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In a stationary lift, time period of a simple pendulum is ' $T$ '.' The lift starts accelerating downwards with acceleration $\left(\frac{\mathrm{g}}{4}\right)$, then the time period of the pendulum will be
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The correct answer is:
$\frac{2}{\sqrt{3}} \mathrm{~T}$
Time period of pendulum: $T=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$
$\therefore \quad$ When lift is accelerated downward with acceleration $\frac{\mathrm{g}}{4}$,
$$
\begin{aligned}
& g=g-\frac{g}{4} \\
& g=\frac{3 g}{4}
\end{aligned}
$$
$\therefore \quad$ New Time period will be
$$
\begin{aligned}
& \mathrm{T}_1=2 \pi \sqrt{\frac{4 \mathrm{~L}}{3 \mathrm{~g}}} \\
& \mathrm{~T}_1=2 \pi \frac{2}{\sqrt{3}} \sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \\
& \mathrm{T}_1=\frac{2}{\sqrt{3}} \mathrm{~T}
\end{aligned}
$$
$\therefore \quad$ When lift is accelerated downward with acceleration $\frac{\mathrm{g}}{4}$,
$$
\begin{aligned}
& g=g-\frac{g}{4} \\
& g=\frac{3 g}{4}
\end{aligned}
$$
$\therefore \quad$ New Time period will be
$$
\begin{aligned}
& \mathrm{T}_1=2 \pi \sqrt{\frac{4 \mathrm{~L}}{3 \mathrm{~g}}} \\
& \mathrm{~T}_1=2 \pi \frac{2}{\sqrt{3}} \sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \\
& \mathrm{T}_1=\frac{2}{\sqrt{3}} \mathrm{~T}
\end{aligned}
$$
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