A solid uniform cylinder of mass $M$ and radius $R$ is pivoted at its centre free to rotate about horizontal axis. A massless inextensible string is wrapped around it, and attached to a block of mass $m$ which is initially at a height $h$ above the floor. The acceleration due to gravity is $g$ , directed downward. The block is released from rest. By what total angle $θ$ does it rotate (in radians)?

\sqrt{1 + \frac{2 M}{m}}

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\frac{h}{R}

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\sqrt{1 + \frac{h}{2 R}}

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\frac{M R}{m h}

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Solution

The correct option is B $\frac{h}{R}$
Since the string does not slip on the cylinder,
$R θ = h$
$\Rightarrow θ = \frac{h}{R}$
When cylinder turns through angle $θ$ , length of the string that unfold is $R θ =$ downward displacement of block $(h)$ .

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