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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.

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Solution

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$$T = 2\pi \sqrt{\dfrac{\ell}{g}}$$

$$\tau = I\alpha$$

$$\left(\rho Vg - \dfrac{\rho}{16} Vg\right)\ell \sin \theta = (\rho V)\ell^2 \alpha$$

$$\dfrac{15g}{16\ell} \theta = \alpha$$

$$\omega = \sqrt{\dfrac{15g}{16 \ell}} \Rightarrow T' = 2\pi \sqrt{\dfrac{16\ell}{15g}}$$

$$T' = \dfrac{4}{\sqrt{15}} T$$

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