A rectangular block of mass ' $\mathrm{m}$ ' and crosssectional area A, floats on a liquid of density '…

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)
  1. $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{A} \rho \mathrm{g}}{\mathrm{m}}}$
  2. $2 \pi \sqrt{\frac{\mathrm{A} \rho \mathrm{g}}{\mathrm{m}}}$
  3. $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho \mathrm{g}}}$
  4. $2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho \mathrm{g}}}$

Solution

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 $f=\frac{1}{T}$ $\therefore \quad \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Apg}}{\mathrm{m}}}$

Asked in: MHT CET 2023 (11 May Shift 2)

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