right)}{\mathrm{f}}\) h
28.5 = 20 + \(\frac{(30-5-x)}{20}\) × 10
28.5 = 20 + \(\frac{25-x}{2}\)
⇒ \(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
25 – x = 2 × 8.5 = 17
x = 25 – 17 = 8
from (1) x + y = 15
8 + y = 15
y = 15 – 8 = 7
∴ x = 8, y = 7

Question 13.
Find the Median of 30 students.

Solution:

No. of observations = n = Σf_i = 30
⇒ \(\frac{\mathrm{n}}{2}\) = \(\frac{30}{2}\) = 15
15 lies in the class 50 – 55
∴ Median dass = 50 – 55
∴ l = lower boundary of the median class = 50
f = frequency of the median class = 8
c.f = 5
class size = h = 6
Median = l + \(\frac{\left(\frac{\mathrm{n}}{2}-\mathrm{cf}\right)}{\mathrm{f}}\) × h
=50 + \(\left(\frac{15-5}{8}\right)\) × 6
= 50 + 125 × 6
= 50 + 7.5
= 57.5
∴ Median weight = 57.5 kg.

Question 14.
The mean of x + y observations Is x – y. find the sum of all the observations. Give mean of x + y observations is x – y. We know x = \(\frac{\Sigma f_i u_i}{f_i}\)
Solution:
Given x = x – y; Σf_i = x + y
so x – y = \(\frac{\Sigma f_i u_i}{x+y}\)
Σf_iu_i = (x + y) (x – y)
Σf_iu_i = x² – y²

Question 15.
Find the median and mode of the following observation. 12, 5, 9, 6, 14, 9 and 8.
Solution:
Given observations are 12, 5, 9, 6, 14, 8
= 5, 6, 8, 9, 12, 14
Median = \(\frac{8+9}{2}\)
= \(\frac{17}{2}\)
= 8.5
Mean x = \(\frac{5+6+8+9+12+14}{6}\)
= \(\frac{54}{6}\) = 9
Mode = 3(median) – 2(mean)
= 3(8.5) – 2(9)
= 25.5 – 18 = 43.5

Question 16.
Write the formula for calculating ‘Arithmetic mean’. In step deviation method and explain each letter in it.
Solution:
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right]\) × h

Question 17.
For the following data if the Median of 60 observations Is 28.5 find the values of x and y.

Solution:

Here. given total observations are = 60
∴ 45 + x + y = 60
x + y = 60 – 45
x + y = 15
From the Table l = 30 ; f = 15; cf = 25 + x ; h = 60; M = 28.5
we know median M = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}}\) × h
∴ 30 + \(\frac{30-(25+x)}{15}\) × 60 = 28.5
30 + (30 – 25 – x) 4 = 28.5
30 + (5 – x )4 = 28.5
30 + 20 – 4x = 28.5
50 – 4x = 28.5
-4x = 28.5 – 50
-4x = -21.5
x = \(\frac{21.5}{4}\)
x = 5.03
we apply x = 5.03, In x + y = 15
5.03 + y = 15
y = 15 – 5.03
y = 9.97

Telangana SCERT 10th Class Physics Study Material Telangana 4th Lesson Refraction of Light at Curved Surfaces Textbook Questions and Answers.

TS 10th Class Physical Science 4th Lesson Questions and Answers Refraction of Light at Curved Surfaces

Improve Your Learning
I. Reflections on concepts

Question 1.
How do you verify experimentally that the focal length of a convex lens is increased when it is kept in water?
Answer:
Aim: To prove that the focal length of a convex lens is increased when it is kept in water.
Apparatus: Convex lens of known focal length, circular lens holder, tall cylindrical glass tumbler, black stone, water.

Procedure:

• Take a cylindrical glass tumbler whose height is much greater than the focal length of the lens and fill it with water.
• Keep a black stone at the bottom of the vessel.
• Now dip the lens Into water using circular lens holder such that it is at a distance which Is less than or equal to focal length of the lens in air.
• Now see through the lens to have a view of the black stone.
• Now increase the height of the lens till you are not able to see the stone’s image.
• When the lens is dipped to a height which is greater than the focal length of lens in air, we are able to see the image. Showing that focal length of the lens has Increased in water.
• From this we conclude that the focal length of a convex lens is increased when it Is kept In water.

Question 2.
How do you find the focal length of a lens experimentally?
Answer:

• Take a v-stand and place it on a long table at the middle.
• Place a convex lens on the v-stand. Imagine the principal axis of the lens.
• Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
• Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.
• Measure the distance of the image from the v-stand of lens (image distance v) and also measure the distance between the candie and stand of lens (object distance ‘u’). Record the values in the table.

• Now place the candle at a distance of 60 cm from the lens, try to get an image of the candle flame on the other side on a screen. Adjust the screen till You get a clear image.
• Measure the image distance ‘y’ and object distance ‘u’ and record the values in table.
• Repeat the experiment for various object distances like 50 cm, 40cm, 30cm etc. Measure the image distances in all cases and note them in table.
• Using the formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \) find f in all the cases. We will observe the value f’ is equal in all cases. This value of ‘f is the focal length of the given lens.

Question 3.
Draw ray diagrams for the following positions and explain the nature and position of image.
(i) Object is placed at C₂
(ii) Object is placed between F₂ and optic centre P.
Answer:
(i)
When an object is placed at the centre of curvature (C₂) on the principal axis, we will get an Image at (C₁) which is real, inverted and of the same size as that of the object.

If the object is placed between focus and optic centre, we will get an image which is virtual, erect and magnified.

Question 4.
Write the lens maker’s formula and explain the terms in it.
Answer:
Lens maker’s formula is = \(\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)
f = focal length of the lens
n = refractive index of the lens
R, and R₂ are the radii of curvatures of two surfaces of the lens.

Application On Concepts

Question 1.
Two converging lenses are to be placed in the path of parallel rays so that the rays remain parallel after passing through both lenses. How should the lenses be arranged? Explain with a neat ray diagram.
Answer:

• A parallel beam of light rays will converge on focal point of the lens.
• light rays passing through focal point will emerge parallel to principal axis, the two lenses should be arranged as shown.
• The two lenses are arranged on a common principal axis such that their focal point coincides with each other.

Question 2.
The focal length of a converging lens is 20cm. An object Is 60cm from the lens. Where will the image be formed and what kind of ¡mage is It?
Answer:
f = 20 cm (convex lens, f = + ve)
u = – 60 cm [object distance = – ve]
v = ?
Lens formula : \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{20}=\frac{1}{60}+\frac{1}{v}=\frac{1}{20}-\frac{1}{60} \Rightarrow \frac{1}{v}=\frac{3-1}{60}=\frac{2}{60}=\frac{1}{30} \)
∴ v= 30 cm.
∴ The image distance is 30 cm.
f = 20 cm; hence, R = 40 cm, Object distance = 60 cm
∴ The object is placed beyond centre of curvature. Hence the image formed is real. inverted and diminished at 30 cm from the Lens.

Question 3.
A double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5. Find the focal length ‘f.
Answer:
n = 1.5
R₁ = R₂ = R
Lens maker’s formula for convex lens : \(\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)
= \(\frac{1}{f}=(1.5-1)\left(\frac{1}{R}+\frac{1}{R}\right) \)
= 0.5 × \(\frac{2}{R}=\frac{1}{R}\)
∴ f = R

Question 4.
Find the refractive index of the glass which is a symmetrical convergent lens it its local length is equal to the radius of curvature of its surface.
Answer:
Given that lens is convergent symmetrical We know that
∴ R₁ = R = f
R₂ = -R= -f
We know that \(\frac{1}{f}=(n-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]\)

∴ Refractive index of glass = 1.5

Question 5.
Man wants to get a picture of a zebra. He photographed a white donkey after fitting a glass with black stripes, onto the lens of his camera. What photo will he get? Explain.
Answer:

1. He will get a photograph which consists of black and white stripes.
2. As the reflected light rays from the white donkey entered into camera through the lens having black stripes, these black stripes do not allow the rays Inside.
3. So, the rays which pass through the transparent part of a camera lens only forms the corresponding image of donkey on the film i.e., white lines as it is white in colour.

Question 6.
Harsha tells Slddhu that the double convex lens behaves ‘like a convergent lens. But Slddhu knows that Harsha’s assertion Is wrong and corrected Harsha by asking some questions. What are the questions asked by Siddhu?
Answer:
Siddhu may ask the following questions

1. What is the shape of the lens If two convex lenses are attached?
2. What happens when a light ray passes through a double convex lens?
3. Is there any convergent point available, if the light ray passes through a double convex lens.

Question 7.
Can a virtual Image be photographed by a camera?
Answer:

• Yes, a virtual Image can be photographed by a camera.
• A plane mirror forms a virtual Image, we can able to take photographs of that Image In plane mirror.
• In the same way, human eyes forms a virtual image, which can able to take a photograph.

Question 8.
How do you appreciate the coincidence of the experimental facts with the results obtained by a ray diagram in terms of behaviour of Images formed by lenses?
Answer:

1. By using a ray diagram, the reflected ray must be placed at a particular point by the principal axis. That means we have to find the images, shorter or longer.
2. By using ray diagrams, we are able to find the focal length from lens maker’s formula in many optical instruments, some lens combinations are used to magnification (or) diminished of the image.
3. When white light passes through a prism, then VIBGYOR is formed on the principal axis that means converging take place.
4. So, I appreciate the coincidence of the experimental facts with the results obtained by a ray diagram in terms of behaviour of images formed by lenses.

Question 9.
Find the radii of curvature of a convexo-concave convergent lens made of glass with refractive Index n = 1.5 having focal length of 24cm. One of the radii of curvature is double the other.
Answer:
n=1.5, f=24cm, R₁=R, R₂=2R
\(\frac{1}{f}=(n-1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\)
R₁ = positive, R₂ = positive
\(\frac{1}{24}=(1.5-1)\left(\frac{1}{R}-\frac{1}{2 R}\right) \)
\(\frac{1}{24}=(0.5)\left(\frac{2-1}{2 \mathrm{R}}\right)\)
\(\frac{1}{24}=(0.5)\left(\frac{1}{2 \mathrm{R}}\right)\)
R = 6 cm

R₁ = R = 6cm,
R₂ = 2R = 12 cm

Higher Order Thinking Questions

Question 1.
A convex lens Is made up of three different materials as shown in the figure. How many of Images does It form?
Answer:

• Given convex lens is made up of three different materials.
• The three different materials have three different refractive indices.
• So the given lens have three different focal lengths. Hence it forms three images.

Question 2.
You have a lens. Suggest an experiment to find out the focal length of the lens.
(OR)
Write the experimental method and apparatus required in finding out the image formation, using convex lens.
Answer:
Answer:(i)
When an object is placed at the centre of curvature (C₂) on the principal axis, we will get an Image at (C₁) which is real, inverted and of the same size as that of the object.

If the object is placed between focus and optic centre, we will get an image which is virtual, erect and magnified.

Question 3.
Figure shows ray AB that has passed through a divergent lens. Construct the patti of the ray up to the lens if the position of Its foci is known.
Answer:

A. Given AB is the ray that passes through a diverging lens. F is the focus of the lens.

If AB ray is extended backward It seems to be passing through focal point (F). This is possible only when incident ray is parallel to the principal axis.

Question 4.
Figure shows a point light source and Its Image produced by a lens with an optical axis N₁N₂ Find the position of the lens and its foci using a ray diagram.

Answer:

• The object is in between focus and optic Centre.
• The image is virtual, erect and magnified.
• ‘l’ Is the lens, ‘O’ Is the object and ‘I’ Is the Image.

Question 5.
Find the focus by drawing a ray diagram using the position of source Sand the image S’ given In the figure.

Answer:

• When the object is between curvature 2F₁ and focus (F₂), the image will be formed beyond, centre of curvature.
• The image will be real inverted and magnified.

Question 6.
A parallel beam of rays is incident on a convergent lens with a focal length of 40cm. Where should a divergent lens with a focal length 15cm be placed for the beam of rays to remain parallel after passing through the two lenses? Draw a ray diagram.
Answer:

Focal length of convex lens, f₁ = 40 cm (+ve)
Focal length of concave lens, f₂ = 15 cm (- ve)
For the emergent rays to be parallel to principal axis, the effective focal length of the combination should be zero. Effective focal length of two lenses separated by some distance is given by

Question 7.
Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height? Why?
Answer:
The friend looks taller than what he actually is. Friend AB Is standing on the bank of the lake. The rays of light BP and BQ from the head (B) of the friend, on refraction at the water-air interface, bend towards the normals at points P and Q and appear to come from point B’ Therefore, to me, my friend will appear as AB’ i.e. taller than what his actual height, AB is.

Question 8.
Use the data obtained by activIty -2 In table-1 of this lesson and draw the graphs of u VS v and \( \frac{1}{u}\) VS \(\frac{1}{v} \)
Answer:
Graph of u – v using data obtained by activity 2.
Take lens with focal length 30 cm.

Graph of \(\frac{1}{u}=\frac{1}{v}\)

For these values the graph is straight line which touches the axes as shown in figure.

Question 9.
The distance between two point sources of light is 24cm. Where should a convergent lens with a focal length of f9cm be placed between them to obtain the Images of both sources at the same point?
Answer:

for S₁
\(\frac{1}{v_1}-\frac{1}{-x}=\frac{1}{9}\)
for S₂
\(\frac{1}{v_2}-\frac{1}{-(24-x)}=\frac{1}{9} \)
\(\frac{1}{v_2}=\frac{1}{9}-\frac{1}{(24-x)}\)
Since, sign conventioi for S₁ and S₂ is just opposite
Hence v₁ = v₂
\(\frac{1}{v_1}=-\frac{1}{v_2} \)

Solving x = 6cm. Therefore, the lens should be kept at a distance of 6cm from either of the object.

IV. Multiple choices questions

Question 1.
Which one of the following materials cannot be used to make a lens? ( )
(A) water
(B) glass
(C) plastic
(D) clay
Answer:
(D) clay

Question 2.
Which of the following is true? ( )
(A) The distance of virtual image is always greater than the object distance for convex lens.
(B) The distance of virtual image is not greater than the object distance for convex lens.
(C) Convex lens always forms a real image.
(D) Convex lens always forms a virtual image.
Answer:
(B) The distance of virtual image is not greater than the object distance for convex lens.

Question 3.
Focal length of the piano-convex lens is ……………………… when its radius of curvature of the surface is R and n is the refractive index of the lens. ( )
(A) f = R
(B) f=R/2
(C) f=R/(n-1)
(D) f=(n-1)/R
Answer:
(C) f=R/(n-1)

Suggested Experiments

Question 1.
Conduct an experiment to find out the focal length of the lens.
(OR)
You have a lens. Suggest an experiment to find out the focal length of the lens.
Answer:

1. Take a V-stand and place it on a long table at the middle.
2. Place a convex lens on the V-stand.
3. Light a candle and place it at a long distance along the principal axis.
4. Adjust the screen which is on other side of lens to get an image on ¡t.
5. Measure the distance of the image from the stand of the lens(v) and also measure the distance between the candle and stand of iens(u) .
6. Repeat the experiment for various object distances(u) like 50cm, 40cm, 30cm and measure distances of images(v) in each case and note in the table.
   
   substituting the values of u, v fn the formula \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) focal length of the lens
   ∴ f = \(\frac{u v}{u+v}\)

Question 2.
Let us assume a system that consists of two lenses with focal length f₁ and f₂ respectively. How do you find the focal length of the system experimentally, when
(i) two lenses are touching each other
(ii) they are separated by a distance d’ with common prlnclpal axis.
Answer:
Consider two lenses A and B of focal lengths f₁ and f₂ placed in contact with each other. Let the object be placed at a point ‘O’ beyond the focus of the first lens
(i) The first lens produces an Image at I₁. Since I₁ Is real, It serves as a virtual image at I.\

for the image formed by the first lens A, we get
\(\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}\) ………………………… (1)
for the mage formed by the second lens B. we get
\(\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2} \) …………………………… (2)
adding the above two equations \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2} \) …………….. (3)
But \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) ……………………….. (4)
from(3) and (4) weget \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \)

(ii) When they are separated by some distance ‘d’.
Two thin lenses are placed coaxially at a distance of separation ‘d The Incident ray AB and the emergent ray CD intersect at E. The perpendicular from E to the principal axis falls at R The equivalent lens should be placed
at this position R A ray ABE going parallel to the principal axis will go through the qulvalent lens and emerge along ECD. The angle of deviation,
from :
θ = θ₁ +θ₂ From ΔBEC.
∴ θ = θ₁ +θ₂

From the above figure. We have

Suggested Projects

Question 1.
Collect the information about the lenses available in an optical shop. Find out how the focal length of a lens may be determined by the given ‘power’ of the lens.
Answer:
Lens is a portion of a transparent retracting medium which is bounded by the coaxial spherical surfaces. There are two types of lenses

• Convergent lenses : These are thick at centre and thin near edges. They bend the rays towards centre of the lens. They are divided Into three categories.

• Divergent lenses: These are thin at the centre and thick near edges. They bend the rays towards the edges of the lens. They are also further divided In to three categories. The power of a lens is defined as reciprocal of focal length of the lens. It is measured In diopters.

Question 2.
Take two watch glasses and affix them. Pour two different liquids (Eg:‘ Water, Navaratan oil) and now It will act like a lens with two dIfferent materials. Put a light source (object) in front of this lens and note the observations and write a report on It.
Answer:

1. Since the refractive indices of water and glass are nearly equal, the portion of the lens filled with water acts as convex lens.
2. Since the refractive Index of Navaratan oil less than the refractive index of glass, the portion of the lens filled with Navaratan oil will acts as diverging lens (concave lens).
3. As there are two types of lenses n a given single lens, It will form two types of images in which one is real Inverted and small Image due to refraction of light through the portion of lens which is filled with water (convex lens).
4. And other is virtual, exerted and small image due to the refraction of light through the portion of lens which is filled wIth Navaratan oil. (Concave lens).
5. The formation of Images is as shown in the fig.

TS 10th Class Physical Science Refraction of Light at Curved Surfaces Intext Questions

Page 57

Question 1.
Have you ever touched a magnifying glass with your hand?
Answer:
Yes

Question 2.
Have you touched the glass in the spectacles used for reading with your hand?
Answer:
Yes

Question 3.
Is It plane or curved surface?
Answer:
Curved

Question 4.
Sitthicker in the middle or at the edge?
Answer:
It is thicker at the edges.

Question 5.
What do you see?
Answer:
We will see a diminished (small-sized) image of the arrow.

Question 6.
Why do you see a diminished Image?
Answer:
When the vessel is empty, light from the arrow refracts at the curved interface, moves through the glass and enters Into air then it again undergoes refraction on the opposite curved surface of the vessel and comes out into the air. In this way light travels through two media and comes out of the vessel and forms a diminished image.

Question 7.
is the Image real or virtual?
Answer:
Virtual

Question 8.
Can you draw a ray diagram showing how ¡t ¡s formed?
Answer:

Page 58

Question 9.
What do you see now?
Answer:
When the vessel is filled with water, there is a curved Interface between two different media.

Question 10.
Do you get an inverted image?
Answer:
When the vessel Is filled with water, light enters the curved surface, moves through water, comes ot4t of the glass and forms an inverted Image.

Question 11.
How could this happen?
Answer:
When the vessel Is filled with water, there is a curved interface between two different media (air and water). Assume that the refractive indices of both water and glass are the same. This setup of air and water separated by a curved surface is shown in the figure 1.

Question 12.
What happens to a ray that is Incident on a curved surface separating the two media?
Answer:
Ught ray gets refracted

Question 13.
Are the laws of refraction still valid?
Answer:
Yes, laws of refraction are valid in the case of refraction at curved surfaces.

Question 14.
How do rays bend when they are incident on a curved surface?
Answer:
As In the case of plane surfaces, a ray will bend towards the normal if it travels from a rarer medium to denser medium and bends away from the normal if It travels from a denser medium to rarer medium.

Page 59

Question 15.
What happens to a ray that travels along the principal axis? Similarly, a ray that travels through the centre of curvature?
Answer:
According to Snell’s law the ray which travels along the normal drawn to the surface does not deviate from its path. Hence both rays mentioned above travel along the normal, so they do not deviate.

Question 16.
What happens to a ray travelling parallel to the principal axis?
Answer:
Observe the figures a, b, c, and d. In all the cases as represented by the diagrams, the incident ray is parallel to the principal axis.
Casi: A ray travelling parallel to the principal axis strikes a convex surface and passes from a rarer medium to denser medium.

(a) Case2: A ray travelling parallel to the principal axis strikes a convex surface and passes from a denser medium to rarer medium.

Case3: A ray travelling parallel to the principal axis strikes a concave surface and passes from a rarer medium to rarer medium.

Case 4: A ray travelling parallel to the principal axis strikes a concave surface and passes from a rarer medium to denser medium.

Question 17.
What difference do you notice in the refracted rays in 4a and 4b?
Answer:
In the case of 4(a) refracted ray reaches a particular point on the principal axis.

Page 60

Question 18.
What could be the reason for that difference?
Answer:
One is convex curved surface and other is concave curved surface.

Question 19.
What difference do you notice In retracted rays in figure 4(c) and 4(d)?
Answer:
In the case of 4(c) refracted ray travels towards principal axis and in case 4(d) refracted ray travels away from the principal axis, but the extended rays intersect the axis and in this case at focal point.

Question 20.
What could be the reasons for that difference?
Answer:
One is convex curved surface and other is concave curved surface

Question 21.
How can you explain this change In the size of the lemon?
Answer:
As the light travels from glass vessel (curved surface) to the air(rarer medium) light bends away from the normal and lemon appears bigger In size.

Question 22.
Is the lemon that appears bigger in size an image of lemon or is it the real lemon?
Answer:
It is the image of the lemon only but not the real lemon.

Question 23.
Can you draw a ray diagram to explain this phenomenon?
Answer:
When a ray of light travels from one transparent medium to another, it bends at the surface, thereby separating the two media. hence, the lemon appears larger than its actual size. This happens because different media have different optical densities. The phenomenon of bending of light as it travels from one medium to another is known as refraction of light.

Page 64

Question 24.
What happens to the light ray when a transparent material with two curved surfaces is placed In Its path?
Answer:
When two transparent curved surfaces are placed In the path of light ray It becomes a lens and light gets refracted.

Question 25.
Have you heard about lenses?
Answer:
Yes, I heard about lenses.

Question 26.
How does a light ray behave when it Is passed through a lens?
Answer:
Depending on the type of lens light ray diverges from a point or converges at a point.

Page 66

Question 27.
How does the lens from an Image?
Answer:
As lens has two surfaces, we can consider the lens as a single surface element because we assume that the thickness of the lens is very small and show the net refraction at only one of the surfaces.

Question 28.
If we allow a light ray to pass through the focus, which path does it take?
Answer:
The ray passing through the focus will take a path parallel to principal axis after refraction.

Page 67

Question 29.
What happens when parallel rays of light fall on a lens making some angle with the principal axis?
Answer:
When parallel rays, making an angle with principal axis, fall on a lens, the rays converge at a point or appear to diverge from a point lying on the focal plane.

Question 30.
What do you mean by an object at infinity?
Answer:
The light rays incident on the lens from the object which is very far off from the lens.

Question 31.
What type of rays fall on the lens?
Answer:
The rays which are parallel to the principal axis are incident of the lens.

Page 69

Question 32.
What do you notice?
Answer:
Irrespective of the position of object, on the principal axis, we get an erect, virtual image, diminished In size in between the focal point and optic centre for concave lens.

Page 70

Question 33.
Can we realise In practice the results obtained in the ray diagrams when we perform experiments with a lens?
Answer:
Yes, when lens is used we get the ¡mage at the same positions.

Question 34.
Why are we using a screen to view this Image? Why don’t we see It directly with our eye?
Answer:
On a screen we get a real image but when we see with our eyes we get virtual image.

Page 71

Question 35.
Could you get an image on the screen for every object’s distance?
Answer:
We are image on the screen when object is placed at any position except when the object is placed between focus and optic centre.

Question 36.
Why don’t you get an image for certain object distances?
Answer:
When the object Is placed between the focus and optic centre the Image formed virtually and also it Is formed on the same side of the object.

Question 37.
Can you find the minimum limiting object distance fur obtaining a real image?
Answer:
Yes, the minimum distance required to get real image must the greater than are equal to focal length.

Question 38.
Answer:
It is called least distance of distinct vision.

Question 39.
Could you see the image?
Answer:
Yes, we can see a magnified image on the same side when we kept the object as indicated.

Question 40.
What type of image do you see?
Answer:
This is a virtual image of the object which we cannot capture on the screen.

Question 41.
Can you find the Image distance of this virtual Image?
Answer:
No, we cannot measure the distances

Question 42.
Could you find focal length of the lens from the values recorded in table -1?
Answer:
Yes we can find the focal length from the values obtained from the table.

Question 43.
Can we establish a relation between u’, v’ and ‘f’?
Answer:
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) and this is known as lens formula.

Question 44.
How Is the image formed?
Answer:
Image formed is real.

Page 73

Question 45.
Is the focal length same for each set of values?
Answer:
Irrespective of object distance and image distance we get same focal length.

Page 74

Question 46.
On what factors does the focal length of the lens depend?
Answer:
The focal length of lens depends upon the surrounding medium in which it is kept.

Question 47.
Can you see the Image of the stone?
Answer:
Yes, we can see the Image of the stone when distance between the lens and stone is less than the focal length of the lens( in air).

Question 48.
If Yes/Not, why? Give your reasons.
Answer:
We can see the ¡mage of the stone ¡f the distance between stone is less than the focal length of the Iens(in air). Now increase the distance between lens and stone until you cannot see the Image of the stone.

Question 49.
Whatdoyouconcludefromthlsacttvlty?
Answer:
When we dipped the lens to a certain height which ¡s greater than the focal length of lens In air, we can see the Image. This shows that the focal length of lens has increased in water.

Question 50.
Does the focal length of the lens depend on surrounding medium?
Answer:
Focal length of the lens depends upon the surrounding medium In which It is kept.

TS 10th Class Physical Science Refraction of Light at Curved Surfaces Activities

Activity 1

Question 1.
Write an activity to observe the refraction of light at curved surfaces.
Answer:
Procedure and observation:

1. Draw an arrow of length 4 cm using a black sketch pen on a thick sheet of paper.
2. Take an empty cylindrical-shaped transparent vessel.
3. Keep it on the table.
4. Ask your friend to bring the sheet of paper on which arrow was drawn behind the vessel while you look at it from the other side.
5. We will see a diminished image of the arrow.
6. Ask your friend to fill vessel with water.
7. Look it the arrow from the same position as before.
8. We Can Observe an inverted Image.

Explanation:

1. At the first time ‘light travels through two media, i.e., glass arid air and conies out of the vessel. So it forms a dimlnishcd imaoe.
2. At the second time, light enters the curved surface, moves through water, comes out of the glass. So it forms an inverted image.

Lab Activity 1

Question 2.
Write an activity to know the types of images and measuring the object distance and image distance from the lens,
Answer:
Procedure:

• Take a v-stand and place a convex lens on this stand.
• Imagine the principal axis of the lens.
• Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
• We use a secreen because It forms a real image generally which will form on a screen. Real Images cannot be seen with an eye.
• Adjust the screen, on other side of lens until clear image forms on it.
• Measure the distance of the image from the y- stand and also measure the distance between the candle arid stand of lens.
• Now place the candle at a distance of 60 cm from the lens. Such that the flame of the candle lies on the principal axis of the lens.
• Try to get an image of candle flame on the other side on a screen.
• Adjust the screen till you get a clear image.
• Measure the distance of image (v) from lens and record the value’s of u and v in the table.
• Repeat this for various distances of mages in all cases and note them in the table.

Observation:

Conclusion:
From this table, we conclude that a convex lens forms both real and virtual images when object is placed at various positions.

Activity 2

Question 3.
How can you show that the focal length of a convex lens will increase in water or In the surrounding medium?
Answer:
Procedure:

• Take a convex lens.
• Note the average focal length of the lens that calculated in the activity.
• Take a cylindrical vessel such as glass tumbler.
• Its height must be 3 or 4 times greater than the focal length of the lens.
• Keep a black stone inside the vessel at its bottom.
• Now pour water into the vessel up to a height such that the height of the water level from the top of the stone is greater than focal length of lens.
• Now dip the lens horizontally using a circular lens holder as shown in the figure above the stone.
• Set the distance between stone and lens that is equal to or less than focal length of lens
• Now look at the stone through the lens.
• You can see the image of the stone if the distance between lens and stone Is less than the focal length of the lens (in air).
• Now increase the distance between lens and stone until you cannot see the image of the stone.
• You have dipped the lens to a certain height which is greater than the focal length of lens in air.
• But you can see the image.
• This shows that the focal length of lens has increased In water Thus we conclude that the focal length of lens depends upon the surrounding medium in which it is kept.