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A body of density $1.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ is dropped from rest from a height $1 \mathrm{~m}$ into a liquid of density $2.4 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$. Neglecting all dissipative effects, the maximum depth to which the body sinks before returning to float on the surface is
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Verified Answer
The correct answer is:
$1 \mathrm{~m}$
$\mathrm{d}=1.2 \times 10^{3}, \mathrm{~h}=1 \mathrm{~m}, \rho=2.4 \times 10^{3}$
By work Energy Theorem,
$$
\begin{array}{l}
\mathrm{Mg}(\mathrm{h}+\mathrm{d})=\mathrm{Bd} \\
\mathrm{V}\left(1.2 \times 10^{-3}\right) \mathrm{g}(1+\mathrm{d})=\mathrm{V}\left(2.4 \times 10^{-3}\right) \mathrm{g} \mathrm{d} \\
\therefore \mathrm{d}=1 \mathrm{~m}
\end{array}
$$
By work Energy Theorem,
$$
\begin{array}{l}
\mathrm{Mg}(\mathrm{h}+\mathrm{d})=\mathrm{Bd} \\
\mathrm{V}\left(1.2 \times 10^{-3}\right) \mathrm{g}(1+\mathrm{d})=\mathrm{V}\left(2.4 \times 10^{-3}\right) \mathrm{g} \mathrm{d} \\
\therefore \mathrm{d}=1 \mathrm{~m}
\end{array}
$$
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