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A rectangular solid box of length $0.3 m$ is held horizontally, with one of its sides on the edge of a platform of height $5 m .$ When released, it slips off the table in a very short time $τ = 0.01 s,$ remainirg essentially horizontal. The angle (in radian) by which it would rotate when it hits the ground will be close to _.

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$0.3$

$0.4$

$0.5$

$0.6$

answer is C.

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Detailed Solution

The box will rotate about the corner of the platform.

$τ \times t = I ω; m g \times \frac{l}{2} \times t = \frac{m l^{2}}{3} \times ω$

$ω = \frac{3 \times g \times t}{2 \times l} = \frac{3 \times 10 \times 0.01}{2 \times 0.3} = 0.5 r a d / s$

Time taken to fall $= \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 s$

Angle rotated in $1 s = ω \times 1 = 0.5 r a d i a n$

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