Question

# The time period of a simple pendulum is air is $$T$$. Now the pendulum is submerged in a liquid of density $$\dfrac{\rho}{16}$$ where $$\rho$$ is density of the bob of the pendulum. The new time period of oscillation is.

A
$$\dfrac{4}{\sqrt{15}} T$$
B
$$\sqrt{\dfrac{4}{15}} T$$
C
$$\sqrt{\dfrac{15}{4}} T$$
D
$$\dfrac{\sqrt{15}}{4} T$$
Solution
Verified by Toppr

#### Correct option is A. $$\dfrac{4}{\sqrt{15}} T$$$$T = 2\pi \sqrt{\dfrac{\ell}{g}}$$$$\tau = I\alpha$$$$\left(\rho Vg - \dfrac{\rho}{16} Vg\right)\ell \sin \theta = (\rho V)\ell^2 \alpha$$$$\dfrac{15g}{16\ell} \theta = \alpha$$$$\omega = \sqrt{\dfrac{15g}{16 \ell}} \Rightarrow T' = 2\pi \sqrt{\dfrac{16\ell}{15g}}$$$$T' = \dfrac{4}{\sqrt{15}} T$$

0
Similar Questions
Q1

A simple pendulum with a brass bob has a time period T. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is 18 that of brass, the time period of the same pendulum will be

View Solution
Q2

The metallic bob of a simple pendulum has the relative density ρ. The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by

View Solution
Q3

A simple pendulum with a brass bob has a time period T. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is 18 that of brass, the time period of the same pendulum will be

View Solution
Q4

A simple pendulum oscillating in air has period $$T.$$ The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $$\dfrac{1}{16}~th$$ of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is

View Solution
Q5

A simple pendulum vibrates in a non- viscous liquid. The density of the liquid is p/8, where p is the density of the material of the bob. What is the time period if the time period of the pendulum in air is T.

View Solution
Solve
Guides