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 0.05 kg what is the mass of the disc $(g = 10 {ms}^{- 2})$

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$1 .2 kg$

$0 .5 kg$

$20 kg$

$3 kg$

answer is A.

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Detailed Solution

Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction.
$F = nmv = 40 \times 0 .05 \times 6 = Mg \Rightarrow M = \frac{40 \times 0 .05 \times 6}{10} = 1 .2 kg$

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