TS Polycet (Polytechnic) 2018 Previous Question Paper with Answers And Model Papers With Complete Analysis

41)If the ratio of radii of two spheres is 2 : 3, then the ratio of their surface areas is

A) 3 : 2

B) 4 : 9

C) 9 : 4

D) 8 : 27

View Answer

B) 4 : 9
Explanation:r1: r2 = 2:3
⇒ LSA1: LSA2
= 4 πr12 : 4 πr22
= 22:32
= 4:9:MsgIncorrect:

42)Ratio of volumes of a Cone, a Cylinder and a Hemi-sphere of same base radius and equal heights is ……………..

A) 1 : 2 : 3

B) 3 : 2 : 1

C) 1 : 3 : 2

D) 3 : 1 : 2

View Answer

C) 1 : 3 : 2
Explanation:Required ratio of volumes of a cone, cylinder and a hemi-sphere
= \frac13 πr2h : πr2h : \frac23 πr2h
=1/3 :1/1:2/3
= 1:3:2

43)The area of equilateral triangle is 36 √3 cm2 , then the perimeter is…………………. cm

A) 16

B) 32

C) 36

D) 72

View Answer

C) 36
Explanation:Area of equilateral △le \frac{\sqrt3}4a^2 = 36√3
⇒ a2 = 4 x 36
⇒a = \sqrt{4x36} =2 x 6 = 12 cm
Perimeter of equilateral triangle
= 3 x side = 3 x 12 = 36 cm

44)If 3 tan A = 4 then sin A =

A) 3/4

B) 4/3

C) 4/5

D) 3/5

View Answer

C) 4/5
Explanation:tan A = 4/3
⇒ AC2 = 42 + 32
= 16 + 9 = 25
⇒ AC = 5
∴ Sin A = BC/CA= 4/5

45)\frac{2\tan\left(30^\circ\right)}{\;1-\tan^2\left(30^\circ\right)} =

A) 1/√3

B) √3

C) 1/2

D) 3

View Answer

B) √3
Explanation:\frac{2\tan\theta}{1-\;\tan^2\theta}
= tan2θ = tan.2.30°
= tan 60° = √3

46)tan 48° . tan 42° =

A) 0

B) 1

C) 2

D) 3

View Answer

B) 1
Explanation:tan 48° . tan (90° -48°)
= tan 48° . cot 48°
= tan 48° . 1/ tan 48°
= 1

47)(1 -cosθ) (1 + cosθ) =

A) sin2θ

B) cos2θ

C) tan2θ

D) cot2θ

View Answer

A) sin2θ
Explanation:(1 – cosθ) (1 + cosθ)
= 1 – cos2θ = sin2θ

48)If tanθ -cotθ = 1 then tan2θ + cot2θ =

A) 1

B) 2

C) -1

D) 3

View Answer

D) 3
Explanation:Given that tan θ – cot θ = 1
We know that a2 + b2 = (a -b) 2 + 2ab tan2θ + cot2θ
= (tan θ – cot θ)2 + 2 tanθ . 1/tanθ
= 1+2.1
= 1+2
= 3

49)If x = a cosθ, y = b secθ ⇒

A) xy = a2b2

B) x2 y2 = ab

C) x2 y2 = a2 b2

D) None

View Answer

C) x2 y2 = a2 b2
Explanation:x = a cosθ ⇒ cosθ = x/a
y = b secθ ⇒ secθ = y/b
cos2θ . sec2θ =1
⇒(x/a)2 (y/b)2 = 1
⇒ x2 y2 = a2 b2

50)Length of shadow of a 12 m high pole is 4√3 metres in the morning. What is the angle of elevation of the sun rays with the ground at that time?

A) 30°

B) 45°

C) 15°

D) 60°

View Answer

D) 60°
Explanation:From the figure
tanθ= AB/BC = 12/4√3 = 3/√3=√3
tanθ = tan 60° ⇒θ = 60°
∴Angle of elevation = θ = 60°
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