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 $T = 0.01 s$ , remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to

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0.02

0.28

0.3

0.5

answer is D.

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Detailed Solution

Angle impulse is equal to an angular momentum change.

$τΔt = ΔL$

$mg \frac{l}{2} \times 0.01 = \frac{{ml}^{2}}{3} ω$

$ω = \frac{3 g \times 0.01}{2 l} = \frac{3 \times 10 \times . 01}{2 \times 0.3} = 0.5 rad / s$

the time duration it took the rod to land

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 \sec$

Angle is now rotated by a rod.

$θ = ωt = 0.05 \times 1 = 0.5 radian$

Hence the correct answer is 0.5.

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