• question_answer A man measures time period of a pendulum (T) in stationary lift. If the lift moves upwards with acceleration g/4, then the new time period will be: A) $\frac{2T}{\sqrt{5}}$                                   B) $\left( \frac{\sqrt{5T}}{2} \right)$C) $\frac{\sqrt{5}}{(2T)}$                    D)  $\frac{2}{\sqrt{5T}}$

Let the new time period be T? The effective value of g for the accelerating lift will be $g+g/4=5g/4$ $T=2\pi \sqrt{\frac{\ell }{g}}$and $T'=2\pi \sqrt{\frac{\ell }{\frac{5g}{4}}}$ $\therefore$    $\frac{T'}{T}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}$or $T'=\frac{2T}{\sqrt{5}}$ Hence, the correction option is [a].