QJEEM14Ph9d70: JEE Main 2014 Physics - Class 11 JEEM14Ph9d70 Solution: A mass ′ m ′ is supported by a massless string wound around a…

Here string is not slipping over pulley.

From figure,

Acceleration $$a=R\alpha ...(i)$$

Applying Newton's second law on the block we get,

$$mg-T=ma$$ $$...(ii)$$

Torque on cylinder due to tension in string about the fixed point,

$$T\times R=I\alpha$$

$$T\times R=m{R}^2\alpha =m{R}^2\left(\frac{a}{R}\right)$$

[ as $$I$$ = MR² for hollow cylinder]

or $$T=ma$$

$$\Rightarrow mg-ma=ma$$

$$\Rightarrow a=\frac{g}{2}$$