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Two bodies $\mathrm{A}$ (of mass $1 \mathrm{~kg}$ ) and $\mathrm{B}$ (of mass $3 \mathrm{~kg}$ ) are dropped from heights of $16 \mathrm{~m}$ and $25 \mathrm{~m}$, respectively. The ratio of the time taken by them to reach the ground is:
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The correct answer is:
$4 / 5$
We know that time taken by a body from height $h$ to reach the ground is.
$$
\begin{aligned}
t & =\sqrt{\frac{2 h}{g}} \\
\therefore \quad \frac{t_{\mathrm{A}}}{t_{\mathrm{B}}} & =\frac{\frac{2 h_{\mathrm{A}}}{g}}{\frac{2 h_{\mathrm{B}}}{g}}=\sqrt{\frac{h_{\mathrm{A}}}{h_{\mathrm{B}}}} \\
& =\sqrt{\frac{16}{25}}=\frac{4}{5} .
\end{aligned}
$$
$$
\begin{aligned}
t & =\sqrt{\frac{2 h}{g}} \\
\therefore \quad \frac{t_{\mathrm{A}}}{t_{\mathrm{B}}} & =\frac{\frac{2 h_{\mathrm{A}}}{g}}{\frac{2 h_{\mathrm{B}}}{g}}=\sqrt{\frac{h_{\mathrm{A}}}{h_{\mathrm{B}}}} \\
& =\sqrt{\frac{16}{25}}=\frac{4}{5} .
\end{aligned}
$$
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