MHT CET · Physics · Mechanical Properties of Fluids
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]
- A \(\frac{(\delta-\rho)}{2 h \rho}\)
- B \(\frac{2 h \rho}{(\delta-\rho)}\)
- C \(\frac{h \rho}{2(\delta-\rho)}\)
- D \(\frac{h \rho}{(\delta-\rho)}\)
Answer & Solution
Correct Answer
(D) \(\frac{h \rho}{(\delta-\rho)}\)
Step-by-step Solution
Detailed explanation
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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