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A body of density ' $\rho$ ' is dropped from rest at a height ' $h$ ' into a lake of density ' $\sigma$ ' $(\sigma>\rho)$. The maximum depth to which the body sinks before returning to float on the surface is (neglect air dissipative forces)
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Verified Answer
The correct answer is:
$\frac{h \rho}{(\rho-\sigma)}$
Initial velocity of the ball $=\sqrt{2 \text { gh }}$
Upward force:
$$
\begin{aligned}
& F=\sigma V g-\rho V g \\
& (\rho V) a=V g(\sigma-\rho) \\
\therefore \quad & a=\frac{g(\sigma-\rho)}{\rho}
\end{aligned}
$$
Final velocity is zero when it sinks.
$$
\begin{aligned}
\therefore \quad & v^2-u^2=2 \text { as } \\
& 0-(\sqrt{2 g h})^2=2 \frac{g(\sigma-\rho)}{\rho} H \\
& H=\frac{h \rho}{(\rho-\sigma)}
\end{aligned}
$$
Upward force:
$$
\begin{aligned}
& F=\sigma V g-\rho V g \\
& (\rho V) a=V g(\sigma-\rho) \\
\therefore \quad & a=\frac{g(\sigma-\rho)}{\rho}
\end{aligned}
$$
Final velocity is zero when it sinks.
$$
\begin{aligned}
\therefore \quad & v^2-u^2=2 \text { as } \\
& 0-(\sqrt{2 g h})^2=2 \frac{g(\sigma-\rho)}{\rho} H \\
& H=\frac{h \rho}{(\rho-\sigma)}
\end{aligned}
$$
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