Question

A pendulum has a time period T in air when it is made to oscillate in water, it acquired a time period \[T' = \sqrt 2 T\] . The density of the pendulum bob is equal to (density of water = 1):
A. \[\sqrt 2 \]
B. 2
C. \[2\sqrt 2 \]
D. None of these

Answer 1

Hint: The time period of the pendulum will change in water as the acceleration due to gravity will change due to the buoyant force. Find the value of $g$ in water.

Complete step by step solution:
Let $\sigma $and$\rho $ be the densities of the water and the bob respectively. As we know that the time period of a simple pendulum in air is given by $T = 2\pi \sqrt {\dfrac{l}{g}} $

Time period of the same pendulum in water is given by $T = 2\pi \sqrt {\dfrac{l}{{g'}}} $
Where, g’ is the acceleration due to gravity on the bob in water. Its value is given by
$g' = g\left[ {1 - \dfrac{\sigma }{\rho }} \right] \Rightarrow g' = g\left[ {1 - \dfrac{1}{\rho }} \right]\,As\,\sigma = 1$
Now, $\dfrac{T}{{T'}} = \sqrt {\dfrac{{g'}}{g}} = \sqrt {1 - \dfrac{1}{\rho }} $
Putting $\dfrac{T}{{T'}} = \dfrac{1}{{\sqrt 2 }}$ we get $1 - \dfrac{1}{\rho } = \dfrac{1}{2} \Rightarrow \rho = 2$

Hence, the correct option is (B).

Additional Information: A simple pendulum consists of a mass m hanging from a string of length l and fixed at a pivot point. When it is displaced to an initial angle and released, the pendulum will swing back and forth with periodic motion. The time taken by the pendulum to complete one back and forth motion is called its time period. The time period of a simple pendulum does not depend on the mass or the initial angular displacement, but depends only on the length of the string and the value of the acceleration due to gravity.

Note: When we put a pendulum in water, gravity can pull the bob of the pendulum down through the water as an equal volume of water is allowed to go up against the force of gravity due to the buoyant force.