Explanation

(d) As retardation $= bv$
$\therefore$ retarding force $= mbv$
$\therefore$ net restoring torque when angular displacement is $\theta$ is given by
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$= - mg\mathcal{l}\sin{\theta + mbv\mathcal{l}}$
$\therefore I\alpha = - mg\mathcal{l}\sin\theta + mbv\mathcal{l}$
where, $I = m\mathcal{l}^{2}$
$\therefore\frac{d^{2}\theta}{dt^{2}} = \alpha = - \frac{g}{\mathcal{l}}\sin\theta + \frac{bv}{\mathcal{l}}$
for small damping, the solution of the above differential equation will be
$\therefore\theta = \theta_{0}e^{- \frac{bt}{2}}\sin{(\omega t + \phi)}$
$\therefore$ angular amplitude will be $= \theta.e^{\frac{- bt}{2}}$
According to question, in $\tau$ time (average life-time),
Angular amplitude drops to $\frac{1}{e}$ value of its original value $(\theta)$
$\therefore\frac{\theta_{0}}{e} = \theta_{0}e^{- \frac{b\tau}{2}} \Rightarrow \frac{b\tau}{2} = 1$
$\therefore\tau = \frac{2}{b}$