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.

Answer 1

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