Question

A simple pendulum with a bob of mass \( m \) and density \( \rho \) and length \( l \) is immersed in a liquid of density \( \sigma \). If it makes small oscillations, then the time period of the pendulum is:

• A. \( T = 2\pi \left( \frac{1}{\sqrt{g - \rho}} \right) \)
• B. \( T = 2\pi \left( \frac{1}{\sqrt{g(1 - \rho)}} \right) \)
• C. \( T = 2\pi \left( \frac{1}{\sqrt{g(\rho - \sigma)}} \right) \)
• D. \( T = 2\pi \left( \frac{1}{\sqrt{g(1 - \frac{\rho}{\sigma})}} \right) \)

Hint:

For pendulums immersed in liquids, the time period is affected by the buoyant force. The effective gravitational acceleration is modified by the ratio of the densities of the bob and the liquid.

Answer 1

The Correct Option is D

The time period of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{l}{g}} \] However, when the pendulum is immersed in a liquid, the effective force acting on the bob is affected by the buoyancy force exerted by the liquid. This reduces the effective gravitational acceleration to: \[ g_{{effective}} = g \left( 1 - \frac{\rho}{\sigma} \right) \] where \( \rho \) is the density of the bob, and \( \sigma \) is the density of the liquid. Therefore, the time period for small oscillations becomes: \[ T = 2\pi \sqrt{\frac{l}{g_{{effective}}}} = 2\pi \sqrt{\frac{l}{g \left( 1 - \frac{\rho}{\sigma} \right)}} \] Thus, the correct answer is: \[ T = 2\pi \left( \frac{1}{\sqrt{g(1 - \frac{\rho}{\sigma})}} \right) \]