A solid disc of mass $M$ is just held in air horizontal 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 $0.05$ kg. What is the mass of the disc?
$(g = 10 m s^{- 2})$

$1.2$ kg

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$0.5$ kg

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$20$ kg

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$3$ kg

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Solution

The correct option is A
$1.2$ kg

Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction
$F = n m v = 40 \times 0.05 \times 6 = M g M = \frac{12}{10} = 1.2 k g$

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A solid disc of mass M is just held in air horizontal 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)

The correct option is A 1.2 kg Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction F=nmv=40×0.05×6=MgM=1210=1.2kg

A solid disc of mass $M$ is just held in air horizontal 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 $0.05$ kg. What is the mass of the disc?
$(g = 10 m s^{- 2})$

The correct option is A
$1.2$ kg

Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction
$F = n m v = 40 \times 0.05 \times 6 = M g M = \frac{12}{10} = 1.2 k g$