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

\frac{4 m g}{M + 2 m}

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

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

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

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Solution

The correct option is B $\frac{4 m g}{M + 4 m}$
Tension in the both string will be same.
$m g - T = m a - (1)$
also Inertia solid cylinder is $M R^{2} / 2$
Torque at centre of cylinder due to both string
$2 T \times R = I α = M R^{2} α / 2 - (2)$
also, $a = α R - (3)$

Solving we get Tension $T = \frac{M m g}{M + 4 m}$
Putting value of tension in equation $(2)$ ,
$α R = a = \frac{4 m g}{M + 4 m}$

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