d
(d) By putting the dimensions of each quantity both the sides we get $[{T^{ - 1}}] = {[M]^x}{[M{T^{ - 2}}]^y}$
Now comparing the dimensions of quantities in both sides we get $x + y = 0\;{\rm{and }}\,2y = 1$
$\therefore $ $x = - \frac{1}{2},\,\,y = \frac{1}{2}$