Question 224 of 280

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

A$0.5$B$0.3$C$0.02$D$0.28$

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Answer: (A) $0.5$

Angular impulse = change in angular momentum
$\tau \Delta t=\Delta L$
$mg\dfrac{l}{2}\times 0.01=\dfrac{m{l}^{2}}{3}\omega$
$\omega =\dfrac{3g\times 0.01}{2l}=\dfrac{3\times 10\times 0.01}{2\times 0.3}=0.5 rad/sec$
Time taken by rod to hit the ground = $\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times 5}{10}}=1s$
Angle rotated by rod = $\omega t=0.5\times 1=0.5rad$