(A) For the falling body of mass $m$,the equation of motion is:$mg - T = ma$ --- $(i)$For the rotating disc of mass $M$ and radius $R$,the torque equ

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(A) For the falling body of mass $m$,the equation of motion is:$mg - T = ma$ --- $(i)$For the rotating disc of mass $M$ and radius $R$,the torque equation is:$RT = I\alpha$Since $I = \frac{1}{2}MR^2$ and $\alpha = \frac{a}{R}$,we have:$RT = (\frac{1}{2}MR^2)(\frac{a}{R}) = \frac{1}{2}MRa$$T = \frac{Ma}{2}$ --- $(ii)$Substituting equation $(ii)$ into equation $(i)$:$mg - \frac{Ma}{2} = ma$$mg = ma + \frac{Ma}{2} = a(m + \frac{M}{2}) = a(\frac{2m + M}{2})$$a = \frac{2mg}{2m + M}$

Class 11 Physics System of Particles and Rotational Motion Rotation Motion Basic, Motion of Connected Mass

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

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System of Particles and Rotational Motion

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Rotation Motion Basic, Motion of Connected Mass

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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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$A$ tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by the liquid on the tube at the other end is

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