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A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :

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

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

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

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

answer is B.

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Detailed Solution

From the figure,

$ma = mg - T______(1)$

The MI of uniforms disc

$I = \frac{1}{2} {MR}^{2}$

also, we know that acceleration of the disc,

$a = αR$

$Torque = Iα = RT$

$∴ RT = \frac{1}{2} {MR}^{2} \times \frac{a}{R} = \frac{MaR}{2}$

$T = \frac{Ma}{2}$

substituting the value in equation 1,

$ma = mg - \frac{Ma}{2}$

$a (m + \frac{Ma}{2}) = mg$

$a = \frac{2 mg}{2 m + M}$

Hence the correct answer is $\frac{2 mg}{2 m + M} .$

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