Explanation

(d) Let $a$ be acceleration of fall of the thread, then net force acting downwards, balances the force due to tension $(T)$ in the thread.
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$mg - T = ma$
$\Rightarrow$ $mg - ma = T$ …(i)
Also torque (also known as moment or couple acts on the system).
$\tau = force \times perpendicular\ distance\ axis\ of\ rotation$
$\tau = T \times R$
From Eq. (i),
$\tau = m(g - a) \times R$ …(ii)
Let $I$ is moment of inertia of reel and $\alpha$ the angular acceleration, then torque is
$\tau = I\alpha$ …(iii)
where, $I = \frac{1}{2}MR^{2},\ \alpha = \frac{a}{R}$
$\therefore$ $\tau = \frac{1}{2}MR^{2} \times \frac{a}{R} = \frac{MRa}{2}$ …(iv)
Equating Eqs. (ii) and (iv), we get
$\tau = m(g - a)R = \frac{mRa}{2}$
$\Rightarrow$ $g - a = \frac{a}{2}$
$\Rightarrow$ $a = \frac{2}{3}\ g$