Question

Time period of a simple pendulum in air is \( T \). If the pendulum is in water and executes SHM, its time period is \( t \). The value of \( \frac{T}{t} \) is [Density of bob is \( \frac{5000}{3} \) kg/m\(^3\)]

• A. \( \frac{2}{5} \)
• B. \( \sqrt{\frac{2}{5}} \)
• C. \( \frac{5}{2} \)
• D. \( \sqrt{\frac{5}{2}} \)

Hint:

When a pendulum oscillates in a fluid, its effective acceleration due to gravity is reduced by buoyancy, leading to a modified time period.

Answer 1

The Correct Option is B

The time period of a simple pendulum in a fluid is given by: \[ t = T \sqrt{\frac{\rho_b}{\rho_b - \rho_f}} \] where:
\( \rho_b \) = Density of the bob \( = \frac{5000}{3} \) kg/m\(^3\)
\( \rho_f \) = Density of the fluid (water) \( = 1000 \) kg/m\(^3\)
Substituting: \[ \frac{T}{t} = \sqrt{\frac{\rho_b - \rho_f}{\rho_b}} \] \[ = \sqrt{\frac{\frac{5000}{3} - 1000}{\frac{5000}{3}}} \] \[ = \sqrt{\frac{\frac{5000 - 3000}{3}}{\frac{5000}{3}}} \] \[ = \sqrt{\frac{2000}{5000}} \] \[ = \sqrt{\frac{2}{5}} \] Thus, the correct answer is \( \sqrt{\frac{2}{5}} \).