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A body of density $\rho$ is dropped from (at rest) height ' $h$ ' into a lake of density ' $\delta$ ' $(\delta>\rho)$. The maximum depth to which the body sinks before returning to float on the surface is [Neglect all dissipative forces]
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Verified Answer
The correct answer is:
$\frac{h \rho}{(\delta-\rho)}$
The velocity of the body when it reaches the surface of the lake is
$$
\begin{aligned}
& \frac{\mathrm{a}}{\mathrm{g}}=\frac{\delta-\rho}{\rho} \\
& \therefore \mathrm{a}=\left(\frac{\delta-\rho}{\rho}\right) \mathrm{g}
\end{aligned}
$$
If $a$ is the acceleration and retardation in the liquid then $\mathrm{v}^2=2 \mathrm{ad}$
by eq. (1) and (4)
$$
\begin{aligned}
& 2 \mathrm{ad}=2 \mathrm{gh} \\
& \therefore \mathrm{d}=\frac{\mathrm{g}}{\mathrm{a}} \mathrm{h} \quad \therefore \mathrm{d}=\frac{\rho}{\delta-\rho} \mathrm{h}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\mathrm{a}}{\mathrm{g}}=\frac{\delta-\rho}{\rho} \\
& \therefore \mathrm{a}=\left(\frac{\delta-\rho}{\rho}\right) \mathrm{g}
\end{aligned}
$$
If $a$ is the acceleration and retardation in the liquid then $\mathrm{v}^2=2 \mathrm{ad}$
by eq. (1) and (4)
$$
\begin{aligned}
& 2 \mathrm{ad}=2 \mathrm{gh} \\
& \therefore \mathrm{d}=\frac{\mathrm{g}}{\mathrm{a}} \mathrm{h} \quad \therefore \mathrm{d}=\frac{\rho}{\delta-\rho} \mathrm{h}
\end{aligned}
$$
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