Explanation

(d) Time period of simple pendulum is given by
$T_{1} = 2\pi\sqrt{\frac{l}{g}}$
and time period of uniform rod in given position is given by
$T_{2} = 2\pi\sqrt{\frac{Inertia\ factor}{spring\ factor}}$
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Here, inertia factor=moment of inertia of rod at one end
$= \frac{ml^{2}}{12} + \frac{ml^{2}}{4} = \frac{ml^{2}}{3}$
Spring factor=restoring torque per unit angular displacement
$= mg \times \frac{l}{2}\ \frac{\sin\theta}{\theta}$
$= mg \times \frac{l}{2}$ (if $\theta$ is small)
$\therefore$ $T_{2} = 2\pi\sqrt{\frac{\frac{ml^{2}}{3}}{\frac{mgl}{2}}} = 2\pi\sqrt{\frac{2}{3}\ \ \frac{l}{g}}\ $
Hence, $\frac{T_{1}}{T_{2}} = \sqrt{\frac{3}{2}}$