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A drum of mass $m_{1}$ and radius $r_{1}$ rotates freely with initial angular velocity $ω_{0} .$ A second drum with mass $m_{2}$ and radius $r_{2} (r_{2} > r_{1})$ is mounted on same axle and is at rest although it is free to rotate. A thin layer of sand with mass $m$ is distributed on inner surface of smaller drum. At $t = 0,$ small perforations in the inner drum are opened. The sand starts to fly out at a constant rate $λ k g / s$ and sticks to the outer drum. Ignore the transit time of the sand. Choose the correct alternatives.

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Difference in final angular speeds of two drums is $0$

Difference in final angular speeds of two drums is $(\frac{m (r_{2}^{2} - r_{1}^{2}) + m_{2} r_{2}^{2}}{(m + m_{2}) r_{2}^{2}}) ω_{0}$

Angular speed of outer drum at time is $\frac{λ t ω_{0}}{m_{2} + λ t} {(\frac{r_{1}}{r_{2}})}^{2}$

Angular speed of inner drum remains constant

answer is A, C, D.

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Detailed Solution

As the sand leaves the inner drum through the open holes, it does not exert any force of the drum, angular momentum remains conserved.

At time t,

$m_{1} r_{1}^{2} ω_{0} = (m_{1} - λ t) r_{1}^{2} ω_{0} + (m_{2} + λ t) r_{2}^{2} ω_{2}$

$ω_{2} = \frac{λ t r_{1}^{2} ω_{0}}{(m_{2} + λ t) r_{2}^{2}}$

The speed of inner drum does not change.

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