Q 11-13-048JEE MainJEE Main 2025 (23 Jan, Shift 1)Medium
A light hollow cube of side length $10\ \text{cm}$ and mass $10\ \text{g}$ is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y\pi\times10^{-2}\ \text{s}$, where the value of $y$ is (Acceleration due to gravity $g = 10\ \text{m/s}^2$, density of water $= 10^3\ \text{kg/m}^3$)
Answer: (B) $2$
Pushing the cube down by $x$ adds an upthrust $\rho A g x$, which acts as a restoring force, so the effective spring constant is
$$k = \rho A g = 10^3\times(0.1)^2\times10 = 100\ \text{N/m}$$
$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.01}{100}} = 2\pi\times10^{-2}\ \text{s}$$
So $y = 2$.
Solution by Sreeraj P, M.Sc Physics