Class 10

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 – \(\frac{3}{4}\)
= \(\frac{4-3}{4}\)
= \(\frac{1}{4}\)

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2

Question 1.
Find the co-ordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2 : 3.
(A.P. Mar. 16)
Solution:
Given points P(-1, 7) and Q (4, -3).
Let ‘R’ be the required point which divides \(\overline{\mathrm{PQ}}\) in the ratio 2 : 3 then

Question 2.
Find the co-ordinates of the points of tri-section of the line segment joining (4, -1) and (-2, -3). (A.P.June ’15)
Solution:
Given points A(4, -1) and B(-2, -3)
Let P and Q be the points of trisection of \(\overline{\mathrm{AB}}\), then AP = PQ = QB

∴P divides AB internally in the ratio 1:2.

P(x, y) = (2, -5/3)
Also, Q divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1 internally.
Q(x, y) = \(\left[\frac{-4+4}{3}, \frac{-6-1}{3}\right]\) = (0, -7/3)
Q(x, y) = (0, -7/3)

Question 3.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Given : A(-3, 10) and B(6, -8)
Let P(-1, 6) divides \(\overline{\mathrm{AB}}\) in the ratio k : 1 internally.
By section formula

6k – 3 = -k – 1 and -8k + 10 = 6k + 6
-7k = – 2 and 14k = 4 4 14
∴ The ratio is 2/7 : 1 or 2 : 7

Question 4.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution:

Given : ABCD is a parallelogram where
A(1, 2), B(4, y), C(x, 6) and D(3, 5).
In a parallelogram, diagonals bisect each other, i.e., the mid points of the diagonals coincide with each other.
i.e., mid point of AC = mid point of BD

1 + x = 7 and ⇒ 8 = y + 5
x = 6 and y = 8 – 5 = 3
∴ x = 6 and y = 3.

Question 5.
Find the co-ordinates of a point A, where AB is the diameter of a circle. Whose centre is (2, -3) and B is (1, 4).
Solution:

Given : A circle with centre ‘C’ (2, -3)
\(\overline{\mathrm{AB}}\) is diameter where B = (1, 4); A = (x, y)
∴ C is the mid point of AB
[∵ centre of a circle is the midpoint of the diameter].

Question 6.
If A and B are (-2, -2) and (2, -4) respectively. Find the co-ordinates of such that AP = \(\frac{3}{7}\) AB and P lies on the segment AB.
Solution:
Given : A(-2, -2) and B(2, -4)
P lies on AB such that AP = \(\frac{3}{7}\) AB

⇒ \(\frac{A P}{A B}\) ⇒ \(\frac{3}{7}\) = \(\frac{A P}{P B}\) = \(\frac{3}{7-3}\) = \(\frac{3}{4}\)
i.e., P divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 4 By section formula.

Question 7.
Find the co-ordinates of points which divide the line segment joining A(-4, 0) and B(0, 6) into four equal parts.
Solution:
Given, A(-4, 0) and B(0, 6)
Let R Q and R be the points which divide \(\overline{\mathrm{AB}}\) into four equal parts.

P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 3
Q → 1 : 1 and R → 3 : 1 use section formula to find P, Q and R.
The Q is the mid point of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 8.
Find the co-ordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution:
Given : A(-2, 2) and B(2, 8)
Let P, Q and R be the points which divide AB into four equal parts.

Then Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 9.
Find the co-ordinates of the point which divide the line segment joining the points (a + b, a – b) and (a – b, a + b) in the ratio 3 : 2 internally.
Solution:
Given : A(a + b, a – b) and B(a – b, a + b)
let P(x, y) divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 2 inter¬nally.
By using section formula,

Question 10.
Find the co-ordinates of centroid of the triangle with vertices following.

i) (-1, 3), (6, -3) and (-3, 6).
Solution:
Given : ΔABC in which A(-1, 3), B(6, -3) and C(-3, 6).
Centroid

ii) (6, 2), (0, 0) and (4, -7).
Solution:
Given : The three vertices of a triangle are A(6, 2), B(0, 0) and C(4, -7)
Centroid (x, y)

iii) (1, -1) (0, 6) and (-3, 0)
Solution:
Given: (1, -1), (0, 6) and (-3, 0) are the vertices of a triangle.
Centroid (x. y)

These TS 10th Class Maths Chapter Wise Important Questions Chapter 7 Coordinate Geometry given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 7 Coordinate Geometry

Previous Exams Questions

Question 1.
Where do the points (0, -3) and (-8, 0) lie on co-ordinate axis ? (A.P. June ’15)
Solution:
The point (0, -3) lie on OY’
∴ Its X-co-ordinate is zero and Y-co-ordinate is negative and the point (-8, 0) lie on OX’
∵ Its Y-co-ordinate is zero and X-co-ordinate is negative.

Question 2.
Find the centroid triangle whose vertices are (3, 4) (-7, -2) and (10, -5). (T.S. Mar. 15)
Solution:
Centroid of the triangle whose vertices are (x₁, y₁) (x₂, y₂) (x₃, y₃)
= \(\left[\frac{X_1+X_2+X_3}{3}, \frac{Y_1+Y_2+Y_3}{3}\right]\)
Here x₁ = 3, y₁ = 4, x₂ = -7, y₂ = -2, and x₃ = 10, y₃ = -5
Now G = \(\left(\frac{3-7+10}{3}, \frac{4-2-5}{3}\right)\)
= \(\left(\frac{6}{3}, \frac{3}{3}\right)\) = (2, 1)

Question 3.
Show that the points A = (4, 2), B (7, 5) and C (9, 7) are collinear. (A.P. Mar. 15)
Solution:
To show that three points are collinear the area formed by the triangle is zero. Area of triangle
Δ = \(\frac{1}{2}\) |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
Here x₁ = 4, y₁ = 2, x₂ = 7, y₂ = 5,
x₃ = 9, y₃ = 7
Now Δ = \(\frac{1}{2}\) |4(5 – 7) + 7(7 – 2) + 9(2 – 5)|
= \(\frac{1}{2}\)|4(-2) + 7(5) + 9(-3)|
= \(\frac{1}{2}\)|-8 + 35 – 27|
= \(\frac{1}{2}\)|-35 + 35| = \(\frac{1}{2}\) |0| = 0
So the above three are collinear.

Question 4.
Name the type of the quadrilateral formed by joining the points A (-1, -2), B (1, 0) C (-1, 2) and D (-3, 0) are graph paper, justify your answer. (A.P. Mar. 15)
Solution:
Given points are A (-1, -2); B (1, 0); C (-1, 2) and D (-3, 0)

Distance between the two points

∴ ABCD is a square.
(OR)

From (1) and (2) ABCD is a square.
(OR)
From the graph
1) Diagonals are equal \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{BD}}\) = 4 units
2) Also diagonals \(\overline{\mathrm{AC}}\) and \(\overline{\mathrm{BD}}\) bisect each other. \(\overline{\mathrm{BD}}\) is a part of X-axis and \(\overline{\mathrm{AC}}\) is parallel to Y-axis.
∴ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\)
From (i), (ii) and (iii) ABCD is a square.

Question 5.
Find the mid point of the line segment formed by the points (-5, 5) and (5, -5) (A.P. Mar. 16)
Solution:
Formula for the mid points of line segment formed by (x₁, y₁) and (x₂, y₂) is
\(\left[\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right]\)
here x₁ = -5, y₁ = 5 and x₂ = 5, y₂ = -5
∴ mid point
= \(\left[\frac{-5+5}{2}, \frac{5-5}{2}\right]\) = \(\left[\frac{0}{2}, \frac{0}{2}\right]\) = [0, 0]

Question 6.
Show that the points A(-3, 3) B (0, 0) C(3, -3) are collinear. (T.S.Mar. 16)
Solution:
To show them as collinear the area formed by the triangle should be zero. Formula for area of triangle
∆ = \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁ + x₃(y₁ – y₂)|
Here x₁ = -3, y₁ = 3; x₂ = 0, y₂ = 0 and x₃ = 3, y₃ = -3
So ∆ = \(\frac{1}{2}\) | (-3) (3 – 0) + 0 (-3 – 3) + 3(3 – 0) |
= \(\frac{1}{2}\) |(-3)(3) + 0(-6) + 3(3) |
= \(\frac{1}{2}\) |-9 + 0 + 9|
= \(\frac{1}{2}\) |0| = 0 sq. units
Hence the above three points are collinear.

Question 7.
Find the trisection points of the line segment joined by the points (-3, 3) and (3, -3). (A.P. Mar. ’16)
Solution:
The points which divide the line segment by 1:2 and 2:1 ratio (internally) are called trisection points.
Formula for the points of trisection of the line segment joined by (x₁, y₂) and (x₂, y₂) are
= \(\left[\frac{\mathrm{mx}_2+\mathrm{nx}_1}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_2+\mathrm{ny}_1}{\mathrm{~m}+\mathrm{n}}\right]\)
Where m = 2 and n = 1.
Here x₁ = -3, y₁ = 3 and x₂ = 3, y₂ = -3, then the point in the ratio 2 : 1 is

So (1, -1) is one trisection point. The point which is at 1 : 2 ratio is another trisection point.
So m = 1, n = 2.
Here x₁ = -3, = 3, x₂ = 3 and y₂ = -3
then \(\left[\frac{1(-3)+2(3)}{1+2}, \frac{1(-3)+2(3)}{1+2}\right]\)
= \(\left[\frac{-3+6}{3}, \frac{-3+6}{3}\right]\) = \(\left[\frac{3}{3}, \frac{3}{3}\right]\)
= (1, 1) is another trisection point.

P, Q are trisection points.

Additional Questions

Question 1.
Find the distance between the following pairs of points. (Each 1 Mark)

i) (3, 4) and (6, 2)
ii) (-4, 6) and (0, 2)
iii) (-a, b) and (a, -b)

i) (3, 4) and (6, 2)
Solution:
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(6-3)^2+(2-4)^2}\)
= \(\sqrt{9+4}\) = \(\sqrt{13}\) Units

ii) (-4, 6) and (0, 2)
Solution:
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(0+4)^2+(2-6)^2}\)
= \(\sqrt{16+(-4)^2}\)
= \(\sqrt{16+16}\) = \(\sqrt{32}\)
= \(\sqrt{16 \times 2}\)
= \(\sqrt{16}\) × \(\sqrt{2}\) = 4\(\sqrt{2}\) Units

iii) (-a, b) and (a, -b)
Solution:
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(a+a)^2+(-b-b)^2}\)
= \(\sqrt{(2 a)^2+(-2 b)^2}\)
= \(\sqrt{4 a^2+4 b^2}\) = \(\sqrt{4\left(a^2+b^2\right)}\)
= \(\sqrt{4} \sqrt{a^2+b^2}\) = \(2 \sqrt{a^2+b^2}\) Units

Question 2.
Verify Whether the points A(1, 3), B(3, 5) and C(-4, 2) are collinear or not.
Solution:
Given A = (1, 3), B = (3, 5) and C = (-4, 2)
Slope of AB = m₁ = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
= \(\frac{5-3}{3-1}\) = \(\frac{2}{2}\) = 1
Slope of BC = m₂ = \(\frac{2-5}{-4-3}\) = \(\frac{-3}{-7}\) = \(\frac{3}{7}\)
Since m₁ ≠ m₂
∴ A, B, C are not collinear.

Question 3.
Show that (5, 3), (1, 2) and (-3, 1) are the vertices of an isosceles triangle.
Solution:
Let A = (5, 3), B = (1, 2) and C = (-3, 1)
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Since AB = BC
∴ ΔABC is an isosceles triangle.
Hence given points are the vertices of an isos- celes triangle.

Question 4.
Show that the points (0, -2), (3, 2), (0, 6) and (-3, 2) are the vertices of a square.
Solution:
Let A = (0, -2), B = (3,2), C = (0,6), D = (-3, 2)
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)


Sin AB = BC = CD = DA
∴ Given vertices form a square.

Question 5.
Show that the points (2, 4), (3, 7), (6, 8) and (5, 5) are the vertices of a Rhombus.
Solution:
Let A = (2, 4), B = (3, 7), C = (6, 8) and D = (5, 5)
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Since AB = BC = CD = DA = \(\sqrt{10}\) and AC ≠ BD
∴ Given points form a Rhombus.

Question 6.
Find the co-ordinates of the point which divides the line segment joining the points (-2, 6) and (3, -4) in the ratio 3 : 5.
Solution:
Given points P (-2, 6) and Q (3, -4)
Let R be the required point which divides \(\overline{\mathrm{PQ}}\) in the ratio 3 : 5, then

Here P = (-2, 6) = (x₁, y₁),
Q = (3, -4) = (x₂, y₂)
and m₁ : m₂ = 3 : 5
Then R (x, y)

Question 7.
Find the ratio in which the X-axis divides the line segment joining the points (6, -2) and (4, 1) also find the point of intersection.
Solution:
Let the ratio be k : 1
given points are A = (6, -2) and B = (4, 1)
Then by section formula, the co-ordinates of the point which divide AB in the ratio k : 1 are
\(\left(\frac{\mathrm{k} \times 4+1 \times 6}{\mathrm{k}+1}, \frac{\mathrm{k} \times 1+1 \times-2}{\mathrm{k}+1}\right)\)
i.e., \(\left(\frac{4 k+6}{k+1}, \frac{k-2}{k+1}\right)\) ……. (1)
Since the point lies on the X-axis
∴ Its Y-co-ordinate is zero
i.e., \(\frac{k-2}{k+1}\) = 0 ⇒ k – 2 = 0 ⇒ k = 2
∴ The ratio is k : 1 = 2 : 1 ⇒ k = 2
Putting k = 2 in (1), we get, the point of intersection
= \(\left(\frac{4 \times 2+6}{2+1}, \frac{2-2}{2+1}\right)\)
= \(\left(\frac{8+6}{3}, \frac{0}{3}\right)\)
= \(\left(\frac{14}{3}, 0\right)\)

Question 8.
Find the ratio in which the Y-axis divides the line segment joining the points (9, -4) and (3, -2). Also find the point of intersection.
Solution:
Let the ratio be k : 1
Given points are A = (9, -4) and B = (3, -2)
Then by section formula, the co-ordinates of the point which divides AB in the ratio k : 1 are
\(\left(\frac{\mathrm{k} \times 3+1 \times 9}{\mathrm{k}+1}, \frac{\mathrm{k} \times(-2)+1 \times(-4)}{\mathrm{k}+1}\right)\)
i.e., \(\left(\frac{3 k+9}{k+9}, \frac{-2 k-4}{k+1}\right)\) —– (1)
Since the point lies on the Y-axis
∴ Its abscissa = 0
i.e., \(\frac{3 \mathrm{k}+9}{\mathrm{k}+9}\) = 0 ⇒ 3k + 9 = 0 ⇒ 3k = -9
⇒ k = \(\frac{-9}{3}\) = -3
∴ The ratio is k : 1 = -3 : 1 ⇒ k = -3
Putting k = -3 in (1), we get, the point of intersection
= \(\left(\frac{3 \times(-3)+9}{-3+9}, \frac{-2(-3)-4}{-3+1}\right)\)
= \(\left(\frac{-9+9}{6}, \frac{6-4}{-2}\right)\)
= \(\left(\frac{0}{6}, \frac{2}{-2}\right)\)
= (0, -1)

Question 9.
Find the centroid of the triangle whose vertices are given below.
i) (1, 6), (7, 4), (1, 5)
ii) (1, 0), (-3, -3), (-4, -3)

i) (1, 6), (7, 4), (1, 5)
Solution:
Given A = (1, 6) = (x₁, y₁)
B = (7, 4) = (x₂, y₂)
C = (1, 5) = (x₃, y₃)
Centroid of the triangle
= \(\left(\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{3}, \frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}\right)\)
∴ Centroid of the triangle
= \(\left(\frac{1+7+1}{3}, \frac{6+4+5}{3}\right)\)
= \(\left(\frac{9}{3}, \frac{15}{3}\right)\) = (3, 5)

ii) (1, 0), (-3, -3), (-4, -3)
Solution:
Given A = (1, 0), B = (-3, -3), C = (-4, -3)
Centroid of the triangle
= \(\left(\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{3}, \frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}\right)\)
= \(\left(\frac{1-3-4}{3}, \frac{0-3-3}{3}\right)\)
= \(\left(\frac{-6}{3}, \frac{-6}{3}\right)\)
= (-2, -2)

Question 10.
Find the area of the triangle whose vertices are
i) (3, 4), (0, 1), (3, -3)
ii) (2, 10), (4, 7), (0, -1)
iii) (3, 4), (0, 1), (3, -3)
Solution:
i)
Given A = (3, 4) = (x₁, y₁)
B = (0, 1) = (x₂, y₂)
C = (3, -3) = (x₃, y₃)
are the vertices of ∆ABC.
Area of ∆ABC
= \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
= \(\frac{1}{2}\)|3(1 + 3) + 0(-3 – 4) + 3(4 – 1)|
= \(\frac{1}{2}\)|3(4) + 0(-7) + 3(3)|
= \(\frac{1}{2}\)|12 + 0 + 9|
= \(\frac{1}{2}\)|21|
= \(\frac{21}{2}\) sq. units

ii) (2, 10), (4, 7), (0, -1)
Solution:
Given A = (2, 10), B = (4, 7), C = (0, -1)
Area of ∆ ABC
= \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
= \(\frac{1}{2}\)|(7 + 1) + 4(-1 – 10) + 0(10 – 7)|
= \(\frac{1}{2}\)|2(8) + 4(-11) + 0(3)|
= \(\frac{1}{2}\)|16 – 44 + 0|
= \(\frac{1}{2}\)|-28| = |-14|
= 14sq. units

Question 11.
Find the value of k’ for which the points are collinear.
i) (k, 9), (7, 7), (1, 5)
ii) (6, -1), (k, -6), (0, -7)
iii) (k, k), (1, 2), (3, -2)

i) (k, 9), (7, 7), (1, 5)
Solution:
Given A(k, 9) = (x₁, y₁)
B(7, 7) = (x₂, y₂)
C(1, 5) = (x₃, y₃) are collinear
∴ Area of Δ ABC = 0
⇒ \(\frac{1}{2}\) x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
⇒| \(\frac{1}{2}\)|k(7 – 5) + 7(5 – 9) + 1(9 – 7)| = 0
⇒ |k(2) + 7(-4) + 1(2)| = 0
⇒ 2k – 28 + 2 = 0
⇒ 2k – 26 = 0
⇒ 2k = 26
k = \(\frac{26}{2}\) = 13
∴ k = 13

ii) (6, -1), (k, -6), (0, -7)
Solution:
Given A(6, -1), B(k, -6), C(0, -7) are collinear
∴ Area of Δ ABC = 0
⇒ \(\frac{1}{2}\)|6(-6 + 7) + k(-7 + 1) + 0(-1 + 6)| = 0
⇒ |6(1) + k(-6) + 0(5)| = 0
⇒ 6 – 6k + 0 = 0
⇒ -6k = -6
⇒ k = \(\frac{-6}{-6}\) = 1
∴ k = 1

iii) (k, k), (1, 2), (3, -2)
Solution:
Given A = (k, k), B(1, 2), C(3, -2) are collinear
∴ Area of Δ ABC = 0
⇒ \(\frac{1}{2}\)|k(2 + 2) + 1(-2 – k) + 3(k – 2)| = 0
⇒ \(\frac{1}{2}\)|k(4) + 1(-2 – k) + 3(k – 2)| = 0
⇒ 4k – 2 – k + 3k – 6 = 0
⇒ 6k – 8 = 0
⇒ 6k = 8
⇒ k = \(\frac{8}{6}\) = \(\frac{4}{3}\)
∴ k = \(\frac{4}{3}\)

Question 12.
Find the slope of the line joining the points.
i) (7, -5) and (8, 1)
ii) (4a, 6b) and (a, -b)
iii) (5, -1) and (-8, 0)

i) (7, -5) and (8, 1)
Solution:
Given A(7, -5) = (x₁, y₁)
B(8, 1) = (x₂, y₂)
Slope of line AB = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{1-(-5)}{8-7}\)
= \(\frac{1+5}{1}\) = \(\frac{6}{1}\) = 6

ii) (4a, 6b) and (a, -b)
Solution:
Given A(4a, 6b), B(a, -b)
Slope of line AB = \(\frac{-b-6 b}{a-4 a}\) = \(\frac{7 b}{3 a}\)

iii) (5, -1) and (-8, 0)
Solution:
Given A(5, -1) and B(-8, 0)
Slope of line AB = \(\frac{0-(-1)}{-8-5}\)
= \(\frac{0+1}{-13}\) = \(\frac{-1}{13}\)

Question 2.
Find the slope of a line whose inclination is

i) 60°
ii) 135°
iii) 120°
iv) 135°

i) 60°
Solution:
Given 60° = θ
Slope of the line = m
= tan θ = tan 60° = \(\sqrt{3}\)

ii) 135°
Solution:
Given 135° = θ
Slope of the line = m = tan θ = tan 135°
= tan (90° + 45°)
= -cot 45° = -1

iii) 120°
Solution:
Given 120° = θ
Slope of the line = m = tan θ = tan 120°
= tan (90° + 30°)
= -cot 30°
= –\(\sqrt{3}\)

iv) 135°
Slope of the line = m = tan θ = tan 135°
= tan (90° + 45°)
= -cot 45°
= -1

Question 3.
If (6, 7) is the mid point of the line segment joining A(7, 6) and B(5, y), find ‘y’.
Solution:
Given Mid point = (6, 7) = m and
A(7, 6) = (x₁, y₁)
B(5, y) = (x₂, y₂)
Mid point of A and B = (6, 7)
⇒ \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) = (6, 7)
⇒ \(\left(\frac{7+5}{2}, \frac{6+y}{2}\right)\) = (6, 7)
Equations Y-co-ordìnate on both sides
⇒ \(\frac{6+\mathrm{y}}{2}\) = 7 ⇒ 6 + y = 2 × 7
⇒ 6 + y = 2 × 7
⇒ 6 + y = 14
⇒ y = 14 – 6
y = 8
∴ y = 8

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°

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.4

Question 1.
Find the slope of the line passing the two given points.

i) (4, -8) and (5, – 2)
Solution:
Slope = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\) = \(\frac{-2+8}{5-4}\) = \(\frac{6}{1}\) = 6

ii) (0, 0) and (\(\sqrt{3}\), 3)
Solution:
Slope = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{3-0}{\sqrt{3}-0}\)

iii) (0, 0) and (\(\sqrt{3}\), 3)
Solution:
Slope = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{-b-3 b}{a-2 a}\)

iv) (a, 0) and (0, b)
Solution:
Slope = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\) = \(\frac{b-0}{0-a}\) = \(\frac{-b}{a}\)

v) A(-1.4, -3.7) and B(-2.4, 1.3)
Solution:
Slope = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\) = \(\frac{1.3+3.7}{-2.4+1.4}\)
= \(\frac{5.0}{-1}\) = -5

vi) A(3,-2) and B(-6,-2)
Solution:
Slope = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{-2+2}{-6-3}\) = 0

vii)

Solution:

viii) A(0, 4), B(4, 0)
Solution:
Slope = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
= \(\frac{0-4}{4-0}\)

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.

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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.

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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.

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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.
  

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3

Question 1.
Find the area of the triangle whose vertices are

i) (2, 3), (-1, 0), (2,-4)
Solution:
Given : A(2, 3), B(-1, 0), C(2, -4) are the vertices of a ΔABC
Area of the ΔABC
= \(\frac{1}{2}\)|(y₂ – y₃) + x₂ (y₃ – y₁) + x₃(y₁ – y₂)|
= \(\frac{1}{2}\) |8 + 7 + 6|
= \(\frac{21}{2}\) = 10\(\frac{1}{2}\) sq. units.

ii) (- 5, – 1), (3, -5), (5, 2) (A.P. Mar. 15)
Solution:
Given A(-5, -1), B(3, -5), C(5, 2) are the ver-tices of a ΔABC
Area of the Δ ABC
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) | 135 + 9 + 20|
= \(\frac{64}{2}\) = 32 sq.units.

iii) (0, 0), (3, 0) and (0, 2)
Solution:
Given 0(0, 0), A(3, 0), B(0, 2) are the vertices of a ΔABC
Area of the ΔAOB
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) |0(0 – 2) + 3(2 – 0) + 0(0 – 0)|
= \(\frac{1}{2}\) |6| = 3 sq. units. (or)

ΔAOB = \(\frac{1}{2}\) × 3 × 2 = 3 sq. units.

Question 2.
Find the value of ‘K’ for which the points are collinear. (A.P. June ’15)

i) (7, -2), (5, 1), (3, K)
Solution:
Given : A(7, -2), B(5, 1), C(3, K) are collinear
∴ Area of ΔABC = 0
But area of triangle.
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
⇒ \(\frac{1}{2}\) | 7 – 7K + 5K + 10 – 9| = 0
⇒ |-2K + 8| = 0
⇒ – 2K = – 8
⇒ K = 4

ii) (8, 1), (K, -4), (2, -5)
Solution:
Given A(8, 1), B(k, -4), C(2, -5) are collinear
∴ Area of ΔABC = 0
⇒ \(\frac{1}{2}\) |8(- 4 + 5) + K(-5 – 1) + 2 (1 + 4) 1= 0
⇒ | 8 – 6K + 10| = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
∴ K = 3

iii) (K, K) (2, 3) and (4, -1)
Solution:
A(K, K), B(2, 3) & C(4, -1) are collinear
∴ Area of ∆ABC = O
⇒ \(\frac{1}{2}\) | K(3 + 1) + 2 (-1 – K) + 4(K – 3)| = 0
⇒ | 4K – 2 – 2k + 4K – 12| = 0
⇒ |6K – 14| = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\)
= \(\frac{7}{3}\)
⇒ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are (0, -1), (2, 1)
and (0, 3). Find the ratio of this area to the area of the given triangle. (A.P. Mar.16)
Solution:

Given : A(0, -1), B(2, 1), C(0, 3) are the vertices of ∆ABC. Let D, E & F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) & \(\overline{\mathrm{AC}}\)
midpoint (x, y) = \(\left[\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}\right]\)
∴ D = \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)\) = (1, 0)
E = \(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)\) = (1, 2)
F = \(\left(\frac{0+0}{2}, \frac{-1+3}{2}\right)\) = (0, 1)
Area of ∆DEF
= \(\frac{1}{2}\) |1(2 – 1) + 1(1 – 0) + 0(0 – 2)|
= \(\frac{1}{2}\) |1 + 1| = \(\frac{2}{2}\) = 1 sq. units.
Ratio of areas = ∆ABC : ∆DEF = 4 : 1
∆ADF \(\cong\) ∆BED \(\cong\) ∆DEF \(\cong\) ∆CEF
∴ ∆ABC : ∆DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3)
Solution:

Given : A(-4, -2), B(-3, -5), C(3, -2) & D(2, 3) are the vertices of the quadrilateral ABCD.
Area of
ABCD = ∆ABC + ∆ACD
Area of triangle
= \(\frac{1}{2}\) |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
Area of ∆ABC
= \(\frac{1}{2}\) |(-4)(-5 + 2)(-3)(-2 + 2) + 3(-2 + 5) |
= \(\frac{1}{2}\)|12 – 0 + 9| = \(\frac{21}{2}\) sq.units.
Area of ∆ACD
= \(\frac{1}{2}\)|-4(-2 – 3) + 3(3 + 2) + 2(-2 + 2) |
= \(\frac{1}{2}\)|20 + 15 + 0 | = \(\frac{35}{2}\) sq. units.
∴ Area of ABCD = \(\frac{21}{2}\) + \(\frac{35}{2}\) = \(\frac{21+35}{2}\)
= \(\frac{56}{2}\) = 28 sq.units.

Question 5.
Find the area of the triangle formed by the points (8, -5), (-2, -7) & (5, 1) by using Herons formula.
Solution:
Given : A(8, -5), B(-2, -7) & C(5, 1) are the vertices of ∆ABC
Distance formula
= \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
AB = \(\sqrt{100+4}\) = \(\sqrt{104}\) = 10.19 = a
BC = \(\sqrt{49+64}\) = \(\sqrt{113}\) = 10.63 = b

We are offering TS 10th Class Maths Notes Chapter 11 Trigonometry to learn maths more effectively.

TS 10th Class Maths Notes Chapter 11 Trigonometry

→ Trigonometry is the study of relationship between the sides and angle of a triangle.

→ Ratios of the sides of a right triangle with respect to its acute angle are called trigonometric ratios of the angle.

→ An equation involving trigonometric ratios of an angle is called a trigonometric identity. If it is true of all values of the angle.

→ Let us consider ΔABC in which ∠B = 90°, A and C are acute angles. Let us study the ratios of the sides of ΔABC with respect to the acute angle A.

sine of ∠A = sin A = \(\frac{\text { Side opposite to angle } \mathrm{A}}{\text { Hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
cosine of ∠A = cos A = \(\frac{\text { Side adjacent to angle } \mathrm{A}}{\text { Hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
tangent of ∠A =tan A = \(\frac{\text { Side opposite to angle } A}{\text { Side adjacent to angle } A}=\frac{B C}{A B}\)

→ cosec A = \(\frac{1}{\sin A}\)
sec A = \(\frac{1}{\cos A}\)
cot A = \(\frac{1}{\tan A}\)

→ If one of trigonometric ratios of an acute angle is known the remaining trigonometric ratios of the angle can be easily determined.

→ Trigonometric ratios of 0°, 30°, 45°, 60° and 90°.

Note : From the above table we can observe that ∠A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0.

→ Trigonometric ratios of complementary angles : Two angles are said to be complementary angle if their sum equals to 90°.

• sin (90° – A) = cos A;
• cosec(90° – A) = sec A
• cos (90° – A) = sin A;
• sec(90°- A) = cosec A
• tan (90° – A) = cot A;
• cot(90° – A) = tan A

→ Trigonometric identities :

• sin²A + cos²A = 1
• sec²A – tan²A = 1 for 0°< A < 90°
• cosec²A – cot²A = 1 for 0° < A < 90°

Note : sin²θ = (sin θ)² but sinθ² ≠ (sin θ)²

Important Formula:

• sin(90° – A) = cos A;
• cosec (90° – A) = sec A
• cos(90° – A) = sin A
• sec(90° – A) = cosec A
• tan(90° – A) = cot A
• cot(90° – A) = tan A
• sin²A + cos²A = 1
• sec²A – tan²A = 1
• cosec²A – cot²A = 1

Flow Chat:

Aryabhatta (476 – 550 A.D):

• The first use of the idea of ‘sine’ in the way we use it today was in the book Aryabhatiyam by Aryabhatta, in A.D. 500.
• Aryabhatta used the word ‘ardhajya1 for the half-chord, which was shortened to jya or jiva in due course.
• When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is. The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin.
• Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe. An English Professor of astronomy Edumund Gunter (1581¬1626), first used the abbreviated notation ‘sin’.

We are offering TS 10th Class Maths Notes Chapter 10 Mensuration to learn maths more effectively.

TS 10th Class Maths Notes Chapter 10 Mensuration

Cuboid :

l, b and h denote respectively the length, breadth and height of a cuboid then

• Lateral surface area = 2h(l + b)
• Total surface area = 2(lb + bh + hl)
• Volume = lbh
• Diagonal of the cuboid = \(\sqrt{l^2+b^2+h^2}\)

Cube :

If the length of each edge of cube is “a” units then

• Lateral surface area = 4a²
• Total surface area = 6a²
• Volume = a³
• Diagonal of the cube = √3 × a = a√3

Right Prism :

• Lateral surface area = Perimeter of base × height
• Total Surface area = Lateral Surface area + 2(Area of end Surface)
• Volume = Area of base × height

Right Circular Cylinder:

If r is the radius of the base and ‘h’ is the height, then

• Lateral surface area = 2πrh
• Total surface area = 2πr(h + r)
• Volume = πr²h

Right Pyramid :

• Lateral surface area = \(\frac{1}{2}\) × perimeter of base × slant height
• Total Surface area = Lateral surface area + area of base
• Volume = \(\frac{1}{3}\) × area of base × height

Right Circular Cone :

If ‘r’ is the radius of the base, ‘h’ is the height and is slant height, then

• Lateral surface area = πrl.
• Total Surface area = πr(l + r)
• Volume = \(\frac{1}{3}\)πr²h
• l² = h² + r²

Sphere :

If ‘r’ is radius of sphere, then

• Lateral surface area = 4πr²
• Total surface area = 4πr²
• Volume = \(\frac{4}{3}\) πr³

Hemisphere :

If Y is the radius of hemi-sphere, then

• Lateral surface area = 2πr²
• Total surface area = 3πr²
• Volume = \(\frac{4}{3}\)πr³

Right Circular Hollow Cylinder:

• Area of each end = π(R² – r²)
• Curved surface area of hollow Cylinder = External area + Internal area
  = 2πrh + 2πRh = 2πh(R + r)
• Total surface area = 2πRh + 2πrh + 2(πR² – πr²)
  = 2πh(R + r) + 2π(R + r)(R – r)
  = 2π(R + r) (R + h – r)
• Volume of the material = External volume – Internal volume
  = πR²h – πr²h = 7th(R² – r²)

Spherical Shell:

If R and r are the outer and inner radii of a spherical shell, then

• Outer surface area = 4πR²
• Volume of material = \(\frac{4}{3}\) π

→ The volume of the solid-formed by joining two basic solids is the sum of the volumes of the constituents.

→ In calculating the S.A of a combination of solids, we cannot add the surface area of the two constituents, because some part of the surface area disappears in the process of joining them.

Flow Chat:

Brahmagupta (598 – 668):

• Brahmagupta was born in the state of Rajasthan.
• He worked in the great astronomical centre of ancient India – Ujjain.
• He made significant contributions to Trigonometry.

Telangana SCERT TS 10th Class Physical Science Study Material Pdf 11th Lesson Principles of Metallurgy Textbook Questions and Answers.

TS 10th Class Physical Science 11th Lesson Questions and Answers Principles of Metallurgy

Improve Your Learning
I. Reflections on concepts

Question 1.
List three metals that are found In nature as Oxide ores.
Answer:
Zinc, Ferrous, Aluminium, and Magnesium are metals which are found in nature as oxide ores.
They are :

Question 2.
List three metals that are found in nature n uncombined form.
Answer:

1. Gold
2. Platinum
3. Silver
4. Copper

Question 3.
Write a note on dressing of ore in metallurgy.
Answer:

1. Dressing is the first step in extraction of metals.
2. Ores that are mined from earth are usually contaminated with impurities such as soil and sand etc.
3. Dressing means, simply getting rid of as much of the unwanted rocky material as possible before the ore is converted into the metal.
4. Physical methods are used to enrich the ore.
5. These methods adopted in dressing the ore depend upon difference between physical properties of ore and gangue.
6. The following physical methods involved in dressing are 1) Hand picking 2) Washing 3) Froth floatation 4) Magnetic separation.

Question 4.
How do metals occur in nature? Give examples to any two types of minerals.
Answer:
The earth’s crust is the major source of metals. Sea water also contains some soluble salts such as sodium chloride and magnesium chloride etc. Some metals like Gold (Au), Silver (Ag) etc., are available in nature in a free state (native) as they are least reactive.

Other metals mostly are found in nature in the combined form due to their reactivity. The elements or compounds of the metals which occur in nature ¡n the earth’s crust are called ‘minerals’. Minerals in oxide form: Bauxite, Zincite, Magnetite, etc.

Question 5.
What is the difference between roasting and calcination? Give one example for each.
Answer:

Question 6.
Draw the diagram showing i) Froth floatation ii) Magnetic separation.
Answer:
i) Froth floats tian

ii) Magnetic separation

Question 7.
Draw a neat diagram of the Reverberatory furnace and label it neatly.
Answer:

Question 8.
What is an ore? On what basis a mineral is chosen as an ore?
Answer:
Ore: A mineral from which a metal can be extracted economically and conveniently is called ‘ore’.
To choose a mineral as an ore the following are considered

• The percentage of the metal in that mineral.
• Whether metal can be profitably extracted from it or not.
• The convenience of extraction of metal.

Question 9.
Write the names of any two ores of iron.
Answer:

1. Haematite – Fe₂O₃
2. Magnetite – Fe₃O₄

Question 10.
How do metals occur In nature? Give examples of any two types of minerals.
Answer:
The earth’s crust is the major source of metals. Sea water also contains some soluble salts such as sodium chloride and magnesium chloride etc. Some metals like Gold (Au), Silver (Ag) etc., are available in nature in free state (native) as they are least reactive. Other metals mostly are found in nature in the combined form due to their reactivity.

The elements or compounds of the metals which occur in nature in the earth’s crust are called ‘minerals’.
Minerals in oxide form: Bauxite, Ziricite, Magnetite, etc.
Minerals in sulphide form: Copper iron pyrites, Gatena, etc.

Question 11.
Write short notes on froth floatation process.
Answer:
Froth Floatation process:

Froth Floatation process for the concentration of sulphide ores.

1. This method is mainly useful for sulphide ores which have no wetting property whereas impurities get wetted.
2. The ore with impurities is finely powdered and kept ¡n water taken in a floatation cell.
3. Air under pressure is blown to produce froth in water.
4. Froth so produced takes the ore particles to the surface whereas, impurities settle at the bottom.
5. Froth is separated and washed to get ore particles.

Question 12.
When do we use magnetic separation method for concentration of an ore? Explain with an example.
Answer:
Magnetic Separation Method:
If the ore contains impurities such that one, of them, is magnetic and the other is non – magnetic they are separated by magnetic separation method.
Eg: Magnetic ores like iron pyrites, (FeS) and magnetite (Fe₃O₄) are concentrated by this method. The crushed ore is allowed to pass through electromagnetic belts. The mineral particles are retained and gangue particles are thrown away as a separate heap.

Question 13.
Write short notes on each of the following:
(i) Roasting.
(ii) Calcination.
(iii) Smelting.
Answer:
(i) Roasting:

• Roasting is a pyrochemical process in which the ore is heated in the presence of oxygen or air below its melting point.
• The products obtained in the process are also produced in solid state.
• Generally, reverberatory furnace is used for roasting.
  Eg : 2ZnS + 3O₂ → 2ZnO + 2SO₂

(ii) Calcination:

• Calcination is a pyrochemical process in which the ore is heated in the absence of air.
• The ore gets generally decomposed n the process.
  Eg: MgCO₃ → MgO + CO₂

(iii) Smelting:

• Smelting is a pyrochemical process ¡n which the ore is mixed with flux and fuel and strongly heated.
• The heat is so strong that the ore is reduced to even metal and the metal is obtained in molten state.
• During smelting, the impurities (gangue) in the ore react with flux to form slag which is removed.
• The smelting is carried out n a specially built furnace known as blast furnace.

Question 14.
What is gangue and slag?
Answer:

• Gangue: The impurity present in the ore is called gangue.
• Slag: A flux is a chemical substance added to convert gangue into fusible mass. This fusible mass is called slag.
  Gangue + Flux Slag

Application of concepts

Question 1.
Magnesium is an active metal, if it occurs as a chloride in nature, which method of reduction is suitable for its extraction?
Answer:
Magnesium is an active metal. If it occurs as chloride in nature, the only method viable to extract magnesium ( any active metal) is electrolysis of fused Magnesium chloride. In electrolysis of fused Magnesium chloride, magnesium is deposited at cathode and chlorine gas is liberated at the anode.
MgCl₂ → Mg⁺² + 2Cl^–
At cathode, Mg⁺²+ 2e^– → Mg
At anode, 2Cl^– → Cl₂ + 2e^–

Question 2.
Mention two methods which produce very pure metals from impure metals.
Answer:
Electrolysis and reduction are the two methods which produce pure metals.

Question 3.
Which method do you suggest for extracting of high reactivity metals? Why?
Answer:
High reactivity metals like K. Na, Ca, Mg, etc., can be extracted by electrolysis.
Reasons:

1. Simple reduction methods like heating with C, CO, etc., to reduce the ores of these metals are not feasible.
2. The temperature required for the reduction is too high and more expensive.,
3. Hence electrolysis is the suggestable method to extract high reactive metals.

Question 4.
Explain Thermite process and mention its applications in our daily life.
Answer:
Thermite process

1. When highly reactive metals such as Na, Ca, Al are used as reducing agents they displace metals of lower reactivity from their compounds.
2. These displacement reactions are highly exothermic. The amount of heat evolved s so large that the metals produced will be in molten state.
3. The reaction of iron oxide (Fe₂O₃), with aluminium is used to join railings of railway tracks or cracked machine parts. This reaction is known as the thermite reaction.
   2Al + Fe₂O₃ → Al₂O₃ + 2Fe + Heat

Applications in daily life
1. To join railings of railway tracks.
2. Used to join cracked machine parts.

Question 5.
Where do we use handpicking and washing methods in our daily life? Give examples. How do you correlate these examples with enrichment of ore?
Answer:
We use hand picking in separating stones from rice and da I. We use washing methods to separate dust from rice, dal, vegetables, fruits etc.

Hand-picking: The colour and size of impurities is different from rice or dal. So we can easily separate them by hand-picking. In the same way if the ore particles and the impurities are n different sizes, colour etc., we can see this hand-picking method to separate ore from impurities.

Washing: Less-density particles like dust is separated from more density particles like rice, vegetables etc., by washing. In the same way ore particles are crushed and kept on a slopy surface. They are washed with a controlled flow of water. Less sensitive impurities are carried away by water flow, leaving the more sensitive ore particles behind.

Question 6.
What is activity series? How it helps in extraction of metals?
Answer:
Activity series: The arrangement of the metals in decreasing order of their reactivity is known as ‘activity series’.
Use of Activity series in extraction of metals
1. The method used for a particular metal for the reduction of its ore to the metal depends mainly on the position of the metal in the activity series.
Eg:
The metals at the top of the activity series (highly reactive) can be extracted by electrolysis.
The metals at the middle of the activity series can be extracted by

• reduction of metal oxide with carbon
• reduction of oxide ores with Co,
• self-reduction of sulphide ores
• reduction of ores with more reactive metals (thermite process).

3. The metals at the bottom of the activity series (less reactive) can be extracted by heating along, and displacement from their aqua solution.

Multiple choice questions

Question 1.
The impurity present in the ore is called as [ ]
(a) Ganguc
(b) fluid
(c) Slag
(d) Mineral
Answer:
(a) Ganguc

Question 2.
Which of the following is a carbonate ore? [ ]
(a) Magncsitc
(b) Bauxitc
(c) Gypsum
(d) tiaicna
Answer:
(a) Magncsitc

Question 3.
Which of the following is the corrcct formula of Gypsum [ ]
(a) CuSO₄. 2H₂O
(b) CaSO₄. 1/2H₂O
(c) CuSO₄. 5H₂O
(d) CaSO₄. 2H₂O
Answer:
(d) CaSO₄. 2H₂O

Question 4.
The oil used in the froth floatation process is [ ]
(a) kerosene oil
(b) pencil
(c) coconut oil
(d) olive coil.
Answer:
(b) pencil

Question 5.
Froth floatation is method used for the purification of …………………….. ore. [ ]
(a) sulphide
(b) oxide
(c) carbonate
(d) nitrate
Answer:
(a) sulphide

Question 6.
Galena is an ore of [ ]
(a) Zn
(b) Pb
(c) Hg
(d) Al
Answer:
(b) Pb

Question 7.
The metal that occurs in the native form is [ ]
(a) Pb
(b) Au
(c) Fc
(d) Hg
Answer:
(b) Au

Question 8.
The most abundant metal in th earth’s cruat is [ ]
(a) Silver
(b) Aluminium
(c) zinc
(d) iron
Answer:
(b) Aluminium

Question 9.
The reducing agent in thermite process is [ ]
(a) Al
(b) Mg
(c) Fe
(d) Si
Answer:
(a) Al

Question 10.
The purpose of smelting an ore is [ ]
(a) Oxidisc
(b) Reduce
(c) Neutndisc
(d) one of these
Answer:
(b) Reduce

Suggested Experiments

Question 1.
Suggest an experiment to prove that the presence of air and water are essential for corrosion. Explain the procedure.
Answer:
Corrosion: Corrosion is the deterioration of a metal, as a result of chemical reaction between it and the surrounding environment.

Experiment:
Aim: To prove that the presence of air and water are essential for corrosion. Materials required : 3 test tubes, 3 iron nails, oil, water, anhydrous calcium chloride, rubber corks.

Procedure:
1. Take three test tubes and place clean iron nails in each of them.
2. Label these tests tubes A, B and C. Pour some water in test tube A and cork it.
3. Pour boiled distilled water in test tube B, add about 1 ml of oil and cork it. The oil will float on water and prevent the air from dissolving in the water.
4. Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture.
5. Leave these test tubes for a few days and then observe.

6. We will observe that ¡ron nails rust in test tube A, but they do not rust in test tubes B and C.
7. In the test tube A the nails are exposed to both air and water whereas in the test tube ‘B’ the nails are exposed to only water and in the test tube ‘C’ the nails are exposed to dry air.
8. This shows air and water are essential for corrosion.

Suggested Projects

Question 1.
Collect information about extraction of metals of low reactivity silver, platinum and gold and prepare a report.
Answer:
Extraction of Silver:

1. Silver occurs both in combined state as well as In free state. The important ores of silver are Argentite (or) Silver glance (Ag₂S) Pyrargyrite (or) Ruby silver 3 Ag₂S Sb₂S₃ silver copper glance (CuAg)₂S
2. Silver is extracted from the ore-Argentite (Ag₂S).
3. The process of extraction of silver is called cyanide process, as sodium cyanide solution is used.
4. The ore Is crushed, concentrated and then treated with sodium cyanide solution.
5. This reaction forms sodium argent cyanide [Na[Ag(CN)₂]]
   Ag₂S + 4 NaCN → 2Na[Ag(CN)₂] + Na₂S
6. This solution of sodium argent cyanide combines with zinc dust and forms tetra cyano zincate and precipitated silver. This precipitated silver is called spongy silver.
   Zn + 2Na[Ag(CN)₂] → Na₂[Zn(CN)₄)] + 2Ag
7. This spongy silver is fused with potassium nitrate to obtain pure silver. Then the silver obtained is purified by electrolytic process.

Extraction of platinum

• Platinum is rarely found on its own, but In combination with other base and precious metals.
• The extraction process of platinum is a complex process which includes milling the ore and smelting at high temperatures. This removes base metals notably sulphur and concentrates PGM (Platinum Group Metals) – Gold, Platinum and Palladium.
• The PGM matter is further processed by electrolysis to remove Nickel, Cobalt and Copper.
• The high-grade concentrate is treated by solvent extraction, distilling, and ion- exchange treatments to separate the PGMs Into its separated metals.

Extraction of Gold:

1. Gold is usually found alone or alloyed with mercury or silver.
2. In all methods of gold ore refining, the ore is usually washed and filtered at the mine, then sent to the mill. At the mill, the ore is ground into smaller particles with water, then ground again in a ball mill to further pulverize the ore.
3. Several processes can be used to separate the Gold from its ore. They are:

a) Cyanide process:

• The ground ore Is put In a tank containing a weak cyanide solution and zinc is added.
• The zinc causes a chemical reaction which separates the gold from the ore.
• The gold is then removed from the solution with a filter press.

b) Carbon-in-pulp method:

1. In this method, the ground ore is mixed with water before cyanide Is added. Then carbon is added to bond with the gold.
2. The carbon-gold particles are put into a caustic carbon solution, separating out the gold.

c) Heap leaching:

1. The ore is placed on open-air pads and cyanide is sprayed over It, taking several weeks to leach down to an imperious base.
2. The solution then pours off and pad into a pond and is pumped from there to a recovery plant, where the gold is recovered.
3. Heap-leaching helps recover gold from ore that would otherwise be to expensive to process.

TS 10th Class Physical Science Principles of Metallurgy Intext Questions

Page 237

Question 1.
Can you mention some articles that are made up of metals?
Answer:
Utensils in kitchen, window gnUs, pots, chairs, iron gates, bodies of motor cars, etc.

Question 2.
Do metals exist in nature in the same form as that we use ¡n our daily life?
Answer:
No, metals do not exist in nature ¡n the form same as that we use in our daily life.

Question 3.
Have you ever heard the words like ore, mineral and metallurgy?
Answer:
Yes.

Question 4.
Do you know how these metals are obtained?
Answer:
The metals are extracted from their ores mainly in three stages.

1. The concentration of ore.
2. Extraction of crude metal.
3. Refining of the metal.

Question 5.
How the metals are present In nature?
Answer:
The metals are present in nature In combined form as their compounds.

Page 239

Question 6.
What metals can we get from the ore mentioned in Table-1?
Answer:

Question 7.
Can you arrange these metals in the order of their reactivity.
Answer:
K>Na>Ca>Mg>Al >Zn>Fe>Pb>Cu>Ag>Au

Question 8.
What do you notice In table-2?
Answer:
I noticed that the ores of many metals are oxides and sulphides.

Question 9.
Can you think how do we get these metals from their ores?
Answer:
These metals are obtained from their ores by suitable metallurgical processes.

Question 10.
Does the reactivity of a metal and form of its ore (oxides, sulphides, chlorides, carbonates, sulphates) has any relation with process of extraction?
Answer:
Yes, highly reactive metals are obtained by electrolysis from their molten salts, moderately reactive metals are obtained by reducing with suitable reagents whereas least reactive metals are available in native form.

Question 11.
How are metals extracted from mineral ores?
Answer:
Metals are extracted from mineral ores In three steps.

1. Concentration of ore or Dressing.
2. Extraction of crude metal.
3. Refining of the metal.

Question 12.
What methods are to be used?
Answer:
Hand-picking, Washing, Froth floatation and Magnetic Separation methods are to be used.

Page 247

Question 13.
Do you know why corrosion occurs?
Answer:
Metals are stable in the form In which they are available in nature. So by means of corrosion, they change to the form as they occur n nature.

Question 14.
What does this tell us about the conditions under which iron articles rust?
Answer:
Corrosion of iron ( commonly known as rusting ) occurs in presence of moisture and air.

Page 250

Question 15.
What is the role of furnace in metallurgy?
Answer:
A furnace is used for heating the ores and crude metals to required temperatures in metallurgical operations.

Question 16.
How they bear large amounts of heat?
Answer:
The furnaces are lined inside with refractory materials and hence they can bear large amounts of heat.
The substances which are capable of withstanding very high temperatures without melting or becoming soft are called refractory materials.

Question 17.
Do all furnaces have same structure?
Answer:
No, all furnaces do not have same structure.

TS 10th Class Physical Science Principles of Metallurgy Activities

Activity 1

Question 1.
How do you classify ores based on their formula?
1) Look at the following ores.
2) Identify the metal present in each ore.

Now classify them as shown in the table.
Answer:
Classification of ores as oxides, sulphides, chlorides, carbonates, and sulphates is done as follows.

Activity 2

Question 2.
Show that both air and water are necessary for corrosion of iron.
Answer:

1. Take three test tubes and place clean iron nails in each of them.
2. Label these test tubes as A, B and C. Pour cork it.
3. Pour boiled distilled water in test tube. B Add about 1 ml of oil and cork it. The oil will float on water and prevent the air from dissolving in the water.
4. Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture if any from the air. Leave these test tubes for a few days and then observe.

Observation: Iron nails rust in test tube A but they do not rust in test tubes B and C. In the test tube A the nails are exposed to both air and water. In the test tube B the nails are exposed to distilled water and the nails in test tube C are exposed to dry air only.

Inference: From this we can conclude that both air and water are necessary for corrosion of iron.