c
(c)When the block moves vertically downward with acceleration \(\frac{g}{4}\) then tension in the cord

\(T = M\left( {g - \frac{g}{4}} \right)\; = \frac{3}{4}Mg\)

Work done by the cord = \(\overrightarrow F .\overrightarrow s = Fs\cos \theta \)

= \(Td\cos (180^\circ )\)\( = - \left( {\frac{{3Mg}}{4}} \right) \times d\)\( = - \frac{3}{4}Mgd\)