Question

A hydrometer executes simple harmonic motion when it is pushed down vertically in a liquid of density \(\rho\). If the mass of hydrometer is \(m\) and the radius of the hydrometer tube is \(r\), then the time period of oscillation is

• A. \(\displaystyle T=2\pi\sqrt{\frac{m}{\pi r^2\rho g}}\)
• B. \(\displaystyle T=2\pi\sqrt{\frac{\pi r^2\rho g}{m}}\)
• C. \(\displaystyle T=\frac1{2\pi}\sqrt{\frac{m}{\pi r^2\rho g}}\)
• D. \(\displaystyle T=\frac1{2\pi}\sqrt{\frac{\pi r^2\rho g}{m}}\)

Hint:

A floating body performs SHM because a small displacement changes the buoyant force proportionally to displacement, producing a restoring force.

Answer 1

The Correct Option is A

Step 1: Find the restoring force.
Suppose the hydrometer is pushed down by a small distance \(x\).
The additional volume of liquid displaced is
\[ \Delta V=A x \] where \(A\) is the cross-sectional area of the tube.
Since the radius of the tube is \(r\),
\[ A=\pi r^2 \]
Hence,
\[ \Delta V=\pi r^2x \]

Step 2: Calculate the additional buoyant force.
Additional buoyant force is equal to the weight of the displaced liquid.
Therefore,
\[ F=\rho g\Delta V \]
\[ F=\rho g(\pi r^2x) \]
\[ F=\pi r^2\rho gx \]
This force acts upward and opposes the displacement. Hence, restoring force is
\[ F=-\pi r^2\rho gx \]

Step 3: Compare with SHM equation.
For simple harmonic motion, restoring force is
\[ F=-kx \]
Comparing,
\[ k=\pi r^2\rho g \]
The time period of SHM is
\[ T=2\pi\sqrt{\frac{m}{k}} \]
Substituting \(k=\pi r^2\rho g\),
\[ T=2\pi\sqrt{\frac{m}{\pi r^2\rho g}} \]

Step 4: Final conclusion.
Hence, the time period of oscillation is
\[ \boxed{T=2\pi\sqrt{\frac{m}{\pi r^2\rho g}}} \]