A rectangular solid box of length 0.3   m is held horizontally, with one of its sides on the edge of a…

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

1. $0.5$
2. $0.3$
3. $0.02$
4. $0.28$

Solution

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

Asked in: JEE Main 2019 (08 Apr Shift 2)

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