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A mass falls from a height ' $h$ ' and its time of fall ' $t$ ' is recorded in terms of time period $T$ of a simple pendulum. On the surface of earth it is found that $t=2 \mathrm{~T}$. The entire set $u$ is taken on the surface of another planet whose mass is half of earth and radius the same. Same experiment is repeated and corresponding times noted as $\mathrm{t}^{\prime}$ and $\mathrm{T}^{\prime}$.
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Verified Answer
The correct answer is:
$t^{\prime}=2 T^{\prime}$
The distance covered by the mass falling from height ' $h$ ' during its time of fall ' $t$ ' is given by
$$
\begin{array}{r}
\mathrm{s}=\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 \\
\text { As, } \mathrm{u}=0 \Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}
\end{array}
$$
The time period of simple pendulum is
$$
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}}
$$
where, lis the length of the pendulum. From Eq. (i) and (ii), since ' $h$ ' and ' $l$ ' are constant so, we can conclude that,
$$
\mathrm{t} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \text { and } \mathrm{T} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \quad \therefore \frac{\mathrm{t}}{\mathrm{T}}=\mathrm{l}
$$
Thus, the ratio of time of fall and time period of pendulum is independent of value of gravity $(\mathrm{g})$ or any other parameter like mass and radius of the planet. Thus, the relation between $\mathrm{t}^{\prime}$ and $\mathrm{T}^{\prime}$ on another planet irrespective of its mass or radius will remains same as it was on earth i.e.
$$
\mathrm{t}^{\prime}=2 \mathrm{~T}^{\prime}
$$
$$
\begin{array}{r}
\mathrm{s}=\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 \\
\text { As, } \mathrm{u}=0 \Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}
\end{array}
$$
The time period of simple pendulum is
$$
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}}
$$
where, lis the length of the pendulum. From Eq. (i) and (ii), since ' $h$ ' and ' $l$ ' are constant so, we can conclude that,
$$
\mathrm{t} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \text { and } \mathrm{T} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \quad \therefore \frac{\mathrm{t}}{\mathrm{T}}=\mathrm{l}
$$
Thus, the ratio of time of fall and time period of pendulum is independent of value of gravity $(\mathrm{g})$ or any other parameter like mass and radius of the planet. Thus, the relation between $\mathrm{t}^{\prime}$ and $\mathrm{T}^{\prime}$ on another planet irrespective of its mass or radius will remains same as it was on earth i.e.
$$
\mathrm{t}^{\prime}=2 \mathrm{~T}^{\prime}
$$
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