Q 11-06-242JEE MainJEE Main 2017 (8 Apr)Easy
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:
Answer: (D) $\dfrac{2mg}{2m + M}$
For the block: $mg - T = ma$.
For the disc: $TR = I\alpha = \dfrac12MR^2\cdot\dfrac aR \Rightarrow T = \dfrac12Ma$.
Adding,
$$mg = \left(m + \frac M2\right)a \;\Rightarrow\; a = \frac{2mg}{2m + M}$$
Solution by Sreeraj P, M.Sc Physics