**1)** The inner diameter of a test tube can be measured accurately using a?

**A.**micrometer screw guage**B.**pair of dividers**C.**metre rule**D.**pair of vernier calipers

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##### Correct Answer: Option D

##### Explanation

A Vernier caliper is an instrument used to measure both the outer and inner diameters of a tube.

**2)** Two bodies have masses in the ratio 3:1. They experience forces which impart to them acceleration in the ratio 2:9 respectively. Find the ratio of forces the masses experienced.

**A.**1 : 4**B.**2 : 1**C.**2 : 3**D.**2 : 5

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##### Correct Answer: Option C

##### Explanation

Let the forces be F1, and F2.

F1 = m1 a1 = 3 x 2 = 6N

F2 = m2 a2 = 1 x 9 = 9N

F1 : F2 = 6:9 = 2:3

Thus ratio of forces is 2: 3

3) Particles of mass 10−2kg is fixed to the tip of a fan blade which rotates with angular velocity of 100rad-1. If the radius of the blade is 0.2m, the centripetal force is

**A.**2 N**B.**20 N**C.**200 N**D.**400 N

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##### Correct Answer: Option B

##### Explanation

Centripetal Force (Fc) = mv2/r

= mω2/r

(Since V = ωr)

F = 100 x 0.2

F =20 N

**4)** A lead bullet of mass 0.05kg is fired with a velocity of 200ms−1

into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is

**A.**50 J**B.**100 J**C.**150 J**D.**200 J

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##### Correct Answer: Option A

##### Explanation

From principle of conservation of linear momentum,

(0.05 x 200) + (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).

10 + 0 = V

Thus V = 10m/s.

Recall Kinetic Energy = 1/2mv2

∴∴ K.E = 1/2 (0.05 + 0.95) x 1022

K.E = 1/2 (1 x 100) = 50 J.

**5)** A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms-1 from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is (take g = 10ms-2)

**A.**5 J**B.**10 J**C.**15 J**D.**20 J

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##### Correct Answer: Option C

##### Explanation

Height h = 1/2 (U2/g) = 102/20 = 5m

Energy before hitting ground is = P.E at the height of 5m.

P.E = mgh = 0.1 x 10 x 10= 10 J

K.E = 1/2 MV2 = 5J

T.E = K.E + P.E = 5 + 10 = 15J