Category: Class 10

Students can download Maths Chapter 6 Trigonometry Ex 6.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Multiple Choice Questions

Question 1.
The value of sin2 θ + \(\frac{1}{1+\tan ^{2} \theta}\) is equal to ………………
(1) tan² θ
(2) 1
(3) cot² θ
(4) 0
Answer:
(2) 1
Hint:
sin² θ + \(\frac{1}{1+\tan ^{2} \theta}\) = sin² θ + \(\frac{1}{\sec ^{2} \theta}\) = sin² θ + cos² θ = 1

Question 2.
tan θ cosec² θ – tan θ is equal to ………………
(1) sec θ
(2) cot² θ
(3) sin θ
(4) cot θ
Answer:
(4) cot θ
Hint:
tan θ cosec² θ – tan θ = tan θ (cosec² θ – 1)
= tan θ × cot² θ = \(\frac{1}{\cot \theta}\) × cot² θ = cot θ

Question 3.
If (sin α + cosec α)² + (cos α + sec α)² = k + tan² α + cot² α, then the value of k is equal to
(1) 9
(2) 7
(3) 5
(4) 3
Solution:
(2) 7
(sin α + cos α)² + (cos α + sec α)²
= sin² α + cosec² α + 2 sin α cosec α + cos² α + sec² α + 2 cos α sec α
= 1 + cosec² α + 2 + sec² α + 2
= 1 + cot² α + 1 + 2 + tan² α + 1 + 2
= 7 + tan² α + cot² α
k = 7

Question 4.
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b (a2 – 1) is equal to ……………
(1) 2 a
(2) 3 a
(3) 0
(4) 2 ab
Answer:
(1) 2 a
Hint:
b (a² – 1) = (sec θ + cosec θ) [(sin θ + cos θ)² – 1]
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\) [sin² θ + cos² θ + 2 sin θ cos θ – 1]

Question 5.
If 5x = sec θ and \(\frac { 5 }{ x } \) = tan θ, then x2 – \(\frac{1}{x^{2}}\) is equal to …………….
(1) 25
(2) \(\frac { 1 }{ 25 } \)
(3) 5
(4) 1
Answer:
(2) \(\frac { 1 }{ 25 } \)
Hint:



Question 6.
If sin θ = cos θ , then 2 tan² θ + sin² θ – 1 is equal to ………………
(1) \(\frac { -3 }{ 2 } \)
(2) \(\frac { 3 }{ 2 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) \(\frac { -2 }{ 3 } \)
Answer:
(2) \(\frac { 3 }{ 2 } \)
Hint:

Question 7.
If x = a tan θ and y = b sec θ then …………..
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
(2) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
(3) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0\)
(4) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=0\)
Answer:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
Hint:
x = a tan θ
\(\frac { x }{ a } \) = tan θ
\(\frac{x^{2}}{a^{2}}\) = tan² θ
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = sec² θ – tan² θ = 1
y = b sec θ
\(\frac{y}{b}\) = sec θ
\(\frac{y^{2}}{b^{2}}\) = sec² θ

Question 8.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) is equal to ……………
(1) 0
(2) 1
(3) 2
(4) -1
Answer:
(3) 2
Hint:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Question 9.
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p² – q² is equal to
(1) a² – b²
(2) b² – a²
(3) a² + b²
(4) b-a
Solution:
(2) b² – a²
(a cot θ + b cosec θ)² = p²
(b cot θ + a cosec θ )² = q²
p² – q² = a² cost²θ + a² cot²θ + 2ab cot θ cosec θ – (b²cot²θ + a² cosec²θ + 2ab cot θ cosec θ) = (a² – b²) cot²θ + (b² – a²)cosec²θ = (a² – b²) (cosec²θ – 1) + (b² – a²) (cosec²θ)
= (a² – b²)cosec²θ – (a² – b²) – (a² – b²) cosec²θ
= b² – a²

Question 10.
If the ratio of the height of a tower and the length of its shadow is \(\sqrt { 3 }\) : 1, then the angle of elevation of the sun has a measure
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Answer:
(4) 60°
Hint:
Ratio of length of the tower : length of the shadow = \(\sqrt { 3 }\) : 1

Let the tower be \(\sqrt { 3 }\) x and the shadow be x
tan C = \(\frac { AB }{ BC } \) ⇒ tan C = \(\frac{\sqrt{3} x}{x}\) = \(\sqrt { 3 }\)
tan C = tan 60° ⇒ ∴ ∠C = 60°

Question 11.
The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘6’ metres above the first, the depression of the foot of the tower is 60° . The height of the tower (in metres) is equal to ……………
(1) \(\sqrt { 3 }\) b
(2) \(\frac { b }{ 3 } \)
(3) \(\frac { b }{ 2 } \)
(4) \(\frac{b}{\sqrt{3}}\)
Answer:
(3) \(\frac { b }{ 2 } \)
Hint:
Let the height of the pole BC be h
AC = b + h
Let CD be x
In the right ∆ BCD, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ x } \)
x = \(\sqrt { 3 }\) h ………. (1)

In the right ∆ ACD, tan 60° = \(\frac { AC }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { b+h }{ x } \)
x = \(\frac{b+h}{\sqrt{3}}\) ………(2)
From (1) and (2) we get
\(\sqrt { 3 }\) h = \(\frac{b+h}{\sqrt{3}}\) ⇒ 3 h = b + h
2 h = b ⇒ h = \(\frac { b }{ 2 } \)

Question 12.
A tower is 60 m height. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30° , then x is equal to
(1) 41. 92 m
(2) 43. 92 m
(3) 43 m
(4) 45. 6 m
Answer:
(2) 43. 92 m
Hint:
In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \) = \(\frac { 60 }{ x+y } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 60 }{ x+y } \) ⇒ x + y = 60 \(\sqrt { 3 }\)

y = 60 \(\sqrt { 3 }\) – x …….(1)
In the right ∆ ABD, tan 45° = \(\frac { AB }{ BD } \)
1 = \(\frac { 60 }{ y } \) ⇒ y = 60 ………..(2)
From (1) and (2) we get
60 = 60 \(\sqrt { 3 }\) – x
x = 60 \(\sqrt { 3 }\) – 60 = 60 (\(\sqrt { 3 }\) – 1) = 60 (1.732 – 1)
= 60 × 0.732
x = 43.92 m

Question 13.
The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is …………….
(1) 20,10\(\sqrt { 3 }\)
(2) 30, 5 \(\sqrt { 3 }\)
(3) 20, 10
(4) 30, 10\(\sqrt { 3 }\)
Answer:
(4) 30, 10\(\sqrt { 3 }\)
Hint:
Let the height of the multistoried building AB be “h”
AE = h – 20
Let BC be x
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \) ⇒ \(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ………..(1)

In the right ∆ ABC, tan 30° = \(\frac { AE }{ ED } \) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
x = (h – 20) \(\sqrt { 3 }\) ………(2)
From (1) and (2) we get,
\(\frac{h}{\sqrt{3}}\) = (h – 20) \(\sqrt { 3 }\)
h = 3h – 60 ⇒ 60 = 2 h
h = \(\frac { 60 }{ 2 } \) = 30
Distance between the building (x) = \(\frac{h}{\sqrt{3}}=\frac{30}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}=10 \sqrt{3}\)

Question 14.
Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is ……………….
(1) \(\sqrt { 2 }\)x
(2) \(\frac{x}{2 \sqrt{2}}\)
(3) \(\frac{x}{\sqrt{2}}\)
(4) 2 x
Answer:
(2) \(\frac{x}{2 \sqrt{2}}\)
Hint:
Consider the height of the 2^nd person ED be “h”
Height of the second person is 2 h
C is the mid point of BD
In the right ∆ ABC, tan θ = \(\frac { AB }{ BC } \)

Question 15.
The angle of elevation of a cloud from a point h metres above a lake is β . The angle of depression of its reflection in the lake is 45° . The height of the location of the cloud from the lake is ………….

Students can download Maths Chapter 6 Trigonometry Ex 6.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 \(\sqrt { 3 }\) m.
Answer:
Height of the tower (AC) = 10 \(\sqrt { 3 }\) m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ
In the right ∆ ABC, tan θ = \(\frac { AC }{ AB } \)
= \(\frac{10 \sqrt{3}}{30}=\frac{\sqrt{3}}{3}\)

tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle of inclination is 30°

Question 2.
A road is flanked on either side by continuous rows of houses of height 4\(\sqrt { 3 }\) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the width of the road.
Answer:
Let the mid point of the road AB is “P” (PA = PB)
Height of the home = 4\(\sqrt { 3 }\) m
Let the distance between the pedestrian and the house be “x”
In the right ∆ APD, tan 30° = \(\frac { AD }{ AP } \)

\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x}\)
x = 4 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 12 m
∴ Width of the road = PA + PB
= 12 + 12
= 24 m

Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the window FE be “h” m
Let FC be “x” m
∴ EC = (h + x) m

In the right ∆ CDF, tan 45° = \(\frac { CE }{ CD } \)
1 = \(\frac { x }{ 5 } \) ⇒ x = 5
In the right ∆ CDE, tan 60° = \(\frac { CE }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { x+h }{ 5 } \) ⇒ x + h = 5\(\sqrt { 3 }\)
5 + h = 5 \(\sqrt { 3 }\) (substitute the value of x)
h = 5 \(\sqrt { 3 }\) – 5 = 5 × 1.732 – 5 = 8. 66 – 5 = 3.66
∴ Height of the window = 3.66 m

Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40° . Find the height of the pedestal.
(tan 40° = 0.8391, \(\sqrt { 3 }\) = 1.732)
Answer:
Height of the statue = 1.6 m
Let the height of the pedestal be “h”
AD = H + 1.6m
Let AB be x
In the right ∆ ABD, tan 60° = \(\frac { AD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h+1.6 }{ x } \)
x = \(\frac{h+1.6}{\sqrt{3}}\) ……..(1)

In the right ∆ ABC, tan 40° = \(\frac { AC }{ AB } \)
0.8391 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.8391 } \)
Substitute the value of x in (1)
\(\frac{h}{0.8391}=\frac{h+1.6}{\sqrt{3}}\)
(h + 1.6) 0.8391 = \(\sqrt { 3 }\) h
0.8391 h + 1.34 = 1.732 h
1.34 = 1.732 h – 0.8391 h
1.34 = 0.89 h
h = \(\frac { 1.34 }{ 0.89 } \) = \(\frac { 134 }{ 89 } \) = 1.5 m
Height of the pedestal = 1.5 m

Question 5.
A Flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (\(\sqrt { 3 }\) = 1.732)

Answer:
Height of the Flag pole (ED) = h m
AF and AD is the radius of the semi circle (r)
AC = (r + 7)
AB = (r + 7 + 5)
= (r + 12)
In the right ∆ ABD, tan 30° = \(\frac { AD }{ AB } \)

\(\frac{1}{\sqrt{3}}=\frac{r}{r+12}\)
\(\sqrt { 3 }\) r = r + 12
\(\sqrt { 3 }\) r – r = 12 ⇒ r (\(\sqrt { 3 }\) – 1) = 12
r[1.732 – 1] = 12 ⇒ 0.732 r = 12
r = \(\frac { 12 }{ 0.732 } \) ⇒ = 16.39 m
In the right ∆ ACE, tan 45° = \(\frac { AE }{ AC } \)
1 + \(\frac { r+h }{ r+7 } \)
r + 7 = r + h
∴ h = 7 m
Height of the pole (h) = 7 m
Radius of the dome (r) = 16.39 m

Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer:
Let the height of the electric pole AD be “h” m
EC = 15 – h m
Let AB be “x”

In the right ∆ ABC, tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 15 }{ x } \)
x = \(=\frac{15}{\sqrt{3}}=\frac{15 \times \sqrt{3}}{3}\)
= 5\(\sqrt { 3 }\)
In the right ∆ CDE, tan 30° = \(\frac { EC }{ DE } \)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{x}\) ………….(1)
Substitute the value of x = 5 \(\sqrt { 3 }\) in (1)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{5 \sqrt{3}} \Rightarrow \sqrt{3}(15-h)=5 \sqrt{3}\)
(15 – h) = \(\frac{5 \sqrt{3}}{\sqrt{3}}\) ⇒ 15 – h = 5
h = 15 – 5 = 10
∴ Height of the electric pole = 10 m

Question 7.
A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Answer:
Let the first part of the pole be “x” and the second part be “9x”
∴ height of the pole (AC) = x + 9x = 10x
Given ∠CDB = ∠BDA
∴ BD is the angle bisector of ∠ADC
By angle bisector theorem

\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
\(\frac { 9x }{ x } \) = \(\frac { AD }{ 25 } \) ⇒ AD = 9 × 25 = 225
In the right ∆ ACD
AD² = AC² + CD²
(225)² = (10x)² + 25²
50625 = 100x² + 625
∴ 100x² = 50625 – 625 = 50000
x² = \(\frac { 50000 }{ 100 } \) = 500
x = \(\sqrt { 500 }\) = \(\sqrt{5 \times 100}=10 \sqrt{5}\)
∴ AC = 10 × 10\(\sqrt { 5 }\) = 100 \(\sqrt { 5 }\) (AC = 10x)
∴ Height of the pole = 100 \(\sqrt { 5 }\) m

Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405)
Answer:
Let the height of the peak be “h” mile. Let AD be x mile.
∴ AB = (x + 1) mile.
In the right ∆ ADC, tan 8° = \(\frac { AC }{ AC } \)

0.1405 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.1405 } \) ………..(1)
In ∆ ABC, tan 4° = \(\frac { AC }{ AB } \)
0.0699 = \(\frac { h }{ x+1 } \) ⇒ (x + 1) 0.0699 = h
0.0699x + 0.0699 = h
0.0699 x = h – 0.0699
x = \(\frac { h-0.0699 }{ 0.0699 } \) ………(2)
Equation (1) and (2) we get,
\(\frac { h-0.0699 }{ 0.0699 } \) = \(\frac { h }{ 0.1405 } \)
0.0699 h = 0.1405 (h – 0.0699)
0.0699 h = 0.1405 h – 0.0098
0.0098 = 0.1405 h – 0.0699 h
0.0098 = 0.0706 h
h = \(\frac { 0.0098 }{ 0.0706 } \) = \(\frac { 98 }{ 706 } \) = 0.1388
= 0.14 mile (approximately)
Height of the peak = 0.14 mile

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Multiple Choice Questions

Question 1.
The area of triangle formed by the points (-5, 0), (0, – 5) and (5, 0) is …………..
(1) 0 sq.units
(2) 25 sq.units
(3) 5 sq.units
(4) none of these
Answer:
(2) 25 sq.units Hint.
Hint:
Area of the ∆

Question 2.
A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is …………
(1) x = 10
(2) y = 10
(3) x = 0
(4) y = 0
Answer:
(1) x = 10
Hint:

Question 3.
The straight line given by the equation x = 11 is …………….
(1) parallel to X axis
(2) parallel to Y axis
(3) passing through the origin
(4) passing through the point (0,11)
Answer:
(2) parallel to Y axis

Question 4.
If (5,7), (3,p) and (6,6) are collinear, then the value of p is ……………
(1) 3
(2) 6
(3) 9
(4) 12
Answer:
(3) 9
Hint:
Since the three points are collinear. Area of a triangle is 0

5p + 18 + 42 – (21 + 6p + 30) = 0
5p + 60 – (51 + 6p) = 0
5p + 60 – 51 – 6p = 0
-p + 9 = 0
-p = -9
p = 9

Question 5.
The point of intersection of 3x – y = 4 and x + 7 = 8 is ……………
(1) (5,3)
(2) (2,4)
(3) (3,5)
(4) (4, 4)
Answer:
(3) (3, 5)

Substitute the value of x = 3 in (2)
3 + 7 = 8
y = 8 – 3 = 5
The point of intersection is (3, 5)

Question 6.
The slope of the line joining (12, 3), (4, a) is \(\frac { 1 }{ 8 } \). The value of ‘a’ is …………….
(1) 1
(2) 4
(3) -5
(4) 2
Answer:
(4) 2
Hint:
Slope of a line = \(\frac { 1 }{ 8 } \)

Question 7.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (- 8, 8) is ………..
(1) -1
(2) 1
(3) \(\frac { 1 }{ 3 } \)
(4) -8
Answer:
(2) 1
Hint:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-0 }{ -8-0 } \) = \(\frac { 8 }{ -8 } \) = -1
Slope of the Perpendicular = 1

Question 8.
If slope of the line PQ is \(\frac{1}{\sqrt{3}}\) then slope of the perpendicular bisector of PQ is …………..
(1) \(\sqrt { 3 }\)
(2) –\(\sqrt { 3 }\)
(3) \(\frac{1}{\sqrt{3}}\)
(4) 0
Answer:
(2) –\(\sqrt { 3 }\)
Hint:
Slope of a line = \(\frac{1}{\sqrt{3}}\)
Slope of the ⊥r bisector = –\(\sqrt { 3 }\)

Question 9.
If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is ……………
(1) 8x + 5y = 40
(2) 8x – 5y = 40
(3) x = 8
(4) y = 5
Answer:
(1) 8x + 5y = 40
Hint:
Let the point A be (0, 8) and B (5, 0)

Question 10.
The equation of a line passing through the origin and perpendicular to the line lx -3y + 4 = 0 is
(1) 7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Answer:
(3) 3x + 7y = 0
Hint:
Slope of the line of 7x – 3y + 4 = 0
= \(\frac { -7 }{ -3 } \) = \(\frac { 7 }{ 3 } \)
Slope of its ⊥^r = \(\frac { -3 }{ 7 } \)
The line passes through (0,0)
Equation of a line is
y – y₁ = m(x – x₁)
y – 0 = \(\frac { -3 }{ 7 } \) (x – 0)
y = \(\frac { -3 }{ 7 } \) x ⇒ 7y = -3x
3x + 7y = 0

Question 11.
Consider four straight lines
(i) l₁ : 3y = 4x + 5
(ii) l₂ : 4y = 3x – 1
(iii) l₃ : 4y + 3x = 7
(iv) l₄ : 4x + 3y = 2
Which of the following statement is true?
(1) l₁ and l₂ are perpendicular
(2) l₂ and l₄ are parallel
(3) l₂ and l₄ are perpendicular
(4) l₂ and l₃ are parallel
Answer:
(3) l₂ and l₄ are perpendicular
Hint:
Slope of l₁ = \(\frac { 4 }{ 3 } \); Slope of l₂ = \(\frac { 3 }{ 4 } \)
Slope of l₃ = – \(\frac { 3 }{ 4 } \); Slope of l₄ = –\(\frac { 4 }{ 3 } \)
(1) l₁ × l₂ = \(\frac { 4 }{ 3 } \) × \(\frac { 3 }{ 4 } \) = 1 …….False
(2) l₁ = \(\frac { 4 }{ 3 } \); l₄ = – \(\frac { 4 }{ 3 } \) not parallel ………False
(3) l₂ × l₄ = \(\frac { 3 }{ 4 } \) × – \(\frac { 4 }{ 3 } \) = -1 …….True
(4) l₂ = \(\frac { 3 }{ 4 } \); l₃ = – \(\frac { 3 }{ 4 } \) not parallel ………False

Question 12.
A straight line has equation 87 = 4x + 21. Which of the following is true …………………….
(1) The slope is 0.5 and the y intercept is 2.6
(2) The slope is 5 and the y intercept is 1.6
(3) The slope is 0.5 and they intercept is 1.6
(4) The slope is 5 and the y intercept is 2.6
Answer:
(1) The slope is 0.5 and they intercept is 2.6
Hint:
8y = 4x + 21
y = \(\frac { 4 }{ 8 } \) x + \(\frac { 21 }{ 8 } \)
= \(\frac { 1 }{ 2 } \) x + \(\frac { 21 }{ 8 } \)
\(\frac { 1 }{ 2 } \) = 0.5
\(\frac { 21 }{ 8 } \) = 2.625
Slope = \(\frac { 1 }{ 2 } \) = 0.5
y intercept = \(\frac { 21 }{ 8 } \) = 2.6

Question 13.
When proving that a quadrilateral is a trapezium, it is necessary to show
(1) Two sides are parallel.
(2) Two parallel and two non-parallel sides.
(3) Opposite sides are parallel.
(4) All sides are of equal length.
Solution:
(2) Two parallel and two non-parallel sides.

Question 14.
When proving that a quadrilateral is a parallelogram by using slopes you must find …………………
(1) The slopes of two sides
(2) The slopes of two pair of opposite sides
(3) The lengths of all sides
(4) Both the lengths and slopes of two sides
Answer:
(2) The slopes of two pair of opposite sides

Question 15.
(2,1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 7 = 0
Solution:
(2) x + y = 3; 3x + y = 7

Students can download 10th Science Chapter 2 Optics Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 2 Optics

Samacheer Kalvi 10th Science Optics Text Book Back Questions and Answers

I. Choose the correct answer.

Question 1.
The refractive index of four substances A, B, C and D are 1.31,1.43,1.33, 2.4 respectively. The speed of light is maximum in:
(a) A
(b) B
(c) C
(d) D
Answer:
(a) A

Question 2.
Where should an object be placed so that a real and inverted image of same size is obtained by a convex lens:
(a) f
(b) 2f
(c) infinity
(d) between f and 2f
Answer:
(b) 2f

Question 3.
Where should an object be placed so that a real and inverted image of the same size is obtained by a convex lens ______.
(a) f
(b) 2f
(c) infinity
(d) between f and 2f.
Answer:
(b) 2f

Question 4.
Magnification of a convex lens is _____.
(a) positive
(b) negative
(c) either positive or negative
(d) zero.
Answer:
(b) negative

Question 5.
A convex lens forms a real, diminished point sized image at focus. Then the position of the object is at:
(a) focus
(b) infinity
(c) at 2f
(d) between f and 2f
Answer:
(b) infinity

Question 6.
Power of a lens is -4D, then its focal length is:
(a) 4 m
(b) -40 m
(c) -0.25 m
(d) -2.5 m
Answer:
(d) -2.5 m

Question 7.
In a myopic eye, the image of the object is formed _____.
(a) behind the retina
(b) on the retina
(c) in front of the retina
(d) on the blind spot.
Answer:
(c) in front of the retina

Question 8.
The eye defect ‘presbyopia’ can be corrected by:
(a) convex lens
(b) concave lens
(c) convex mirror
(d) Bi focal lenses
Answer:
(d) Bi focal lenses

Question 9.
Which of the following lens would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 5 cm
(b) A concave lens of focal length 5 cm
(c) A convex lens of focal length 10 cm
(d) A concave lens of focal length 10 cm
Answer:
(d) A concave lens of focal length 10 cm

Question 10.
If V_B, V_G, V_R be the velocity of blue, green and red light respectively in a glass prism, then which of the following statement gives the correct relation?
(a) V_B = V_G = V_R
(b) V_B > V_G > V_R
(c) V_B < V_G < V_R
(d) V_B < V_G > V_R
Answer:
(c) V_B < V_G < V_R

II. Fill in the blanks.

1. The path of the light is called as ………
2. The refractive index of a transparent medium is always greater than ……….
3. If the energy of incident beam and the scattered beam are same, then the ………. scattering of light is called as scattering ……….
4. According to Rayleigh’s scattering law, the amount of scattering of light is inversely proportional to the fourth power of its ……….
5. Amount of light entering into the eye is controlled by ……….

Answer:

1. ray
2. unity
3. elastic
4. wavelength
5. iris

III. True or False. If false correct it.

1. Velocity of light is greater in denser medium than in rarer medium.
2. The power of lens depends on the focal length of the lens.
3. Increase in the converging power of eye lens cause ‘hypermetropia’
4. The convex lens always gives small virtual image.

Answer:

1. False – Velocity of light is greater in rarer medium than in denser medium.
2. True
3. True
4. False – The convex lens does not give small virtual image always.

IV. Match the following.

Answer:
1. d
2. a
3. e
4. b
5. c

V. Assertion and reasoning type.

Mark the correct choice as-
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
1. Assertion: If the refractive index of the medium is high (denser medium) the velocity of the light in that medium will be small
Reason: Refractive index of the medium is inversely proportional to the velocity of the light.

2. Assertion: Myopia is due to the increase in the converging power of eye lens.
Reason: Myopia can be corrected with the help of concave lens.
Answer:
1. (a)
2. (a)

VI. Answer Briefly.

Question 1.
What is refractive index?
Answer:
Refractive index gives us an idea of how fast or how slow light travels in a medium.

Question 2.
State Snell’s law.
Answer:
The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
\(\frac{sin i}{sin r}\) = \(\frac{µ_2}{µ_1}\)

Question 3.
Draw a ray diagram to show the image formed by a convex lens when the object is placed between F and 2F.
Answer:

Question 4.
Define dispersion of light.
Answer:
When a beam of white light or composite light is refracted through any transparent media such as glass or water, it is split into its component colours. This phenomenon is called as ‘dispersion of light’.

Question 5.
State Rayleigh’s law of scattering.
Answer:
Rayleigh’s scattering law states that, “The amount of scattering of light is inversely proportional to the fourth power of its wavelength”.
Amount of scattering ‘S’ ∝\(\frac{1}{λ^4}\)

Question 6.
Differentiate convex lens and concave lens.
Answer:

Question 7.
What is the power of accommodation of the eye?
Answer:

• The ability of the eye lens to focus nearby as well as the distant objects is called the power of accommodation of the eye.
• This is achieved by changing the focal length of the eye lens with the help of ciliary muscles.

Question 8.
What are the causes of ‘Myopia’?
Answer:

1. The lengthening of eye ball.
2. The focal length of eye lens is reduced.
3. The distance between eye lens and retina increases.
4. The far point will not be at infinity.
5. The far point comes closer.

Question 9.
Why does the sky appear blue in colour?
Answer:
When sunlight passes through the atmosphere, the blue colour (shorter wavelength) is scattered to a greater extent than the red colour (longer wavelength). This scattering causes the sky to appear blue in colour.

Question 10.
Why are traffic signals red in colour?
Answer:

• Red light has the highest wavelength.
• It is scattered by atmospheric particles.
• So red light is able to travel the longest distance through a fog, rain etc.

VII. Give the answer in detail.

Question 1.
List any five properties of light?
Answer:

• Light is a form of energy.
• Light always travels along a straight line.
• Light does not need any medium for its propagation. It can even travel through a vacuum.
• The speed of light in vacuum or air is, c = 3 × 10⁸ ms^-1
• Since light is in the form of waves, it is characterized by a wavelength (λ) and a frequency (v), which are related by the following equation: c = vλ (c = velocity of light).
• Different coloured light has a different wavelength and frequency.

Question 2.
Explain the rules for obtaining images formed by a convex lens with the help of ray diagram.
Answer:
Rule-1: When a ray of light strikes the convex or concave lens obliquely at its optical centre, it continues to follow its path without any deviation.

Rule-2: When rays parallel to the principal axis strikes a convex or concave lens, the refracted rays are converged to (convex lens) or appear to diverge from (concave lens) the principal focus.

Rule-3: When a ray passing through (convex lens) or directed towards (concave lens) the principal focus strikes a convex or concave lens, the refracted ray will be parallel to the principal axis.

Question 3.
Differentiate the eye defects: Myopia and Hypermetropia.
Answer:

Question 4.
Explain the construction and working of a ‘Compound Microscope’.
Answer:
Construction : A compound microscope consists of two convex lenses. The lens with the shorter focal length is placed near the object, and is called as ‘objective lens’ or ‘objective piece’. The lens with larger focal length and larger aperture placed near the observer’s eye is called as ‘eye lens’ or ‘eye piece’. Both the lenses are fixed in a narrow tube with adjustable provision.

Working : The object (AB) is placed at a distance slightly greater than the focal length of objective lens (u > F₀). A real, inverted and magnified image (A’B’) is formed at the other side of the objective lens. This image behaves as the object for the eye lens. The position of the eye lens is adjusted in such a way, that the image (B’B’) falls within the principal focus of the eye piece. This eye piece forms a virtual, enlarged and erect image (A”B”) on the same side of the object.

Compound microscope has 50 to 200 times more magnification power than simple microscope.

VIII. Numerical Problems.

Question 1.
An object is placed at a distance 20 cm from a convex lens of focal length 10 cm. Find the image distance and nature of the image.
Answer:
Distance of an object u = 20 cm
Focal length of a convex lens f = 10 cm
Let the image distance be v
We know

v = 20 cm
Magnification m = \(\frac{v}{u}\) = \(\frac{20}{20}\) = 1
Hence a real image of same size is formed at 20 cm.
Image distance = 20 cm

Question 2.
An object of height 3 cm is placed at 10 cm from a concave lens of focal length 15 cm. Find the size of the image.
Answer:
Object distance u = 10 cm
Focal length of a concave lens f= -15 cm
Let v be the image distance,

Distance of image v = 6 cm
Magnification m = \(\frac{v}{u}\) = \(\frac{6}{10}\) = 0.6
And Magnification m = \(\frac{h’}{h}\)
Where h’ – height of image
h – height of object
0.6 = \(\frac{h’}{3}\)
∴ h’ = 3 × 0.6 = 1.8 cm
∴ Height of image = 1.8 cm

IX. Higher order thinking (HOT) questions.

Question 1.
While doing an experiment for the determination of focal length of a convex lens, Raja Suddenly dropped the lens. It got broken into two halves along the axis. If he continues his experiment with the same lens,
(a) can he get the image?
(b) Is there any change in the focal length?
Answer:
(a) He can get the image.
(b) The focal length of the lens will be doubled.

Question 2.
The eyes of the nocturnal birds like owl are having a large cornea and a large pupil. How does it help them?
Answer:

• The large pupil opens wider and allows the maximum amount of light to enter the eye in the dark.
• Their lens is large and situated near the retina. This also allows a lot of light to register on the retina. The retina contains 2 types of light-sensing cells rods and cones.
• Cones are responsible for the coloured vision and require bright, focused light.
• Rods are extremely sensitive to light and have a photosensitive pigment called rhodopsin which plays a vital role in night vision.

Samacheer Kalvi 10th Science Optics Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
The scattering of sun light by the atoms or molecules of the gases in the Earth’s atmosphere is known as:
(a) Mie scattering
(b) Tyndall scattering
(c) Rayleigh scattering
(d) Raman scattering
Answer:
(c) Rayleigh scattering

Question 2.
Mie scattering is responsible for the _____ appearance of the clouds.
(a) red
(b) blue
(c) colourless
(d) white.
Answer:
(d) white

Question 3.
In an inelastic scattering the energy of the incident beam of light is ……….. that of scattering beam.
(a) greater than
(b) less than
(c) equal to
(d) different from
Answer:
(d) different from

Question 4.
As per Rayleigh’s scattering law, amount of scattering is:
(a) directly proportioanl to fourth power of wavelength
(b) inversely proportioanl to fourth power of wavelength
(c) inversely proportioanl to square of wavelength
(d) directly proportional to square of wavelength
Answer:
(b) inversely proportioanl to fourth power of wavelength

Question 5.
The refractive index of a medium is dependent on the _____ of the light.
(a) wavelength
(b) strength
(c) density
(d) refraction.
Answer:
(a) wavelength

Question 6.
The scattering of light by colloidal particles in the colloidal solution is called:
(a) Raman scattering
(b) Tyndall scattering
(c) Mie scattering
(d) Elastic scattering
Answer:
(b) Tyndall scattering

Question 7.
A piece of transparent material bounded by curved surfaces is called:
(a) mirror
(b) prism
(c) slab
(d) lens
Answer:
(d) lens

Question 8.
If the energy of the incident and the scattered beam of light are not the same, then it is called as _____.
(a) Elastic
(b) Raman
(c) Inelastic
(d) Mie.
Answer:
(c) Inelastic

Question 9.
A convex lens does not produce:
(a) real magnified image
(b) virtual magnified image
(c) virtual diminished image
(d) real diminished image
Answer:
(c) virtual diminished image

Question 10.
A lens which is thicker in the middle than at the edges is known as:
(a) concave lens
(b) convex lens
(c) bifocal lens
(d) cylindrical lens
Answer:
(b) convex lens

Question 11.
The object is always placed on the _____ side of the lens.
(a) left
(b) right
(c) top
(d) bottom.
Answer:
(a) left

Question 12.
The parallel rays from the outer edge are deviated towards the middle in a:
(a) convex mirror
(b) concave lens
(c) concave mirror
(d) convex lens
Answer:
(d) convex lens

Question 13.
The light rays passing through the optic centre will:
(a) diverged
(b) scattered
(c) converged
(d) emerge undeviated
Answer:
(d) emerge undeviated

Question 14.
All the distances are measured from the ______ of the lense.
(a) centre of curvature
(b) optical centre
(c) principal focus
(d) infinity.
Answer:
(b) optical centre

Question 15.
A ray passing through the principal focus and incident on the lens will:
(a) converge
(b) diverge
(c) emerge parallel to the principal axis
(d) not emerge out
Answer:
(c) emerge parallel to the principal axis

Question 16.
When the object is placed at infinity from the convex lens, the image is formed at:
(a) F
(b) C
(c) infinity
(d) between F and 2F
Answer:
(a) F

Question 17.
The human eye is ____ in nature.
(a) convex
(b) concave
(c) transparent glass
(d) Plano – concave.
Answer:
(a) convex

Question 18.
The image formed by a concave lens is:
(a) virtual
(b) diminished
(c) virtual and diminished
(d) virtual and enlarged
Answer:
(c) virtual and diminished

Question 19.
To get a real image using convex lens, the object must be placed at:
(a) infinity
(b) principal focus
(c) beyond principal focus and infinity
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 20.
_____ is the centre part of the iris.
(a) cornea
(b) retina
(c) pupil
(d) eye lens.
Answer:
(c) pupil

Question 21.
For a convex lens the point at which the parallel rays converge is called of the lens.
(a) pole
(b) centre of curvature
(c) principal focus
(d) none
Answer:
(c) principal focus

Question 22.
A real image formed by a convex lens is always:
(a) erect
(b) magnified
(c) inverted
(d) diminished
Answer:
(c) inverted

Question 23.
The law of distances is given by:

Answer:
(b) \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)

Question 24.
The unit of focal length is:
(a) dioptre
(b) metre
(c) ohm
(d) ampere
Answer:
(b) metre

Question 25.
The sign of focal length of a convex lens is ………. sign.
(a) negative
(b) positive
(c) negative or positive
(d) none of the above
Answer:
(b) positive

Question 26.
The focal length of concave lens has ……… sign.
(a) positive
(b) negative
(c) positive or negative
(d) none of the above
Answer:
(b) negative

Question 27.
The magnification in terms of object distance u and image distance v is m :
(a) \(\frac{u}{v}\)
(b) u + v
(c) \(\frac{v}{u}\)
(d) uv
Answer:
(c) \(\frac{v}{u}\)

Question 28.
In terms of object distance u and focal length/, magnification is given by m =

Answer:
(b) \(\frac{f}{u-f}\)

Question 29.
The magnification in terms of v and/:
(a) f
(b) v – f
(c) \(\frac{f}{v-f}\)
(d) \(\frac{v-f}{f}\)
Answer:
(d) \(\frac{v-f}{f}\)

Question 30.
The unit of power is:
(a) m
(b) ohm
(C) dioptre
(d) ampere
Answer:
(C) dioptre

Question 31.
If the focal length of a convex lens is 1 m then its power is:
(a) 1 dioptre
(b) 0.1 dioptre
(c) 10 dioptre
(d) 0.01 dioptre
Answer:
(a) 1 dioptre

Question 32.
In a simple microscope, the magnification can be increased by:
(a) lens of long focal length
(b) lens
(c) lens of short focal length
(d) lens of infinite focal length
Answer:
(c) lens of short focal length

Question 33.
Convex lenses are used:
(a) as camera lenses
(b) as magnifying lenses
(c) to correct hypermetropia
(d) all the above
Answer:
(d) all the above

Question 34.
Which lens is used in wide angle spyhole in doors?
(a) convex lens
(b) concave lens
(c) cylindrical lens
(d) parabolic lens
Answer:
(b) concave lens

Question 35.
The mathematical form of lens maker’s formula is:

Answer:
(a)

Question 36.
If f is the focal length of the lens then its power is given by:
(a) P = \(\frac{2}{f}\)
(b) p = \(\frac{1}{f}\)
(c) p = f
(d) p = f₁
Answer:
(b) p = \(\frac{1}{f}\)

Question 37.
Which part of the human eye changes the focal length of the eye lens?
(a) pupil
(b) retina
(c) ciliary muscles
(d) cornea
Answer:
(c) ciliary muscles

Question 38.
On which part of human eye, image is formed?
(a) cornea
(b) iris
(c) retina
(d) pupil
Answer:
(c) retina

Question 39.
For normal human eye the value of near point is:
(a) 25 cm
(b) 25 m
(c) 2.5 m
(d) 25 mm
Answer:
(a) 25 cm

Question 40.
In hypermeteropia, the focal length of the eye lens is:
(a) decreased
(b) remains the same
(c) increased
(d) none of the above
Answer:
(c) increased

Question 41.
Presbyopia can be corrected by using:
(a) convex lens
(b) bifocal lens
(c) concave lens
(d) cylindrical lens
Answer:
(b) bifocal lens

Question 42.
Astigmatism can be corrected by using:
(a) bifocal lens
(b) cylindrical lens
(c) convex lens
(d) concave lens
Answer:
(b) cylindrical lens

Question 43.
The magnifying power of compound microscope is:
(a) 10
(b) 20
(c) 50
(d) 50 to 200
Answer:
(d) 50 to 200

Question 44.
The accuracy of travelling microscope is of the order of:
(a) 0.01 cm
(b) 0.01 mm
(c) 0.1 mm
(d) 0.1 cm
Answer:
(b) 0.01 mm

II. Fill in the blanks.

1. The velocity of light in vacuum is ……….
2. If v is the frequency and λ is the wavelength then velocity of the wave is c = ……….
3. Among colours of visible light ……… colour has the highest wavelength.
4. According to Snell’s law refractive index, µ₂ = ……….
5. In a medium having high value of refractive index then speed of light in that medium is ……….
6. Angle of refraction is the smallest for ……… and the highest for ……….
7. The refractive index depends on ………. of light.
8. Colours having shorter wavelength scattered more than longer wavelength colours according to ……….. law.
9. After passing through a convex lens ……….. rays ………. at the principal focus.
10. For a convex lens, as the object distance increases, the image distance ……….
11. A ray passing through the optic centre of a lens emerges …………
12. ……… is due to irregular curvature of the surface of the eye lens.
13. When a parallel beam of light passes through a convex lens, the rays from the outer edges are …………
14. A ray parallel to the principal axis of a convex lens after refraction passes through …………
15. When the object is placed between ………….. and ………… of a convex lens a virtual image will be formed.
16. For a convex lens, as the object approaches the lens the image becomes …………
17. In a phographic camera ………… lens is used.
18. The shorter the focal length, the ………. is the magnification.
19. The nature of the image formed by a simple microscope is ……….., ………… and …………
20. Real images are formed by a ……….. lens.
21. Concave lens produces ………… images.
22. The value of power of a lens having focal length one metre is ………..
23. For a normal eye the value of far point is …………
24. …………. is known as short sightedness.
25. Hyper metropia is known as ………..
26. The mathematical form of focal length of a concave lens used to correct myopia is f = ……….
27. ……….. lenses are used to correct astigmatism.
28. For a normal eye, the value of least distance of distinct vision is ………..
29. The objective of the compound microscope has …………. focal length.
30. The focal length of ………. is greater in a compound microscope.
31. ……… is an optical instrument to see the distant objects.
32. A terrestrial telescope produces ……… image.
33. Elaborate view of galaxies and planets is obtained by ………
Answer:
1. 3 × 10⁸ m/s
2. vλ
3. red
4. \(\frac{sin i}{sin r}\)
5. low
6. red, violet
7. wavelength
8. Rayleigh scattering
9. parallel, converge
10. will decrease
11. undeviated
12. Astigmatism
13. deviated towards the centre of the lens
14. the principal focus
15. principal focus, optical centre
16. bigger
17. biconvex
18. greater
19. virtual, erect, magnified
20. convex
21. virtual
22. One dioptre
23. infinity
24. Myopia
25. long sightedness
26. xy/x – y
27. Cylindrical
28. 25 cm
29. shorter
30. eye piece
31. Telescope
32. an erect
33. Telescope

III. True or False. If false correct it.

1. Light does not travel along a straight line.
2. All coloured light has same wavelength.
3. In refraction incident ray, refracted ray and normal lie in the same plane.
4. Velocity of light is greater in rarer medium is greater than that in denser medium.
5. For red colour angle of refraction is the least.
6. The refractive index of a medium ¡s independent of wavelength.
7. Tyndall scattering, is the scattering of light by colloids.
8. According to Rayleigh’s scattering law, red colour is scattered to a greater extent than blue colour.
9. Mie scattering takes place when the diameter is larger than the wavelength of the incident light.
10. The lines in Raman scattering having frequencies lower than the incident frequency are called Antistoke’s lines.
11. In front of a convex lens when the object is placed at infinity the formed image is smaller than that of the object.
12. When an object is placed at finite distance from the concave lens a virtual image is formed between optical centre and focus of the concave lens.
13. Pupil of human eye bends the incident light on to the lens.
14. For a normal human eye, the value of far point is 25 cm.
15. Astigmatism is corrected by cylindrical lenses.
Answer:
1. False – Light always travels along a straight lines.
2. False – Different coloured light has different wavelength.
3. True
4. True
5. True
6. False – The refractive index of a medium depends on wavelength.
7. True
8. False – According to Rayleigh’s scattering law, blue colour is scattered to a greater extent than red colour.
9. True
10. False – The lines in Raman scattering having frequencies higher than the incident frequency are called Antistoke’s lines.
11. True
12. True
13. False – Cornea of human eye bends the incident light on to the lens.
14. False – For a normal human eye, the value of near point is 25 cm.
15. True

IV. Match the following.

Question 1.
Match the column I with column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 2.
Position of the object placed infront of a convex lens are given in Column
I. Match them with the natures of the images formed by the convex lens given in column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 3.
Match the following:

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 4.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (ii)
D. (i)

Question 5.
Match the Column I with Column II.

Answer:
A. (iv)
B. (i)
C. (ii)
D. (iii)

Question 6.
Match the column I with column II.

Answer:
A. (iii)
B. (i)
C. (iv)
D. (ii)

Question 7.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (ii)
D. (i)

V. Assertion and Reasoning type.

Question 1.
Assertion : The sun looks bigger in size at sunrise and sunset than during day.
Reason : In detraction light rays bend around the edges of the obstacle.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 2.
Assertion: Colours can be scan in thin layers of oil on the surface water. Reason: White light is composed of several colours.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 3.
Assertion: Raman spectrum of a liquids contains lines whose frequencies are not equal to that of incident radiation.
Reason: If a photon strikes an atom in a liquid that is in existed state photon losses energy.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 4.
Assertion: The refractive index of a prism depends only on the material of the prism.
Reason: The refractive index of a prism depends upon the refracting angle and angel of minimum deviation.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(c) If Assertion is true but Reason is false.

Question 5.
Assertion: A single lens produces a coloured image of an object illuminated by white light.
Reason: The refractive index of material of lens is different for different wavelength of light.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 6.
Assertion: If a convex lens is placed in water, its convergence power decrease.
Reason: Focal length of lens is independent of refractive index of the medium.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(c) If Assertion is true but Reason is false.

Question 7.
Assertion: Light waves travel in straight lines.
Reason: Rectilinear propagation of light confirm the above mentioned properly.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 8.
Assertion: Raman scattering the scattering of monochromatic light by atoms and molecule of a liquid.
Reason: The wavelength of Raman lines is same.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 9.
Assertion: Power of a lens is the reciprocal of its focal length.
Reason: The unit of power is one dioptre when the unit of focal length is one metre.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 10.
Assertion: Presbyopia is due to ageing of human beings.
Reason: For those persons, ciliary muscles of the eye become weak.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

VI. Answer Briefly.

Question 1.
What is meant by refraction?
Answer:
When a ray of light travels from one transparent medium into another obliquely, the path of the light undergoes deviation. This deviation of ray of light is called refraction.

Question 2.
State laws of refraction.
Answer:
First law of refraction: The incident ray, the refracted ray of light and the normal to the refracting surface all lie in the same plane.

Second law of refraction: The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
\(\frac{sin i}{sin r}\) = \(\frac{µ_2}{µ_1}\)

Question 3.
Define refractive index of a medium.
Answer:
The ratio of speed of light in vacuum to the speed of light in a medium is defined as refractive index ‘p’ of that medium.

Question 4.
What is meant by monochromatic source?
Answer:
If a source of light produces a light of single colour, it is known as a monochromatic source.

Question 5.
When white light is refracted by a transparent medium what will you get? Why?
Answer:

1. When white light is refracted by a transparent medium, a spectrum is obtained.
2. This is because, different coloured lights are bent through different angles.

Question 6.
What is scattering of light?
Answer:
When sunlight enters the Earth’s atmosphere, the atoms and molecules of different gases present in the atmosphere refract the light in all possible directions. This is called as ‘Scattering of light’.

Question 7.
State the types of scattering.
Answer:

1. Elastic scattering
2. Inelastic scattering

Question 8.
What is elastic scattering?
Answer:
If the energy of the incident beam of light and the scattered beam of light are same, then it is called as ‘elastic scattering’.

Question 9.
What is inelastic scattering?
Answer:
If the energy of the incident beam of light and the scattered beam of light are not same, then it is called as ‘inelastic scattering’.

Question 10.
How are different types of scattering formed? Mention the types of scattering.
Answer:
The nature and size of the scatterer results in different types of scattering. They are

1. Rayleigh scattering
2. Mie scattering
3. Tyndall scattering
4. Raman scattering

Question 11.
What is Rayleigh scattering?
Answer:
The scattering of sunlight by the atoms or molecules of the gases in the earth’s atmosphere is known as Rayleigh scattering.

Question 12.
Why the colour of the Sun is red at sunrise and sunset?
Answer:
At sunrise and sunset, the light rays from the Sun have to travel a larger distance in the atmosphere than at noon. Hence, most of the blue lights are scattered away and only the red light which gets least scattered reaches us. Therefore, the colour of the Sun is red at sunrise and sunset.

Question 13.
When does Mie scattering take place?
Answer:
Mie scattering takes place when the diameter of the scatterer is similar to or larger than the wavelength of the incident light.

Question 14.
What are the causes of Mie scattering?
Answer:
Mie scattering is caused by pollen, dust, smoke, water droplets, and other particles in the lower portion of the atmosphere.

Question 15.
Why the clouds have white appearance?
Answer:
Mie scattering is responsible for the white appearance of the clouds. When white light falls on the water drop, all the colours are equally scattered which together form the white light.

Question 16.
What is Tyndall Scattering?
Answer:
The scattering of light rays by the colloidal particles in the colloidal solution is called Tyndall Scattering or Tyndall Effect.

Question 17.
What is meant by colloid? State few examples.
Answer:
Colloid is a microscopically small substance that is equally dispersed throughout another material. Eg: Milk, Ice cream, muddy water, smoke.

Question 18.
What is meant by Raman Scattering?
Answer:
When a parallel beam of monochromatic (single coloured) light passes through a gas or liquid or transparent solid, a part of light rays are scattered.

Question 19.
Define Raman Scattering.
Answer:
Raman Scattering is defined as “The interaction of light ray with the particles of pure liquids or transparent solids, which leads to a change in wavelength or frequency.”

Question 20.
What is Rayleigh line?
Answer:
The spectral lines having frequency equal to the incident ray frequency is called ‘Rayleigh line’.

Question 21.
What are Raman lines?
Answer:
The spectral lines which are having frequencies other than the incident ray frequency are called ‘Raman lines’.

Question 22.
What are stokes lines and Antistokes lines?
Answer:
The lines having frequencies lower than the incident frequency is called stokes lines and the lines having frequencies higher than the incident frequency are called Antistokes lines.

Question 23.
What is a lens?
Answer:
A lens is an optically transparent medium bounded by two spherical refracting surfaces or one plane and one spherical surface.

Question 24.
How is lens classified?
Answer:
Lens is basically classified into two types. They are:

1. Convex Lens
2. Concave Lens.

Question 25.
What is biconvex lens?
Answer:
Convex or bi-convex lens: It is a lens bounded by two spherical surfaces such that it is thicker at the centre than at the edges. A beam of light passing through it, is converged to a point. So, a convex lens is also called as converging lens.

Question 26.
What is meant by biconcave lens?
Answer:
Concave or bi-concave Lens: It is a lens bounded by two spherical surfaces such that it is thinner at the centre than at the edges. A parallel beam of light passing through it, is diverged or spread out. So, a concave lens is also called as diverging lens.

Question 27.
What are
(i) Plano-convex lens?
(ii) Plano-concave lens?
Answer:
(i) If one of the faces of a bi-convex lens is plane, it is known as a plano-convex lens.
(ii) If one of the faces of a bi-concave lens is plane, it is known as a plano-concave lens.

Question 28.
State the applications of convex lenses.
Answer:

1. Convex lenses are used as camera lenses.
2. They are used as magnifying lenses.
3. They are used in making microscope, telescope and slide projectors.
4. They are used to correct the defect of vision called hypermetropia.

Question 29.
Draw diagrams of different types converging lenses.
Answer:

Question 30.
Represent different types of lenses by diagram.
Answer:

Question 31.
What is the nature of the image formed by an object is placed behind the centre of curvature of a convex leas. Draw a ray diagram.
Answer:
When an object is placed behind the center of curvature (beyond C), a real and inverted image is formed between the center of curvature and the principal focus. The size of the image is the same as that of the object.

Question 32.
Draw a ray diagram to indicate the nature of the image formed when an object is placed in between the centre of curvature and principal focus of a convex lens.
Answer:
When an object is placed in between the center of curvature and principal focus, a real and inverted image is formed behind the center of curvature. The size of the image is bigger than that of the object.

Question 33.
Draw a ray diagram for the formation of image (by the concave lens) when object is at infinity.
Answer:
When an object is placed at infinity, a virtual image is formed at the focus. The size of the image is much smaller than that of the object.

Question 34.
What are the applications of concave lens?
Answer:

1. Concave lenses are used as eye lens of‘Galilean Telescope’.
2. They are used in wide angle spy hole in doors.
3. They are used to correct the defect of vision called ‘myopia’.

Question 35.
What do you know about lens formula?
Answer:
The lens formula gives the relationship among distance of the object (u), distance of the image (v) and the focal length (f) of the lens. It is expressed as
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\)

Question 36.
Define magnification.
Answer:
It is defined as the ratio of the height of the image to the height of an object. Magnification is denoted by the letter ‘m’. If height of the object is h and height of the image is h’, the magnification produced by lens is,

Question 37.
What is lens formula?
Answer:
The lens formula relates the focal length of a lens with the distance of object and image.

where µ is the refractive index of the material of the lens; R₁ and R₂ are the radii of curvature of the two faces of the lens; f is the focal length of the lens.

Question 38.
What is meant by power of lens?
Answer:
The ability of a lens to converge (convex lens) or diverge (concave lens) light rays is called as its power.

Question 39.
Define power of a lens. State its unit.
Answer:
Power of a lens is numerically defined as the reciprocal of its focal length.
P = \(\frac{1}{f}\)
The SI unit of power of a lens is dioptre.

Question 40.
What is meant by dioptre.
Answer:
Dioptre is the power of a lens, whose focal length is 1 metre.
1 Dioptre = 1 m^-1.

Question 41.
Differentiate convex lens from concave lens.
Answer:

Question 42.
What are
(i) Pupil &
(ii) Retina?
Answer:
(i) Pupil: It is the centre part of the Iris. It is the pathway for the light to retina.
(ii) Retina: This is the back surface of the eye. It is the most sensitive part of human eye, on which real and inverted image of objects is formed.

Question 43.
What is persistence of vision?
Answer:
If the time interval between two consecutive light pulses is less than 0.1 second, human eye cannot distinguish them separately. It is called persistence of vision.

Question 44.
What is least distance of distinct vision?
Answer:
The minimum distance required to see the objects distinctly without strain is called least distance of distinct vision. It is called as near point of eye. It is 25 cm for normal human eye.

Question 45.
What is far point?
Answer:
The maximum distance up to which the eye can see objects clearly is called as far point of the eye.

Question 46.
What is Presbyopia?
Answer:
Due to ageing, ciliary muscles become weak and the eye-lens become rigid (inflexible) and so the eye loses its power of accommodation. Because of this, an aged person cannot see the nearby objects clearly. So, it is also called as ‘old age hypermetropia’.

Question 47.
What is meant by astigmatism?
Answer:
In this defect, eye cannot see parallel and horizontal lines clearly. It may be inherited or acquired. It is due to the imperfect structure of eye lens because of the development of cataract on the lens, ulceration of cornea, injury to the . refracting surfaces, etc. Astigmatism can be corrected by using cylindrical lenses (Torrid lenses).

Question 48.
State the principle of microscope. How is it classified?
Answer:
It works under the principle of angular magnification of lenses. It is classified as

1. Simple microscope
2. Compound microscope

Question 49.
What are the uses of simple microscope?
Answer:
Simple microscopes are used

1. By watch repairers and jewellers.
2. To read small letters clearly.
3. To observe parts of flower, insects etc.
4. To observe finger prints in the field of forensic science.

Question 50.
How is telescope classified?
Answer:
According to optical property, it is classified into two groups:

1. refracting telescope
2. reflecting telescope

Question 51.
Mention the advantages of telescope.
Answer:

1. Elaborate view of the Galaxies, Planets, stars and other heavenly bodies is possible.
2. Camera can be attached for taking photograph for the celestial objects.
3. Telescope can be viewed even with the low intensity of light.

Question 52.
What are the disadvantages of telescope?
Answer:

1. Frequent maintenance is needed.
2. It is not easily portable.

VII. Give the answer in Detail.

Question 1.
State the Laws of Refraction.
Answer:
The incident ray, the refracted ray of light and the normal to the refracting surface all lie in the same plane.
Second law of Refraction:

1. The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
   \(\frac{\sin i}{\sin r}=\frac{\mu_{2}}{\mu_{1}}\)
2. Refractive index gives us an idea of how fast or how slow light travels in a medium. The ratio of the speed of light in a vacuum to the speed of light in a medium is defined as the refractive index ‘µ’ of that medium.
3. The speed of light in a medium is low if the refractive index of the medium is high and vice versa.
4. When light travels from a denser medium into a rarer medium, the refracted ray is bent away from the normal drawn to the interface.
5. When light travels from a rarer medium into a denser medium, the refracted ray is bent towards the normal drawn to the interface.

Question 2.
Describe Raman Scattering.
Answer:
When a parallel beam of monochromatic (single coloured) light passes through a gas or liquid or transparent solid, a part of light rays are scattered.

The scattered light contains some additional frequencies (or wavelengths) other than that of incident frequency (or wavelength). This is known as Raman scattering or Raman Effect.

Raman Scattering is defined as “The interaction of light ray with the particles of pure liquids or transparent solids, which leads to a change in wavelength or frequency.”

The spectral lines having frequency equal to the incident ray frequency is called ‘Rayleigh line’ and the spectral lines which are having frequencies other than the incident ray frequency are called ‘Raman lines’. The lines having frequencies lower than the incident frequency is called stokes lines and the lines having frequencies higher than the incident frequency are called Antistokes lines.

Question 3.
With the help of ray diagram, explain the nature, size and position of the image formed by a convex lens. When object is placed at
(i) infinity
(ii) beyond C
(iii) placed at C
(iv) Placed between F and C,
(v) placed at F
(vi) placed between F and optical centre O.
Answer:
(i) Object at infinity: When an object is placed at infinity, a real image is formed at the principal focus. The size of the image is much smaller than that of the object.

(ii) Object placed beyond C (>2F): When an object is placed behind the center of curvature(beyond C), a real and inverted image is formed between the center of curvature and the principal focus. Th e size of the image is the same as that of the object.

(iii) Object placed at C: When an object is placed at the center of curvature, a real and inverted image is formed at the other center of curvature. The size of the image is the same as that of the object.

(iv) Object placed between F and C: When an object is placed in between the center of curvature and principal focus, a real and inverted image is formed behind the center of curvature. The size of the image is bigger than that of the object.

(v) Object placed at the principal focus F: When an object is placed at the focus, a real image is formed at infinity. The size of the image is much larger than that of the object.

(vi) Object placed between the principal focus F and optical centre O: When an object is placed in between principal focus and optical centre, a virtual image is formed. The size of the image is larger than that of the object.

Question 4.
Explain the formation of images formed by a concave lens.
Answer:
Object at Infinity: When an object is placed at infinity, a virtual image is formed at the focus. The size of the image is much smaller than that of the object.

Object anywhere on the principal axis at a finite distance: When an object is placed at a finite distance from the lens, a virtual image is formed between optical center and focus of the concave lens. The size of the image is smaller than that of the object.

But, as the distance between the object and the lens is decreased, the distance between the image and the lens also keeps decreasing. Further, the size of the image formed increases as the distance between the object and the lens is decreased.

Question 5.
Explain Mie Scattering.
Answer:
Mie scattering:

1. Mie scattering takes place when the diameter of the Scatterer is similar to or larger than the wavelength of the incident light. It is also an elastic scattering.
2. The amount of scattering is independent of wavelength.
3. Mie scattering is caused by pollen, dust, smoke, water droplets, and other particles in the lower portion of the atmosphere.
4. Mie scattering is responsible for the white appearance of the clouds.
5. When white light falls on the water drop, all the colours are equally scattered which, together form the white light.

Question 6.
With the help of a diagram, explain the structure and working of human eye.
Answer:
Structure of the eye:
The eye ball is approximately spherical in shape with a diameter of about 2.3 cm. It consists of a tough membrane called sclera, which protects the internal parts of the eye.

Cornea : This is the thin and transparent layer on the front surface of the eyeball as shown in figure. It is the main refracting surface. When light enters through the cornea, it refracts or bends the light on to the lens.

Iris : It is the coloured part of the eye. It may be blue, brown or green in colour. Every person has a unique colour, pattern and texture. Iris controls amount of light entering into the pupil like camera aperture.

Pupil : It is the centre part of the Iris. It is the pathway for the light to retina.

Retina : This is the back surface of the eye. It is the most sensitive part of human eye, on which real and inverted image of objects is formed.

Ciliary muscles : Eye lens is fixed between the ciliary muscles. It helps to change the focal length of the eye lens according to the position of the object.

Eye Lens : It is the important part of human eye. It is convex in nature.

Working of the eye : The transparent layer cornea bends the light rays through pupil located at the centre part of the Iris. The adjusted light passes through the eye lens. Eye lens is convex in nature. So, the light rays from the objects are converged and a real and inverted image is formed on retina. Then, retina passes the received real and inverted image to the brain through optical nerves. Finally, the brain senses it as erect image.

Question 7.
Describe simple microscope.
Answer:
Simple microscope: It has a convex lens of short focal length. It is held near the eye to get enlarged image of small objects.
Let an object (AB) is placed at a point within the principal focus (u < f) of the convex lens and the observer’s eye is placed just behind the lens. As per this position the convex lens produces an erect, virtual and enlarged image (A’B’), The image formed is in the same side of the object and the distance equal to the least distance of distinct vision (D) (For normal human eye D = 25 cm).

Question 8.
Write short notes on
(i) Astronomical telescope
(ii) Terrestrial telescope.
Answer:
(i) Astronomical Telescope: An astronomical telescope is used to view heavenly bodies like stars, planets, galaxies and satellites.

(ii) Terrestrial Telescope: The image in an astronomical telescope is inverted. So, it is not suitable for viewing objects on the surface of the Earth. Therefore, a terrestrial telescope is used. It provides an erect image. The major difference between astronomical and terrestrial telescope is erecting the final image with respect to the object.

VIII. Numerical Problems.

Question 1.
A needle of size 5 cm is placed 45 cm from a lens produced an image on a screen placed 90 cm away from the lens.
Answer:
(i) Identify the types of lens.
Calculate focal length of the lens.
Height of the object h₁ = 5 cm
Distance of the object u = -45 cm
Distance of the image v = 90 cm
We know that

Focal length of the lens = 30 cm
Since focal length is positive the lens is convex lens.

(ii) Identify the size of the image

∴ h₂ = -10 cm
The negative sign indicates that the image is real and inverted.

Question 2.
A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image 10 cm from the lens?
Answer:
v = -10 cm; f =-15 cm; u = ?
Lens formula:
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\)
Type of lens: Concave lens

u =-30 cm
Thus, the object distance is 30 cm.

Question 3.
An object is placed at a distance of 30 cm from a concave lens of focal length 15 cm. An erect and virtual image is formed at a distance of 10 cm from the lens. Calculate the magnification.
Answer:
Type of lens is Cancave lens.
Formula:
Magnification m = \(\frac{v}{u}\)
Object distance u = -30 cm
Image distance v = -10 cm
m = \(\frac{-10}{-30}\) = \(\frac{1}{3}\) = +0.33

Question 4.
The focal length of a concave lens is 2 cm. Calculate the power of the lens.
Answer:
Formula:
P = \(\frac{1}{f}\)
Type of lens is concave lens.
Focal length of concave lens,
f = -2 m power of the lens.
P = \(\frac{1}{-2 m}\)
P = -0.5 dioptre

Question 5.
A needle placed at 30 cm from the lens forms an image on a screen placed 60 cm on the other side of the lens. Identify the type of lens and determine the focal length.
Answer:
u = -30 cm
v = 60 cm
u is negative because image is formed on the on the other side of the lens.

It is a convex lens.

Question 6.
A 3 cm tall bulb is placed at a distance of 20 cm from a diverging lens having a focal length of 10.5 cm. Determine the distance of the image.
Answer:
u = -20 cm
f = -10.5 cm

The distance of the image is -6.88 cm

Question 7.
A ray from medium 1 is refracted below while passing through medium 2. Find the refractive index of the second medium with respect to medium 1.
Answer:
Refractive index µ

Refractive index = 0.707

Question 8.
The optical prescription of a pair of spectacle is
Right eye: -3.5 D, Left eye: -4.00 D.
(i) Name the defect of the eye.
Answer:
Shortsighted (Myopia)

(ii) Are these lenses thinner at the middle or at the edges?
Answer:
These lenses are thinner in the middle.

(iii) Which lens has a greater focal length?
Answer:
power = \(\frac{1}{focal length}\)
Right eye: power P = -3.5 D

Left eye: Power P = -4 D

Hence the lens having power of -3.5 D has greater focal length.

Question 9.
The radii of curvature of two surfaces of a double convex lens are 10 cm each. Calculate its focal length and power of the lens in air and liquid. Refractive indices of glass and liquid are 1.5 and 1.8 respectively.
Answer:
Radius of curvature of first surface R₁ = 10 cm
Radius of curvature of second surface R₂ = 10 cm
In air

p_l = -3.33 d

Question 10.
An object 2 cm tall is placed 10 cm in front of a convex lens of focal length 15 cm. Find the position, size and nature of the image formed.
Answer:
Focal length of a convex lens f = 15 × 10^-2 m
Weight of the object ho = 2 × 10^-2 m
Let weight of the image be h_v
Distance of the object u = 10 × 10^-2 m
Distance of the image v = 15 × 10^-2 m
We know

∴ v = -30 × 10^-2 m
Distance of the image = 30 × 10^-2 m

Hence a virtual image 6 × 10^-2 m height is formed at a distance of 30 × 10^-2 m from the lens on the same side of the lens.

IX. Higher order thinking (HOT) questions.

Question 1.
Ramu passes white light through a quartz prism. For which colour refractive index is greater?
Answer:
Refractive index is maximum for violet light when white light passes through a quartz prism.

Question 2.
Sita has kept a stud consists of diamond. What will she observe? Give reason.
Answer:
The diamond stud appears bright because of total internal reflection.

Question 3.
Guna passes a ray light through a glass slab. Which optical phenomenon will take place? What can he observe with reference to wavelength?
Answer:
When a ray of light enters a glass slab he can observe refraction of light. He observed that wavelength of light decreases.

Question 4.
A prism is placed in the minimum deviation position. Chari has passed a ray of light at an angle of 45°, then what is the value of angle of emergence? Why?
Answer:
The angle of emergence = 45°.
Since, in the minimum deviation positron, angle of incidence is equal to angle of emergence.

Question 5.
Mani is using a lens of power 2 dioptre. What is the focal length of the lens?
Answer:
Focal length = \(\frac{1}{power}\)
F= \(\frac{1}{2}\) = 0.5 m

Question 6.
Surya has placed a lens of power 1 D in side water. What will happen to power of the lens?
Answer:
The power of the lens will be more than original power.

Question 7.
Sonu has observed some lines in solar spectrum are absorbed by the elements present in the atmosphere. What are the lines?
Answer:
The lines are Fraunhofer lines.

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4

Question 1.
Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x – \(\frac { 3 }{ 17 } \) = 0
Solution:
(i) 5y – 3 = 0
5y = 3 ⇒ y = \(\frac { 3 }{ 5 } \)
Slope = 0

(ii) 7x – \(\frac { 3 }{ 17 } \) = 0 (Comparing with y = mx + c)
7x = \(\frac { 3 }{ 17 } \)
Slope is undefined

Question 2.
Find the slope of the line which is
(i) parallel to y = 0.7x – 11
(ii) perpendicular to the line x = -11
Solution:
(i) y = 0.7x – 11
Slope = 0.7 (Comparing with y = mx + c)
(ii) Perpendicular to the line x = – 11
Slope is undefined (Since the line is intersecting the X-axis)

Question 3.
Check whether the given lines are parallel or perpendicular
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 and \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Solution:
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 ; \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
Slope of the line (m₁) = \(\frac { -a }{ b } \)
= – \(\frac { 1 }{ 3 } \) ÷ \(\frac { 1 }{ 4 } \) = –\(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
Slope of the line (m₂) = – \(\frac { 2 }{ 3 } \) ÷ \(\frac { 1 }{ 2 } \) = –\(\frac { 2 }{ 3 } \) × \(\frac { 2 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
m₁ = m₂ = – \(\frac { 4 }{ 3 } \)
∴ The two lines are parallel.

(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Slope of the line (m₁) = \(\frac { -5 }{ 23 } \)
Slope of the line (m₂) = \(\frac { -23 }{ -5 } \) = \(\frac { 23 }{ 5 } \)
m₁ × m₂ = \(\frac { -5 }{ 23 } \) × \(\frac { 23 }{ 5 } \) = -1
∴ The two lines are perpendicular

Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
Solution:
Slope of the first line 12y = -(p + 3)x +12
y = \(-\frac{(p+3) x}{12}+1\) (Comparing with y = mx + c)
Slope of the second line (m₁) = \(\frac { -(p+3) }{ 12 } \)
Slope of the second line 12x – 7y = 16
(m₂) = \(\frac { -a }{ b } \) = \(\frac { -12 }{ -7 } \) = \(\frac { 12 }{ 7 } \)
Since the two lines are perpendicular
m₁ × m₂ = -1
\(\frac { -(p+3) }{ 12 } \) × \(\frac { 12 }{ 7 } \) = -1 ⇒ \(\frac { -(p+3) }{ 7 } \) = -1
-(p + 3) = -7
– p – 3 = -7 ⇒ -p = -7 + 3
-p = -4 ⇒ p = 4
The value of p = 4

Question 5.
Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
Solution:
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of the line QR = \(\frac { 4+2 }{ -5-3 } \) = \(\frac { 6 }{ -8 } \) = \(\frac { 3 }{ -4 } \) ⇒ – \(\frac { 3 }{ 4 } \)
Slope of its parallel = – \(\frac { 3 }{ 4 } \)
The given point is p(-5, 2)
Equation of the line is y – y₁ = m(x – x₁)
y – 2 = – \(\frac { 3 }{ 4 } \) (x + 5)
4y – 8 = -3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0

Question 6.
Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Let the vertices A (6, 7), B (2, -3), D (6, -2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-7 }{ 2-6 } \) = \(\frac { -10 }{ -4 } \) = \(\frac { 5 }{ 2 } \)
Slope of its perpendicular (CD) = – \(\frac { 2 }{ 5 } \)
Equation of the line CD is y – y₁ = m(x – x₁)
y + 2 = –\(\frac { 2 }{ 5 } \) (x – 6)
5(y + 2) = -2 (x – 6)
5y + 10 = -2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0

Question 7.
A(-3,0) B(10, -2) and C(12,3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Solution:
To find the equation of the altitude from A.
The vertices of ∆ABC are A(-3, 0), B(10, -2) and C(12, 3)

Slope of BC = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 3+2 }{ 12-10 } \) = \(\frac { 5 }{ 2 } \)
Slope of the altitude AD is – \(\frac { 2 }{ 5 } \)
Equation of the altitude AD is
y – y₁ = m (x – x₁)
y – 0 = – \(\frac { 2 }{ 5 } \) (x + 3)
5y = -2x -6
2x + 5y + 6 = 0
Equation of the altitude AD is 2x + 5y + 6 = 0
Equation of the altitude from B

Slope of AC = \(\frac { 3-0 }{ 12+3 } \) = \(\frac { 3 }{ 15 } \) = \(\frac { 1 }{ 5 } \)
Slope of the altitude AD is -5
Equation of the altitude BD is y – y₁= m (x – x₁)
7 + 2 = -5 (x – 10)
y + 2 = -5x + 50
5x + 7 + 2 – 50 = 0 ⇒ 5x + 7 – 48 = 0
Equation of the altitude from B is 5x + y – 48 = 0

Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
Solution:
“C” is the mid point of AB also CD ⊥ AB.

Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { -4-2 }{ 6+4 } \) = \(\frac { -6 }{ 10 } \) = – \(\frac { 3 }{ 5 } \)
Slope of the ⊥^r AB is \(\frac { 5 }{ 3 } \)
Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
= (\(\frac { -4+6 }{ 2 } \),\(\frac { 2-4 }{ 2 } \)) = (\(\frac { 2 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (1,-1)
Equation of the perpendicular bisector of CD is
y – y₁ = m(x – x₁)
y + 1 = \(\frac { 5 }{ 3 } \) (x – 1)
5(x – 1) = 3(y + 1)
5x – 5 = 3y + 3
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Equation of the perpendicular bisector is 5x – 3y – 8 = 0

Question 9.
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Solution:
Given lines are.

x = \(\frac { 43 }{ 43 } \) = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10 ⇒ 3y = 10 – 7
y = \(\frac { 3 }{ 3 } \) = 1
The point of intersection is (1,1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1,1)
13 (1) + 5 (1) + k = 0
13 + 5 + k = 0 ⇒ 18 + k = 0
k = -18
∴ The equation of the line is 13x + 5y – 18 = 0

Question 10.
Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
Solution:
Given lines are.

Substitute the value of x = \(\frac { 16 }{ 7 } \) in (2)
3 × \(\frac { 16 }{ 7 } \) + 2y = 10 ⇒ 2y = 10 – \(\frac { 48 }{ 7 } \)
2y = \(\frac { 70-48 }{ 7 } \) ⇒ 2y = \(\frac { 22 }{ 7 } \)
y = \(\frac{22}{2 \times 7}\) = \(\frac { 11 }{ 7 } \)
The point of intersect is (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
This line passes through (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
7 (\(\frac { 16 }{ 7 } \)) + 4 (\(\frac { 11 }{ 7 } \)) + k = 0 ⇒ 16 + \(\frac { 44 }{ 7 } \) + k = 0
\(\frac { 112+44 }{ 7 } \) + k = 0 ⇒ \(\frac { 156 }{ 7 } \) + k = 0
k = – \(\frac { 156 }{ 7 } \)
Equation of the line is 7x + 4y – \(\frac { 156 }{ 7 } \) = 0
49x + 28y – 156 = 0

Question 11.
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
Solution:
The given lines are.


Substitute the value of x = 0 in (1)
3 (0) + y = -2
y = -2
The point of intersection is (0, -2).
The given equation is

Substitute the value of y = \(\frac { 9 }{ 11 } \) in (6)
– x + 2 (\(\frac { 9 }{ 11 } \)) = 3 ⇒ -x + \(\frac { 18 }{ 11 } \) = 3
-x = 3 – \(\frac { 18 }{ 11 } \) = \(\frac { 33-18 }{ 11 } \) = \(\frac { 15 }{ 11 } \)
x = – \(\frac { 15 }{ 11 } \)
The point of intersection is (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \))
Equation of the line joining the points (0, -2) and (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \)) is


31 × (- 11x) = 11 × 15 (y + 2) = 165 (y + 2)
– 341 x = 165 y + 330
– 341 x – 165 y – 330 = 0
341 x + 165 y + 330 = 0
(÷ by 11) ⇒ 31 x + 15 y + 30 = 0
The required equation is 31 x + 15 y + 30 = 0

Question 12.
Find the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Solution:
Given lines are.
8x + 3y = 18 …..(1)
4x + 5y = 9 …..(2)

x = \(\frac { 63 }{ 28 } \) = \(\frac { 9 }{ 4 } \)
Substitute the value of x = \(\frac { 9 }{ 4 } \) in (2)
4 (\(\frac { 9 }{ 4 } \)) + 5y = 9
9 + 5y = 9 ⇒ 5y = 9 – 9
5y = 0 ⇒ y = 0
The point of intersection is (\(\frac { 9 }{ 4 } \),0)
Mid point of the points (5, -4) and (-7, 6)

Equation of the line joining the points (\(\frac { 9 }{ 4 } \),0) and (-1,1)

-13y = 4x – 9
-4x – 13y + 9 = 0 ⇒ 4x + 13y – 9 = 0
The equation of the line is 4x + 13y – 9 = 0

Students can download Maths Chapter 6 Trigonometry Ex 6.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.3

Question 1.
From the top of a rock 50 \(\sqrt { 3 }\) m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.
Answer:
Let the distance of the car from the rock is “x” m

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \)
\(\frac{1}{\sqrt{3}}=\frac{50 \sqrt{3}}{x}\)
x = 50 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3
= 150 m
∴ Distance of the car from the rock = 150 m

Question 2.
The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building.
Answer:
Let the height of the first building AD be “x” m
∴ EC = 120 – x
In the right ∆ CDE,
tan 45° = \(\frac { CE }{ CD } \)

1 = \(\frac { 120-x }{ 70 } \) ⇒ 70 = 120 – x
x = 50 cm
∴ The height of the first building is 50 m

Question 3.
From the top of the tower 60 m high the anles of depression the top and bottom of a vertical lamp post are observed be 38° and 60° respectively
Find the height of the lamp post. (tan 38° = 0.7813,\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the lamp post be “h”
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ ABC,

tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 60 }{ x } \)
x = \(\frac{60}{\sqrt{3}}\) ……..(1)
In the right ∆ DEC, tan 38° = \(\frac { EC }{ DE } \)
0.7813 = \(\frac { 60-h }{ x } \)
x = \(\frac { 60-h }{ 0.7813 } \) …….(2)
From (1) and (2) we get
\(\frac{60}{\sqrt{3}}\) = \(\frac { 60-h }{ 0.7813 } \)
60 × 0.7813 = 60 \(\sqrt { 3 }\) – \(\sqrt { 3 }\) h
\(\sqrt { 3 }\) h = 60 \(\sqrt { 3 }\) – 46.88
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732 h = 57.04 ⇒ h = \(\frac { 57.04 }{ 1.732 } \)
h = \(\frac { 570440 }{ 1732 } \) = 32.93 m
∴ Height of the lamp post = 32.93 m

Question 4.
An aeroplane at an altitude of 1800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are
60° and 30° respectively. Find the distance between the two boats. (\(\sqrt { 3 }\) = 1.732)
Answer:
C and D are the position of the two boats.
Let the distance between the two boats be “x”
Let BC = y
∴ BD = (x + y)

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BD } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 1800 }{ x+y } \)
x + y = 1800 \(\sqrt { 3 }\)
y = 1800 \(\sqrt { 3 }\) – x ……(1)
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \)
\(\sqrt { 3 }\) = \(\frac { 1800 }{ y } \)
y = \(\frac{1800}{\sqrt{3}}\) ……….(2)
From (1) and (2) we get
\(\frac{1800}{\sqrt{3}}\) = 1800 \(\sqrt { 3 }\) – x
1800 = 1800 × 3 – \(\sqrt { 3 }\)x
\(\sqrt { 3 }\)x = 5400 – 1800
x = \(\frac{3600}{\sqrt{3}}=\frac{3600 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3600 \times \sqrt{3}}{3}\)
= 1200 × 1.732 = 2078.4 m
Distance between the two boats = 2078.4 m

Question 5.
From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \(\frac{4 h}{\sqrt{3}}\) m.
Answer:
A and C be the position of two ships.
Let AB be x and BC be y. Distance between the two ships is x + y

In the right ∆ ABD, tan 60° = \(\frac { BD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ……(1)
In the right ∆ BCD,
tan 30° = \(\frac { BD }{ BC } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ y } \)
y = \(\sqrt { 3 }\) h
Distance between the two ships (x + y) = \(\frac{h}{\sqrt{3}}+\sqrt{3} h\)
= \(\frac{h+3 h}{\sqrt{3}}=\frac{4 h}{\sqrt{3}}\)
Hence it is verified

Question 6.
A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is 30 \(\sqrt { 3 }\) feet from the entrance of the lift, find the speed of the lift which is descending.
Answer:
Let the speed of the lift is “x” feet / minute
Distance AB = 2 x feet (speed × time)
BC = (90 – 2x)
In the right ∆ BCD,

tan 30° = \(\frac { BC }{ DC } \)
\(\frac{1}{\sqrt{3}}=\frac{90-2 x}{30 \sqrt{3}}\)
\(\sqrt { 3 }\) (90 – 2x) = 30\(\sqrt { 3 }\)
(90 – 2x) = \(\frac{30 \sqrt{3}}{\sqrt{3}}\) ⇒ (90 – 2x) = 30
2x = 60
x = \(\frac { 60 }{ 2 } \) = 30
x = 30 feet/minute
Speed of the lift = 30 feet / minute (or) [ \(\frac { 30 }{ 60 } \) second) 0.5 feet / second

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

Question 1.
Find the area of the triangle formed by the points
(i) (1,-1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:
(i) Let the vertices A (1, -1), B (-4, 6) and C (-3, -5)


= \(\frac { 1 }{ 2 } \) [(6 + 20 + 3) – (4 – 18 – 5)] = \(\frac { 1 }{ 2 } \) [29 – (-19)] = \(\frac { 1 }{ 2 } \) [29 + 19]
= \(\frac { 1 }{ 2 } \) × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)

Area of ∆ABC = \(\frac { 1 }{ 2 } \)[(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(50 + 3 + 32) – (12 + 40 + 10)]

= \(\frac { 1 }{ 2 } \) [85 – (62)] = \(\frac { 1 }{ 2 } \) [23] = 11.5
Area of ∆ACB = 11.5 sq.units

Question 2.
Determine whether the sets of points are collinear?
(i) (-\(\frac { 1 }{ 2 } \),3)
(ii) (a,b + c), (b,c + a) and (c,a + b)
Solution:
(i) Let the points be A (-\(\frac { 1 }{ 2 } \),3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₃ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(- 3 – 40 – 24) – (-15 – 48 – 4)]

= \(\frac { 1 }{ 2 } \) [-67 + 67] = \(\frac { 1 }{ 2 } \) × 0 = 0
Area of a ∆ is 0.
∴ The three points are collinear.

(ii) Let the points be A (a, b + c), B (b, c + a) and C (c, a + b)
Area of the triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]

Since the area of a triangle is 0.
∴ The given points are collinear.

Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’

Solution:
(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 20

\(\frac { 1 }{ 2 } \) [(0 + 2p + 0) – (0 + 48 + 0)] = 20
\(\frac { 1 }{ 2 } \) [2p – 48] = 20
2p – 48 = 40 ⇒ 2p = 40 + 48
p = \(\frac { 88 }{ 2 } \) = 44
The value of p = 44

(ii) Let the vertices be A (p, p), B (5, 6) and C (5, -2)
Area of a triangle = 32 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 32

\(\frac { 1 }{ 2 } \) [6p – 10 + 5p) – (5p + 30 – 2p)] = 32
\(\frac { 1 }{ 2 } \) [11 p – 10 – 3p – 30] = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40 ⇒ 8p = 104
p = \(\frac { 104 }{ 8 } \) = 13
The value of p = 13

Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2,3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and (-4 -a, 6 – 2a).
Solution:
(i) Let the points be A (2, 3), B(4, a) and C(6, -3).
Since the given points are collinear.
Area of a triangle = 0
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 0)

\(\frac { 1 }{ 2 } \) [(2a – 12 + 18) – (12 + 6a – 6)] = 0
2a + 6 – (6 + 6a) = 0
2a + 6 – 6 – 6a = 0
-4a = 0 ⇒ a = \(\frac { 0 }{ 4 } \) = 0
The value of a = 0

(ii) Let the points be A (a, 2 – 2a), B (-a + 1, 2a) C (-4 -a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 0

6a² – 2a – 2 – (-2a² – 6a + 2) = 0
6a² – 2a – 2 + 2a² + 6a – 2 = 0
8a² + 4a – 4 = 0 (Divided by 4)
2a² + a – 1 = 0
2a² + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0

(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = \(\frac { 1 }{ 2 } \)
The value of a = -1 (or) \(\frac { 1 }{ 2 } \)

Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8,6), (-1, -2) and (-6, -3)
Solution:
(i) Let the vertices A (-9, -2), B(-8, -4), C(2, 2) and D(1, -3).
Plot the vertices in a graph.

[Note: Consider the points in counter clock wise order]

Area of the Quadrilateral ABDC = \(\frac { 1 }{ 2 } \) [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= \(\frac { 1 }{ 2 } \) [58 – (-12)] – \(\frac { 1 }{ 2 } \)[58 + 12]
= \(\frac { 1 }{ 2 } \) × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

(ii) Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB

= \(\frac { 1 }{ 2 } \) [33 + 35] = \(\frac { 1 }{ 2 } \) × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units

Question 6.
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:
Let the vertices A (-A, -2), B (-3, k), C (3, -2) and D (2, 3)
Area of the Quadrilateral = 28 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 28

-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – \(\frac { 35 }{ 7 } \) = -5
The value of k = -5

Question 7.
If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.
Solution:
Since the three points are collinear
Area of a ∆ = 0

-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0

Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
Let the vertices of the ∆ABC be A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)

Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:

= \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]
= \(\frac { 1 }{ 2 } \) [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= \(\frac { 1 }{ 2 } \) [212 – (-212)]

= \(\frac { 1 }{ 2 } \) [212 + 212] = \(\frac { 1 }{ 2 } \) [424] = 212 sq. units

= \(\frac { 1 }{ 2 } \) [90 – (-90)]

= \(\frac { 1 }{ 2 } \) [90 + 90]
= \(\frac { 1 }{ 2 } \) × 180 = 90 sq. units
Area of the patio = Area of the Quadrilateral ABCD – Area of the Quadrilateral EFGH
= (212 – 90) sq. units
Area of the patio = 122 sq. units

Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(l, 6) and C(7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:
Given the vertices of the triangular glass is A (-5, -4), B (1, 6), and C (7, -4)


= \(\frac { 1 }{ 2 } \) [(20 + 42 – 4) – (-28 – 4 – 30)]
= \(\frac { 1 }{ 2 } \) [58 – (-62)]
= \(\frac { 1 }{ 2 } \) [58 + 62]
= \(\frac { 1 }{ 2 } \) × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = \(\frac { 60 }{ 6 } \) = 10 ⇒ Number of cans = 10

Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG.

Solution:
Area of a triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
(i) Area of ∆AGF = \(\frac { 1 }{ 2 } \) [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= \(\frac { 1 }{ 2 } \) [-22 – (-29.5)]

= \(\frac { 1 }{ 2 } \) [-22 + 29.5]
= \(\frac { 1 }{ 2 } \) × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = \(\frac { 1 }{ 2 } \) [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]
= \(\frac { 1 }{ 2 } \) [5.5 – (-0.5)]

= \(\frac { 1 }{ 2 } \) [5.5 + 0.5] = \(\frac { 1 }{ 2 } \) × 6 = 3 sq.units

(iii)

= \(\frac { 1 }{ 2 } \) [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]
= \(\frac { 1 }{ 2 } \) [15.75 + 12]
= \(\frac { 1 }{ 2 } \) [27.75] = 13.875
= 13.88 sq. units

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

Question 1.
What is the slope of a line whose inclination with positive direction of x -axis is
(i) 90°
(ii) 0°
Solution:
Here θ = 90°
Slope (m) = tan θ
Slope = tan 90°
= undefined.

(ii) Here θ = 0°
Slope (m) = tan θ
Slope = tan 0°
= 0

Question 2.
What is the inclination of a line whose slope is
(i) 0
(ii) 1
Solution:
(i) m = 0
tan θ = 0 ⇒ θ = 0°
(ii) m = 1 ⇒ tan θ = tan 45° ⇒ 0 = 45°

Question 3.
Find the slope of a line joining the points
(i) (5,\(\sqrt { 5 }\)) with the origin
(ii) (sin θ, -cos θ) and (-sin θ, cos θ)
Solution:
(i) The given points is (5,\(\sqrt { 5 }\)) and (0, 0)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = \(\frac{0-\sqrt{5}}{0-5}\)
= \(\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)

(ii) The given points is (sin θ, -cos θ) and (-sin θ, cos θ)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{\cos \theta+\cos \theta}{-\sin \theta-\sin \theta}\)
= \(\frac{2 \cos \theta}{-2 \sin \theta}\) = – cot θ

Question 4.
What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6,4).
Solution:
Mid point of XY = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) = (\(\frac { 4-6 }{ 2 } \),\(\frac { 2+4 }{ 2 } \))
= (\(\frac { -2 }{ 2 } \),\(\frac { 6 }{ 2 } \)) = (-1, 3)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = (\(\frac { 3-1 }{ -1-5 } \))
= \(\frac { 2 }{ -6 } \) = – \(\frac { 1 }{ 3 } \)

Question 5.
Show that the given points are collinear: (-3, -4), (7,2) and (12, 5)
Solution:
The vertices are A(-3, -4), B(7, 2) and C(12, 5)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 2+4 }{ 7+3 } \) = \(\frac { 6 }{ 10 } \) = \(\frac { 3 }{ 5 } \)
Slope of BC = \(\frac { 5-2 }{ 12-7 } \) = \(\frac { 3 }{ 5 } \)
Slope of AB = Slope of BC = \(\frac { 3 }{ 5 } \)
∴ The three points A,B,C are collinear.

Question 6.
If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Solution:
The vertices are A(3, -1), B(a, 3) and C(1, -3)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 3+1 }{ a-3 } \) = \(\frac { 4 }{ a-3 } \)
Slope of BC = \(\frac { 3+3 }{ a-1 } \) = \(\frac { 6 }{ a-1 } \)
Since the three points are collinear.
Slope of AB = Slope BC
\(\frac { 4 }{ a-3 } \) = \(\frac { 6 }{ a-1 } \)
6 (a – 3) = 4 (a – 1)
6a – 18 = 4a – 4
6a – 4a = -4 + 18
2a = 14 ⇒ a = \(\frac { 14 }{ 2 } \) = 7
The value of a = 7

Question 7.
The line through the points (-2, a) and (9,3) has slope –\(\frac { 1 }{ 2 } \) Find the value of a.
Solution:
The given points are (-2, a) and (9, 3)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
– \(\frac { 1 }{ 2 } \) = \(\frac { 3-a }{ 9+2 } \) ⇒ – \(\frac { 1 }{ 2 } \) = \(\frac { 3-a }{ 11 } \)
2(3 – a) = -11 ⇒ 6 – 2a = -11
-2a = -11 – 6 ⇒ -2a = -17 ⇒ a = – \(\frac { 17 }{ 2 } \)
∴ The value of a = \(\frac { 17 }{ 2 } \)

Question 8.
The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8,12) and (x, 24). Find the value of x.
Solution:
Find the slope of the line joining the point (-2, 6) and (4, 8)
Slope of line (m1) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-6 }{ 4+2 } \) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)
Find the slope of the line joining the points (8, 12) and (x, 24)
Slope of a line (m2) = \(\frac { 24-12 }{ x-8 } \) = \(\frac { 12 }{ x-8 } \)
Since the two lines are perpendicular.
m₁ × m₂ = -1
\(\frac { 1 }{ 3 } \) × \(\frac { 12 }{ x-8 } \) = -1 ⇒ \(\frac{12}{3(x-8)}=-1\)
-1 × 3 (x – 8) = 12
-3x + 24 = 12 ⇒ – 3x = 12 -24
-3x = -12 ⇒ x = \(\frac { 12 }{ 3 } \) = 4
∴ The value of x = 4

Question 9.
Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem.
(i) A(1, -4) , B(2, -3) and C(4, -7)
(ii) L(0, 5), M(9,12) and N(3,14)
Solution:
(i) The vertices are A(1, -4), B(2, -3) and C(4, -7)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3+4 }{ 2-1 } \) = \(\frac { 1 }{ 1 } \) = 1
Slope of BC = \(\frac { -7+3 }{ 4-2 } \) = \(\frac { -4 }{ 2 } \) = -2
Slope of AC = \(\frac { -7+4 }{ 4-1 } \) = – \(\frac { 3 }{ 3 } \) = -1
Slope of AB × Slope of AC = 1 × -1 = -1

∴ AB is ⊥^r to AC
∠A = 90°
∴ ABC is a right angle triangle
Verification:

20 = 2 + 18
20 = 20 ⇒ Pythagoras theorem verified

(ii) The vertices are L(0, 5), M(9, 12) and N(3, 14)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of LM = \(\frac { 12-5 }{ 9-0 } \) = \(\frac { 7 }{ 9 } \)

Slope of MN = \(\frac { 14-12 }{ 3-9 } \) = \(\frac { 2 }{ -6 } \) = – \(\frac { 1 }{ 3 } \)
Slope of LN = \(\frac { 14-5 }{ 3-0 } \) = \(\frac { 9 }{ 3 } \) = 3
Slope of MN × Slope of LN = – \(\frac { 1 }{ 3 } \) × 3 = -1
∴ MN ⊥ LN
∠N = 90°
∴ LMN is a right angle triangle
Verification:


130 = 90 + 40
130 = 130 ⇒ Pythagoras theorem is verified

Question 10.
Show that the given points form a parallelogram:
A (2.5,3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5).
Solution:
Let A(2.5, 3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5) are the vertices of a parallelogram.


Slope of AB = Slope of CD = -1
∴ AB is Parallel to CD ……(1)

Slope of BC = Slope of AD
∴ BC is parallel to AD
From (1) and (2) we get ABCD is a parallelogram.

Question 11.
If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Solution:
Let A(2, 2), B(-2, -3), C(1, -3) and D(x, y) are the vertices of a parallelogram.

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-2 }{ -2-2 } \) = \(\frac { -5 }{ -4 } \) = \(\frac { 5 }{ 4 } \)
Slope of BC = \(\frac { -3+3 }{ -2-1 } \) = \(\frac { 0 }{ -3 } \) = 0
Slope of CD = \(\frac { y+3 }{ x-1 } \)
Slope of AD = \(\frac { y-2 }{ x-2 } \)
Since ABCD is a parallelogram
Slope of AB = Slope of CD
\(\frac { 5 }{ 4 } \) = \(\frac { y+3 }{ x-1 } \)
5(x – 1) = 4 (y + 3)
5x – 5 = 4y + 12
5x – 4y = 12 + 5
5x – 4y = 17 ……(1)
Slope of BC = Slope of AD
0 = \(\frac { y-2 }{ x-2 } \)
y – 2 = 0
y = 2
Substitute the value of y = 2 in (1)
5x – 4(2) = 17
5x -8 = 17 ⇒ 5x = 17 + 18
5x = 25 ⇒ x = \(\frac { 25 }{ 5 } \) = 5
The value of x = 5 and y = 2.

Question 12.
Let A(3, -4), B(9, -4) , C(5, -7) and D(7, -7). Show that ABCD is a trapezium.
Solution:
Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7) are the vertices of a quadrilateral.

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -4+4 }{ 9-3 } \) = \(\frac { 0 }{ 6 } \) = 0
Slope of BC = \(\frac { -7+4 }{ 5-9 } \) = \(\frac { -3 }{ -4 } \) = \(\frac { 3 }{ 4 } \)
Slope of CD = \(\frac { -7+7 }{ 7-5 } \) = \(\frac { 0 }{ 2 } \) = 0
Slope of AD = \(\frac { -7+4 }{ 7-3 } \) = \(\frac { -3 }{ 4 } \) = – \(\frac { 3 }{ 4 } \)
The slope of AB and CD are equal.
∴ AB is parallel to CD. Similarly the slope of AD and BC are not equal.
∴ AD and BC are not parallel.
∴ The Quadrilateral ABCD is a trapezium.

Question 13.
A quadrilateral has vertices at A(-4, -2), B(5, -1) , C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram.
Solution:
Let A(-4, -2), B(5, -1), C(6, 5) and D(-7, 6) are the vertices of a quadrilateral.





Slope of EF = Slope of GH = \(\frac { 7 }{ 10 } \)
∴ EF || GH …….(1)
Slope of FG= Slope of EH = – \(\frac { 7 }{ 12 } \)
∴ FG || EH ……(2)
From (1) and (2) we get EFGH is a parallelogram.
The mid point of the sides of the Quadrilateral ABCD is a Parallelogram.

Question 14.
PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.
Solution:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of SM = \(\frac { 1+1 }{ 1-2 } \) = \(\frac { 2 }{ -1 } \) = -2
Slope of PM = \(\frac { 1 }{ 2 } \) (Since SM and PM are ⊥^r)
Let the point p be (a,b)
Slope of PM = \(\frac { 1 }{ 2 } \)
\(\frac { b+1 }{ a-2 } \) = \(\frac { 1 }{ 2 } \) ⇒ a – 2 = 2b + 2

a – 2b = 4
a = 4 + 2b ……(1)
Given QS = 2PR
\(\frac { QS }{ 2 } \) = PR
∴ SM = PR
SM = 2PM (PR = 2PM)

Squaring on both sides

∴ (b + 1)² = \(\frac { 1 }{ 4 } \) ⇒ b + 1 = ± \(\frac { 1 }{ 2 } \)
b = \(\frac { 1 }{ 2 } \) – 1 (or) b = – \(\frac { 1 }{ 2 } \) – 1
= – \(\frac { 1 }{ 2 } \) – 1 (or) b = –\(\frac { 1 }{ 2 } \) – 1
= – \(\frac { 1 }{ 2 } \) (or) – \(\frac { 3 }{ 2 } \)
a = 4 + 2b
a = 4 + 2 (\(\frac { -1 }{ 2 } \))
a = 3
a = 4 + 2 (\(\frac { -3 }{ 2 } \))
a = 4 – 3
a = 1
The point of p is (3,\(\frac { -1 }{ 2 } \)) (or) (1,\(\frac { -3 }{ 2 } \))

Students can download Maths Chapter 2 Numbers and Sequences Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Additional Questions

I. Choose the correct answer.

Question 1.
The sum of the exponents of the prime factors in the prime factorisation of 504 is ……….
(1) 3
(2) 2
(3) 1
(4) 6
Answer:
(4) 6

Hint: 504 = 2³ × 3² × 7¹
Sum of the exponents
= 3 + 2 + 1 = 6

Question 2.
If two positive integers a and 6 are expressible in the form a = pq² and b = p³q ; p, q being prime numbers, then L.C.M. of (a, b) is ……………
(i) pq
(2) P² q²
(3) p³ q³
(4) P³ q²
Answer:
(4) P³ q²
Hint: a = p × q² and b = p³ × q
∴ L.C.M. of (a, b) is p³ q²

Question 3.
If n is a natural number then 7³ⁿ – 4³ⁿ is always divisible by …………….
(1) 11
(2) 3
(3) 33
(4) both 11 and 3
Answer:
(4) both 11 and 3
Hint: 7³ⁿ – 4³ⁿ is of the form a²ⁿ – b²ⁿ which is divisible by both a – b and a + b. So 7³ⁿ – 4³ⁿ is divisible by both 7 – 4 = 3 and 7 + 4 = 11.

Question 4.
The value of x when 200 = x (mod 7) is …………………
(1) 3
(2) 4
(3) 54
(4) 12
Answer:
(2) 4
Hint:

200 ≡ x (mod 7)
200 ≡ 4 (mod 7)
The value of x = 4

Question 5.
The common difference of the A.P.
\(\frac { -2 }{ 2b } \) , \(\frac { 1-6b }{ 2b } \), \(\frac { 1-12b }{ 2b } \) is …………….
(1) 2b
(2) -2b
(3) 3
(4) -3
Answer:
(4) -3
Hint:
\(\frac { 1-12b }{ 2b } \) – \(\frac { 1-6b }{ 2b } \) = \(\frac { 1-12b-1+6b }{ 2b } \) = \(\frac { -6b2 }{ 2b } \) = \(\frac { -6b }{ 2b } \) = -3

Question 6.
Which one of the following is not true?
(1) A sequence is a real valued function defined on N.
(2) Every function represents a sequence.
(3) A sequence may have infinitely many terms.
(4) A sequence may have a finite number of terms.
Answer:
(2) Every function represents a sequence.
Hint: A sequence is a function whose domain is the set of natural numbers.

Question 7.
The 8^th term of the sequence 1,1,2,3,5,8, ………. is ……….
(1) 25
(2) 24
(3) 23
(4) 21
Answer:
(4) 21
Hint: In fibonacci sequence
F_n = F_n-1 + F_n-2
F₈ = F₇ + F₆
8^th term = 6^th term + 7^th term
= 8 + (5 + 8) = 21

Question 8.
The next term of \(\frac { 1 }{ 20 } \) in the sequence \(\frac { 1 }{ 2 } \),\(\frac { 1 }{ 6 } \),\(\frac { 1 }{ 12 } \),\(\frac { 1 }{ 20 } \) is ……………
(1) \(\frac { 1 }{ 24 } \)
(2) \(\frac { 1 }{ 22 } \)
(3) \(\frac { 1 }{ 30 } \)
(4) \(\frac { 1 }{ 18 } \)
Answer:
(3) \(\frac { 1 }{ 30 } \)
Hint:
The general term t_n = \(\frac{1}{n(n+1)}\)
⇒ t₅ = \(\frac{1}{5(6)}\) = \(\frac { 1 }{ 30 } \)

Question 9.
If a, 6, c, l, m are in A.P, then the value of a – 46 + 6c – 4l + m is …………
(1) 1
(2) 2
(3) 3
(4) 0
Answer:
(4) 0
Hint: Given, a, b, c, l, m, are in A.P.
a = a; b = a + d; c = a + 2 d; 1 = a + 3d;
m = a + 4d [The general form of A.P.]
a – 4b + 6c – 4l + m
= a – 4(a + d) + 6(a + 2d) – 4 (a + 3d) + a + 4d
= a – 4a – 4d+ 6a + 12 d – 4a – 12d + a + 4d
= a + 6a + a – 4a – 4a
= 8a – 8a
= 0

Question 10.
If a, b, c are in A.P. then \(\frac { a-b }{ b-c } \) is equal to ……………
(1) \(\frac { a }{ b } \)
(2) \(\frac { b }{ c } \)
(3) \(\frac { a }{ c } \)
(4) 1
Answer:
(4) 1
Hint: a, b, c are in A.P.
b – a = c-b [common difference is same t₂ – t₁ = t₃ – t₂]
\(\frac { b-a }{ c-b } \) = 1 ⇒ \(\frac{-(a-b)}{-(b-c)}\) = 1
∴ \(\frac { a-b }{ b-c } \) = 1

Question 11.
If the n^th term of a sequence is 100n + 10, then the sequence is ……
(1) an A.P.
(2) a G..P.
(3) a constant sequence
(4) neither A.P. nor G.P.
Answer:
(1) an A.P.
Hint: t_n = 100/n + 10
t₁ = 100 + 10 = 110
t₂ = 200+ 10 = 210
t₃ = 300+ 10 = 310
∴ The series 110, 210, 310 …………. are in A.P.

Question 12.
If a₁, a₂, a₃, …… are in A.P. such that \(\frac{a_{4}}{a_{7}}=\frac{3}{2}\), then the 13th term of the A.P. is ……………..
(1) \(\frac { 3 }{ 2 } \)
(2) 0
(3) 12a1
(4) 14a1
Answer:
(2) 0
Hint:
\(\frac{a_{4}}{a_{7}}=\frac{3}{2}\)
2a₄ = 3a₇
2(a + 3d) = 3(a + 6d)
2a + 6d = 3a + 18d
0 = a + 12d
[t_n = a + (n – 1)d]
0 = t₁₃

Question 13.
If the sequence a₁, a₂, a₃ ,… is in A.P., then the sequence a₅, a₁₀, a₁₅, …. is …..,
(1) a G.P.
(2) an A.P.
(3) neither A.P nor G.P.
(4) a constant sequence
Answer:
(2) an A.P.
Hint: a₅, a₁₀, a₁₅, ……….. = a + 4d, a + 9d, a + 14d
t₂ – t₁ = a + 9d – (a + 4d) = 5d
t₃ – t₂ = a + 14d – (a + 9d) = 5d
Common difference is 5d
If terms of an A.P are chosen at equal intervals then they form an A.P.

Question 14.
If k + 2,4k – 6, 3k – 2 are the 3 consecutive terms of an A.P, then the value of K is ……………
(1) 2
(2) 3
(3) 4
(4) 5
Answer:
(2) 3
Hint: Here, a = k + 2, b = 4k-6, c = 3k-2
We know that a, b, c are in A.P.
b – a = c – b ⇒ 2b = a + c
2(4k – 6) = k + 2 + 3k – 2
8k – 12 = 4k ⇒ 4k = 12
K = \(\frac { 12 }{ 4 } \) = 3

Question 15.
If a, b, c, l, m, n are in A.P, then 3a + 7, 3b + 7, 3c + 7, 31 + 7, 3m + 7, 3n + 7 form ………..
(1) a G . P.
(2) an A.P.
(3) a constant sequence
(4) neither A.P. nor G.P.
Answer:
(2) an A.P.
Hint: In an A.P, if each term is multiplied by a constant or added by a constant, the resulting sequence is an A.P.

Question 16.
If the third term of a G.P. is 2, then the product of first 5 terms is ……………..
(1) 5²
(2) 2⁵
(3) 10
(4) 15
Answer:
(2) 2⁵
Hint: Consider the 5 terms of the G.P be \(\frac{a}{r^{2}}\),\(\frac { a }{ r } \), a, ar, ar²
Product of 5 terms = \(\frac{a}{r^{2}} \times \frac{a}{r} \times a \times a r \times ar^{2}\), = a⁵, 2⁵ (Given a = 2)

Question 17.
If a, b, c are in G.P., then \(\frac { a-b }{ b-c } \) is equal to …………..
(1) \(\frac { a }{ b } \)
(2) \(\frac { b }{ a } \)
(3) \(\frac { a }{ c } \)
(4) \(\frac { c }{ b } \)
Ans.
(1) \(\frac { a }{ b } \)
Hint: Let the common ratio be “r”

Question 18.
If x, 2x + 2, 3x + 3, are in G.P., then 5x, 10x + 10, 15x + 15, form …………..
(1) anA.P.
(2) a G.P.
(3) a constant sequence
(4) neither A.P. nor a G.P.
Answer:
(2) a G.P.
Hint: If a G.P. is multiplied by a constant then the sequence is also a G.P.
In the given question each term is multiplied by 5

Question 19.
The sequence – 3, – 3, – 3, …… is ……..
(1) an A.P. only
(2) a G.P. only
(3) neither A.P. nor G.P.
(4) both A.P. and G.P.
Answer:
(4) both A.P. and G.P.
Hint: The given sequence is constant.
The sequence is A.P. and G.P.

Question 20.
If the product of the first four consecutive terms of a G.P is 256 and if the common ratio is 4 and the first term is positive, then its 3^rd term is …….
(1) 8
(2) \(\frac { 1 }{ 16 } \)
(3) \(\frac { 1 }{ 32 } \)
(4) 16
Answer:
(1) 8
Hint: The general form of the G.P. is a, ar, ar², ar³, ar⁴, …………..
By data, a (ar) (ar²) (ar³) = 256
a⁴ × r⁶ = 256
[Given r = 4]

Question 21.
In G.P, t₂ = \(\frac { 3 }{ 5 } \) and t₃ = \(\frac { 1 }{ 5 } \) Then the common ratio is ……….
(1) \(\frac { 1 }{ 5 } \)
(2) \(\frac { 1 }{ 3 } \)
(3) 1
(4) 5
Answer:
(2) \(\frac { 1 }{ 3 } \)
Hint: common ratio is
(r) = \(\frac{t_{3}}{t_{2}}=\frac{1}{5} \times \frac{5}{3} \Rightarrow r=\frac{1}{3}\)

Question 22.
If x ≠ 0, then 1 + sec x + sec² x + sec³ x + sec⁴ x + sec⁵ x is equal to ……………
(1) (1 + sec x) (sec² x + sec³ x + sec⁴ x)
(2) (1 + sec x) (1 + sec² x + sec⁴ x)
(3) (1 – sec x) (sec x + sec³ x + sec⁵ x)
(4) (1 + sec x) (1 + sec³ x + sec⁴ x)
Answer:
(2) (1 + sec x) (1 + sec² x + sec⁴ x).
Hint:
1 + sec x + sec² x + sec³ x
+ sec⁴ x + sec⁵ x
= (1 + sex x) + sec² x (1 + sec x) + sec⁴ x (1 + see x)
= (1 + sec x) + sec² x (1 + sec x) + sec⁴ x (1 + secx)
= (1 + sec x) (1 + sec² x + sec⁴ x)

Question 23.
If the n^th term of an A.P. is t_n = 3 – 5n, then the sum of the first n terms is …………….
(1) \(\frac { n }{ 2 } \) [1 – 5n]
(2) n (1 – 5n)
(3) \(\frac { n }{ 2 } \) (1 + 5n)
(4) \(\frac { n }{ 2 } \) (1 + n)
Answer:
(1) \(\frac { n }{ 2 } \) [1 – 5n]
Hint:
t_n =. 3 – 5(n)
t₁ = 3 – 5(1) = -2 ; t₂ = 3 – 10 = -7
a = -2, d = -7 – (-2) = -5
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
= \(\frac { n }{ 2 } \) [-4 + (n – 1) (-5)]
= \(\frac { n }{ 2 } \) [- 4 -5n + 5] = \(\frac { n }{ 2 } \) [1 – 5n]

Question 24.
The common ratio of the G.P. a^m-n, a^m, a^m+n is …………
(1) a^m
(2) a^-m
(3) aⁿ
(4) a^-n
Answer:
(3) aⁿ

Question 25.
If 1 + 2 + 3 + … + n = k then 1³ + 2³ + ……. + n³ is equal to …………
(1) K²
(2) K³
(3) \(\frac{k(k+1)}{2}\)
(4) (K + 1)³
Answer:
(1) K²

II. Answer the following.

Question 1.
Show that the square of any positive integer of the form 3m or 3m + 1 for some integer m.
Answer:
Let a be any positive integer. Then it is of the form 3q or 3q + 1 or 3q + 2
Case – 1 When a = 3q
a² = (3q)² = 9 q²
= (3q) (3q)
= 3m where m = 3q
Case – 2 When a = 3q + 1
a² = (3q + 1)² = 9q² + 6q + 1
= 3q (3q + 2) + 1
= 3 m + 1
where m = q (3q + 2)
Case – 3 When a = 3q + 2
a² = (3q + 2)² = 9q² + 12q + 4
= 9q² + 12q + 3 + 1
= 3(3q² + 4q + 1) + 1
= 3m + 1
where m = 3q² + 4q + 1
Hence a is of the form 3m or 3m + 1

Question 2.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Solution:
Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with ‘a’ and ‘b’ = 4.
Since 0 < r < 4, the possible remainders are 0, 1, 2, 3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3.

Question 3.
Compute x such that 5⁴ = x (mod 8)
Answer:
5² = 25 = 1 (mod 8)
5⁴ = (5²)² = l² (mod 8)
= 1
5⁴ = 1 (mod 8)

Question 4.
The first term of an A.P. is 6 and the common difference is 5. Find the A.P. and its general term.
Answer:
Given, a = 6, d = 5
General term t_n = a + (n – 1) d
= 6 + (n – 1) 5
= 6 + 5 n – 5
= 5 n + 1
The general form of the A.P. is a, a + d, a + 2d,
The A.P. is 6, 11, 16, 21, ……… 5n + 1.

Question 5.
Which term of the arithmetic sequence 24, 23\(\frac { 1 }{ 4 } \), 22 \(\frac { 1 }{ 2 } \), 21 \(\frac { 3 }{ 4 } \), …….. is 3?
Answer:
Given The A.P is 24, 23 \(\frac { 1 }{ 4 } \), 22 \(\frac { 1 }{ 2 } \), 21 \(\frac { 3 }{ 4 } \), …………..

Question 6.
Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution:
We have
a₃ = a + (3 – 1)d = a + 2d = 5 ……….. (1)
a₇ = a + (7 – 1)d = a + 6d = 9 ………… (2)
(1) – (2) ⇒ -4d = -4 ⇒ d = 1.
Sub, d = 1 in (1), we get
a + 2(1) = 5
a = 3
Hence the required A.P. is 3, 4, 5, 6, 7.

Question 7.
If a, b, c are in A.P. then prove that (a – c)² = 4 (b² – ac).
Answer:
Given a, b, c are in A.P.
n = 10
∴ b – a = c – b
2 b = a + c
Squaring on both sides,
(a + c)² = (2b)²
a² + c² + 2ac = 4b²
(a – c)² + 2ac + 2ac = 4b²
[a² + c² = (a – c)² + 2ac]
(a – c)² = 4b² – 4ac
(a – c)² = 4(b² – ac)
Hence it is proved.
Aliter: Given a, b, c are in A.P.
b – a = c – b
2b = a + c
To prove, (a – c)² = 4(b² – ac)
L.H.S.= (a – c)²
= a² + c² – 2 ac
= (a + c)² – 2 ac – 2ac
= (a + c)² – 4ac
= (2b)² – 4ac (2b = a + c)
= 4b² – 4ac = 4 (b² – ac) = R.H.S
∴ L.H.S = R.H.S., Hence proved.

Question 8.
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution:
Here S₁₄ = 1050
n = 14
a = 10
S_n = \(\frac { n }{ 2 } \) (2a + (n – 1)d)
1050 = \(\frac { 14 }{ 2 } \) (20 + 13d)
= 140 + 91d
910 = 91d
d = 10
a₂₀ = 10 + (20 – 1) × 10
= 20
∴ 20th term = 200.

Question 9.
Find the sum of the first 40 terms of the series 1² – 2² + 3² – 4² + ….
Answer:
The given series is 1² – 2² + 3² – 4² + …. 40 terms
Grouping the terms we get,
(1² – 2²) + (3² – 4²) + (5² – 6²) + ………….. 20 terms
(1 – 4) + (9 – 16) + (25 – 36) + …………… 20 terms
(- 3) + (- 7) + (- 11) + ………………. 20 term
This is an A.P
Here a = – 3, d = – 7 – (-3) = – 7 + 3 = – 4 n = 20
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1) d ]
S₂₀ = \(\frac { 20 }{ 2 } \) [2(-3) + 19(-4)]
= 10 (- 6 – 76) = 10 (- 82) = – 820
∴ Sum of 40 terms of the series is – 820.
Aliter. 1² – 2² + 3² – 4² + …….. + 39² – 40²
= 1² + 3² + 5² + ……. + 39² –
(2² + 4² + 6² + ……….. + 40²)
= 1² + 2² + 3² + 4² + …. + 40²
– (2² + 4² + 6² + …. + 40²)
– (2² + 4² + 6² + …. + 40²)
= 1² + 2² + 3² + …. + 40²
= 2 × 2² (1² + 2² + ….. + 20²)

Question 10.
Find the sum of first 24 terms of the list of numbers whose nth term is given by
a_n = 3 + 2n.
Solution:
a_n = 3 + 2n
a₁ = 3 + 2 = 5
a₂ = 3 + 2 × 2 = 7
a₃ = 3 + 2 × 3 = 9
List of numbers becomes 5, 7, 9. 11, ……….
Here,7 – 5 = 9 – 7 = 11 – 9 = 2 and soon. So, it forms an A.P. with common difference d = 2.
To find S₂₄, we have n = 24, a = 5, d = 2.
S₂₄ = \(\frac { 24 }{ 2 } \) [2 × 5 + (24 – 1) × 2 ]
= 12 [10 + 46] = 672
So, sum of first 24 terms of the list of numbers is 672.

Question 11.
If a clock strikes once at 1 o’clock, twice at 2 o’clock and so on, how many times will it strike in a day?
Answer:
Number of times the clock strikes each hour form an A.P.
Number of times strike in 12 hours is
1 + 2 + 3 + …….. + 12
Here, a = 1, d = 1, n = 12, l = 12
S_n = \(\frac { n }{ 2 } \) (a + l) = \(\frac { 12 }{ 2 } \) (1 + 12)
= 6 × 13 = 78 times
∴ Number of times the clock strike in 24 hours
= 78 × 2 = 156 times.

Question 12.
If the 4^th and 7^th terms of a G.P. are 54 and 1458 respectively, find the G.P.
Answer:
Given, t₄ = 54 and t₇ = 1458

Substituting the value of r = 3 in (1)
a × 27 = 54
a = \(\frac { 54 }{ 27 } \) = 2
∴ The Geometric sequence is 2, 6, 18, 54 ………

Question 13.
Which term of the geometric sequence,
(i) 5, 2, \(\frac { 4 }{ 5 } \), \(\frac { 8 }{ 25 } \), ……… is \(\frac { 128 }{ 15625 } \)?
Answer:
The given G.P. is 5, 2, \(\frac { 4 }{ 5 } \),\(\frac { 8 }{ 25 } \), …….., is \(\frac { 128 }{ 15625 } \)
Here a = 5, r = \(\frac { 2 }{ 5 } \), t_n = \(\frac { 128 }{ 15625 } \)
t_n = a.r^n-1

Question 14.
How many consecutive terms starting from the first term of the series 2 + 6 + 18 + … would sum to 728?
Answer:
The given series is
2 + 6 + 18 + …. + t_n = 728
Here a = 2, r = = 3, S_n = 728

∴ Required number of terms = 6

Question 15.
A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is 9 and sum of the last two terms is 36. Find the series.
Answer:
Let the four terms of the G.P. be a, ar, ar² and ar³ ……….
Given, a + ar = 9 ……(1)
ar² + ar³ = 36
r² (a + ar) = 36
r² (9) = 36
[from (1)]
r² = =4
r = ± 2
then r = 2
(given common positive ratio)
a + a (2) = 9 from (1)
3a = 9
a = 3
∴ The required series
= 3 + 3(2) + 3 (2²) + 3 (2³) + ……
= 3 + 6 + 12 + 24 + ……..

Question 16.
Suppose that five people are ill during the first week of an epidemic and each sick person spreads the contagious disease to four other people by the end of the second week and so on. By the end of 15^th week, how many people will be affected by the epidemic?
Answer:
Number of people affected by the epidemic during each week form a geometric series.
S₁₅ = 5 + (4 × 5) + (4 × 20) + (4 × 80) + …. 15 terms
= 5 + 20 + 80 + 320 + … 15 terms
Here a = 5, r = 4, n = 15

Question 17.
A gardener wanted to reward a boy for his good deeds by giving some mangoes. He gave the boy two choices. He could either have 1000 mangoes at once or he could get 1 mango on the first day, 2 on the second day, 4 on the third day, 8 mangoes on the fourth day and so on for ten days. Which option should the boy choose to get the maximum number of mangoes?
Answer:
I choice
The boy could get 1000 mangoes at once
II choice
The boy receives mangoes daily for ten days
S₁₀ = 1 + 2 + 4 + 8 + ……… 10 terms

Question 18.
Find the value of k if
1³ + 2³ + 3³ + ….. + K³ = 2025
Answer:
Given, 1³ + 2³ + 3³ + … + K³ = 2025

Question 19.
If 1³ + 2³ + 3³ + …… + K³ = 8281, then find 1 + 2 + 3 + … + K.
Answer:
Given, 1³ + 2³ + 3³ + …… + K³ = 8281

1 + 2 + 3 + …… + K = 91

Question 20.
Find the sum of all 11 term of an AP whose middle most term in 30.
Answer:
Let ‘a’ be the first term and ‘d’ be the common difference of the give A.P.
Middle most term = (\(\frac { 11+1 }{ 2 } \))^th = 6^th term
t_n = a + (n – 1)d
t₆ = a + 5d
a + 5d = 30
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S₁₁ = \(\frac { 11 }{ 2 } \) [2a + 10d]
= \(\frac { 11 }{ 2 } \) × 2 [a + 5d]
= 11 × 30
= 330
Sum of 11 terms = 330

III. Answer the following.

Question 1.
Use Euclid’s division algorithm to find the HCF of 867 and 255.
Answer:
Here 867 > 255
Applying Euclid’s Lemma to 867 and 255 we get

867 = (255 × 3) + 102
The remainder 102 ≠ 0
Again applying Euclid’s
Lemma to 255 and 102
255 = 102 × 2 + 51
The remainder 51 ≠ 0
Again applying Euclid’s
Lemma to 102 and 51 we get
102 = 51 × 2 + 0
The remainder is 0
∴ HCF of 867 and 255 is 51

Question 2.
Find the number of integer solutions of 5x = 2 (mod 13)
Answer:
5x ≡ 2 (mod 13) can be written as
5x – 2 = 13 k for some integer
5x = 13 k + 2
x = \(\frac{13 k+2}{5}\)
Since 5k is an integer \(\frac{13 k+2}{5}\) cannot be an inter.
There is no integer solution.

Question 3.
Find the 40^th term of A.P. whose 9^th term is 465 and 20^th term is 388.
Answer:
t_n = a + (n – 1) d
t₉ = a + 8d (t₉ = 465)
a + 8d = 465 ……(1)

Substitute the value of d = -7 in (1)
a + 8(-7) = 465
a – 56 = 465
a = 465 + 56 = 521
a = 521, d = -7, n = 40
t₄₀ = 521 + 39(-7)
= 521 – 273
= 248
40^th term of an A.P. is 248

Question 4.
Find the three consecutive terms in an A.P. whose sum is 18 and the sum of their squares is 140.
Answer:
Let the three consecutive terms in an
A.P. be m – d,m,m + d
By the given data,
Sum of threee terms = 18
m – d + m + m + d = 18
3m = 18
m = \(\frac { 18 }{ 3 } \) = 6
Again by the given data,
Sum of their squares = 140
(m – d)² + m² + (m + d)² = 140
m² + d² – 2md + m² + m² + d² + 2md = 140
3m² + 2d² = 140
3(62) + 2d² = 140
3(36) + 2d² = 140
2d² = 140 – 108
2d² = 32
d² = \(\frac { 32 }{ 2 } \) = 16
∴ d = ± 4
when, m = 6, d = + 4
m – d = 6 – 4 = 2
m = 6
m + d = 6 + 4 = 10
when, m = 6, d = -4
m – d = 6-(-4) = 6 + 4= 10
m = 6
m + d = 6 +(-4) = 6 – 4 = 2
∴ The three numbers are 2, 6 and 10 or 10, 6,2.

Question 5.
If m times the m^th term of an A.P. is equal to n times its n^th term, then show that the (m + n)^th term of the A.P. is zero.
Answer:
Given, mt_m = nt_n
m[a + (m – 1) d] = n [a + (n – 1) d]
[we know that t_n = a + (n – 1)d]
m[a + md – d] = n[a + nd – d]
ma + m²d – md = na + n²d – nd
ma – na + m²d – n²d = md – nd
a (m – n) + d (m² – n²) = d(m – n)
a (m – n) + d(m + n)(m – n) = d(m – n) ÷ by (m – n) on both sides,
a + d (m + n) = d
a + d(m + n) – d = 0
a + d(m + n – 1) = 0 ….. (1)
To prove, t_(m+n) = 0
t_(m+n) = a + (m + n – 1)d
t_(m+n) = 0(from(1))
Hence it is proved.

Question 6.
If a, b, c are in A.P. then prove that (a – c)² = 4 (b² – ac).
Answer:
Given a, b, c are in A.P.
∴ b – a = c – b
2 b = a + c
Squaring on both sides,
(a + c)² = (2b)²
a² + c² + 2 ac = 4 b²
(a – c)² + 2 ac + 2 ac = 4 b²
[a² + c² = (a – c)² + 2 ac]
(a – c)² = 4b² – 4ac
(a – c)² = 4 (b² – ac)
Hence it is proved.
Aliter: Given a, b, c are in A.P.
b – a = c – b
2 b = a + c
To prove, (a – c)² = 4(b² – ac)
L.H.S. = (a – c)²
= a² + c² – 2ab
= (a + c)² – 2ac – 2ac
= (a + c)² – 4ac
= (2b)² – 4ac (2b = a + c)
= 4 b² – 4ac = 4 (b² – ac) = R.H.S
∴ L.H.S = R.H.S., Hence proved.

Question 7.
The ratio of the sums of first m and first n terms of an arithmetic series is m² : n² show that the ratio of the m^th and n^th terms is (2m – 1) : (2n -1).
Answer:

n[2a + (m – 1) d] = tn[2a + nd-d]
2 an + mnd- nd = 2 am + mnd— md 2an-2am = nd- md 2 a (n-m) = d(n- m)
÷ by (n – m) on both sides,
2a = d
To prove, t_m : t_n = (2m – 1) : (2n – 1)
L.H.S = t_m : t_n
= a + (m – 1) d : a + (n – 1)d
= a + (m – 1) 2a : a + (n – 1)2a
[Substitute the value of d = 2a]
= a + 2 am – 2 a: a + 2 am – 2a
= 2am – a : 2an – a
= a (2m – 1) : a (2n – 1)
= (2m – 1) : (2n – 1) = R. H. S
= (2m – 1) : (2n – 1)
∴ t_m : t_n
L.H.S = R.H.S
Hence it is proved.

Question 8.
A construction company will be penalised each day for delay in construction of a bridge. The penalty will be ₹4000 for the first day and will increase by ₹1000 for each following day. Based on its budget, the company can afford to pay a maximum of ₹1,65,000 towards penalty. Find the maximum number of days by which the completion of work can be delayed.
Solution:
Penalty for the first day (a) = ₹ 4000
Increased rate for every day (d) = ₹ 1000
Maximum amount of penalty (S_n)
= ₹ 1,65,000
S_n = \(\frac { n }{ 2 } \) [2a + (n-1) 1000]
165000 = \(\frac { n }{ 2 } \) [2(4000) + (n – 1) 1000]
= \(\frac { n }{ 2 } \) [8000 + 1000 n – 1000]
165000 = \(\frac { n }{ 2 } \) (7000 + 1000n)
330000 = 7000 n + 1000 n²
0 = 1000 n² + 7000 n – 330000 ÷ 1000 on both sides,
n² + 7n – 330 = 0
(n + 22) (n – 15) = 0
n = -22 or n = 15
n = -22 (not possible)
∴ Maximum number of days for which the work can be delayed is 15 days.

Question 9.
If the product of three consecutive terms in G.P. is 216 and sum of their products in pairs is 156, find them.
Let the three consecutive terms of G.P. be \(\frac { a }{ r } \), a, ar.
Product of three terms = 216
\(\frac { a }{ r } \) × a × ar = 216
a³ = 216
a³ = 63
a = 6
Sum of their products in pairs = 156

Question 10.
If a, b, c, d are in a geometric sequence, then show that
(a – b + c) (b + c + d) = ab + bc + cd.
Answer:
Given, a, b, c, d are in a geometric sequence.
Let a = a, b = ar, c = ar², d = ar³
To prove, (a – b + c) (b + c + d) = ab + bc + cd
L.H.S. = (a – b + c)(b + c + d)
= (a – ar + ar²) (ar + ar² + ar³ )
= a (1 – r + r²)ar (1 + r + r²)
= a²r (1 – r + r²) (1 + r + r²)
= a²r (1 + r² + r4)
= a²r + a²r³ + a²r⁵
= a (ar) + ar (ar²) + ar² (ar³)
= ab + bc + cd
= R.H.S.
L.H.S. = R.H.S., Hence proved.

Question 11.
Find the sum of the first n terms of the series 0.4 + 0.94 + 0.994 + ………..
Answer:
Given series is 0.4 + 0.94 + 0.994 + ……………. + n terms.
S_n = 0.4 + 0.94 + 0.994 + ……….. + n terms

Question 12.
Find the total area of 12 squares whose sides are 12 cm, 13 cm,… 23 cm respectively.
Answer:
Given, the sides of 12 squares are 12 cm, 13 cm, 14 cm,… 23 cm
Total area of 12 squares
= 12² + 13² + 14² + … + 23²
= (1² + 2² + 3² … + 23²) – (1² + 2² + … + 11²)

Students can download Maths Chapter 3 Algebra Ex 3.16 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.16

1. In the

write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to

Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are

The possible orders of the matrix having 6 elements are

Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) a_ij = |i – 2j|
Answer:
a_ij = |i – 2j|
The general 3 × 3 matrices is

a₁₁ = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0 = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix

(ii) a_ij = \(\frac{(i+j)^{3}}{3}\)
Answer:
a₁₁ = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)
a₁₂ = \(\frac{(1+2)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₁₃ = \(\frac{(1+3)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₁ = \(\frac{(2+1)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₂₂ = \(\frac{(2+2)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₃ = \(\frac{(2+3)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₁ = \(\frac{(3+1)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₃₂ = \(\frac{(3+2)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₃ = \(\frac{(3+3)^{3}}{3}\) = \(\frac { 216 }{ 3 } \) = 72
The required matrix

Question 4.
If  then find the tranpose of A.
Answer:

transpose of A = (A^T)

Question 5.
If then find the tranpose of – A
Answer:

Transpose of – A = (-A^T) =

Question 6.
If A = then verify (AT)T = A
Answer:

Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)

Answer:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

(ii)

Answer:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = \(\frac { 8 }{ x } \)
Substitute the value of y = \(\frac { 8 }{ x } \) in (1)
x + \(\frac { 8 }{ x } \) = 6
x² + 8 = 6x
x² – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = \(\frac { 8 }{ 4 } \) = 2 or y = \(\frac { 8 }{ 2 } \) = 4

∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

(iii)

Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)

Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3