AP EAPCET 18 May 2024 Shift 1 Solved Paper

Time period of a simple pendulum in air is $T=2\pi\sqrt{\frac{l}{g}}$.In water, the upthrust of water reduces the effective weight of the bob, so the effective acceleration due to gravity becomes $g'=g\left(1-\frac{\sigma}{\rho}\right)$, where $\sigma$ is the density of water and $\rho$ is the density of the bob.Taking the density of the bob, $\rho=\frac{5000}{3}\ \mathrm{kg/m^3}$, and the density of water, $\sigma=1000\ \mathrm{kg/m^3}$.The time period in water is therefore $t=2\pi\sqrt{\frac{l}{g'}}=\frac{T}{\sqrt{1-\frac{\sigma}{\rho}}}$.Substitute the values: $1-\frac{\sigma}{\rho}=1-\frac{1000 \times 3}{5000}=1-0.6=0.4=\frac{2}{5}$.Hence $\frac{T}{t}=\sqrt{1-\frac{\sigma}{\rho}}=\sqrt{0.4}=\sqrt{\frac{2}{5}}$.Note that $\frac{T}{t}<1$ because buoyancy makes the pendulum slower in water, so $\sqrt{\frac{5}{2}}$ (the reciprocal) is incorrect.Thus the value of $\;\frac{T}{t}$ is $\sqrt{\;\frac{2}{5}}$.