SATHEE: Laws Of Motion Question 269

Laws Of Motion Question 269

Question: A solid disc of mass M is just held in air horizontally by throwing 40 stones per sec vertically upwards to strike the disc each with a velocity 6 $m s^{- 1}$ . If the mass of each stone is $0.05 k g$ what is the mass of the disc $(g = 10 m s^{- 2}$ ) [Kerala (Engg.) 2001]

Options:

A) $1.2 k g$

B) $0.5 k g$

C) $20 k g$

D) 3kg

Show Answer

Answer:

Correct Answer: A

Solution:

Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction. $F = n m v = 40 \times 0.05 \times 6 = M g$

therefore $M = \frac{40 \times 0.05 \times 6}{10} = 1.2 k g$

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Laws Of Motion Question 269

Question: A solid disc of mass M is just held in air horizontally by throwing 40 stones per sec vertically upwards to strike the disc each with a velocity 6 ms−1 . If the mass of each stone is 0.05kg what is the mass of the disc(g=10ms−2 ) [Kerala (Engg.) 2001]

Options:

Answer:

Solution:

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Question: A solid disc of mass M is just held in air horizontally by throwing 40 stones per sec vertically upwards to strike the disc each with a velocity 6 $m s^{- 1}$ . If the mass of each stone is $0.05 k g$ what is the mass of the disc $(g = 10 m s^{- 2}$ ) [Kerala (Engg.) 2001]