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

Q). 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

Q). 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

Q). 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

Q). 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

Q). \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

Q). 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

Q). (1 -cosθ) (1 + cosθ) =
A) sin2θ
B) cos2θ
C) tan2θ
D) cot2θ

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

Q). 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

Q). 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

Q). 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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