A uniform solid cylinder of mass $M$ and radius $R$ rotates about a frictionless horizontal axle. Two similar masses suspended with the help two ropes wrapped around the cylinder. If the system is released from rest then the acceleration of each mass will be

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$\frac{2 M g}{m + 2 M}$

$\frac{M g}{2 M + m}$

$\frac{4 m g}{M + 4 m}$

$\frac{M g}{M + 2 m}$

answer is D.

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Detailed Solution

$τ = R . F = I α$
$2 R . T = \frac{M R^{2}}{2} α$
$2 T = \frac{M R α}{2}$
$T = \frac{M a}{4}$
$M g - T = m a$
$m g - \frac{m a}{4} = m a$
$m g = a (\frac{M}{4} + m)$
$a = \frac{4 m g}{M + 4 m}$

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