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A rectangular block of mass ' $\mathrm{m}$ ' and crosssectional area A, floats on a liquid of density ' $\rho$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ' $n$ ' equal to ( $g=$ acceleration 'due to gravity)
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Verified Answer
The correct answer is:
$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{A} \rho \mathrm{g}}{\mathrm{m}}}$
The formula for the time period is given as
$\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$
The mass of displaced fluid is
$\text { Mass }=\text { density } \times \text { volume }$
$\mathrm{m}=\rho \times \mathrm{Al}$
At equilibrium,
Weight of the block = Weight of the displaced liquid
$\begin{aligned}
& \therefore \quad \mathrm{mg}=\mathrm{A} / \mathrm{pg} \\
& \therefore \quad l=\frac{\mathrm{m}}{\mathrm{Ap}}
\end{aligned}$
Substituting the values in the equation
$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \\
& \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho \mathrm{g}}}
\end{aligned}$
The frequency $f=\frac{1}{T}$
$\therefore \quad \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Apg}}{\mathrm{m}}}$
$\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$
The mass of displaced fluid is
$\text { Mass }=\text { density } \times \text { volume }$
$\mathrm{m}=\rho \times \mathrm{Al}$
At equilibrium,
Weight of the block = Weight of the displaced liquid
$\begin{aligned}
& \therefore \quad \mathrm{mg}=\mathrm{A} / \mathrm{pg} \\
& \therefore \quad l=\frac{\mathrm{m}}{\mathrm{Ap}}
\end{aligned}$
Substituting the values in the equation
$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \\
& \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho \mathrm{g}}}
\end{aligned}$
The frequency $f=\frac{1}{T}$
$\therefore \quad \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Apg}}{\mathrm{m}}}$
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