Q66 / 75 · JEE Main 2025 (23 Jan Shift 1) · Physics
A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y \pi \times 10^{-2} \mathrm{~s}$, where the value of $y$ is
(Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$, density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3$ )
- A$6$
- Correct answerB$2$
- C$4$
- D$1$
Solution
Additional buoyant force
$\begin{aligned} & g \rho a^2 x=\sigma a^3 A \\ & A=\frac{\rho}{\sigma} \frac{g}{a} x \\ & T=2 \pi \sqrt{\frac{\sigma a}{\rho g}} \end{aligned}$ Now, $\sigma=\frac{10 \times 10^{-3}}{10^{-3}}=10$ $\begin{aligned} \Rightarrow \quad T & =2 \pi \sqrt{\frac{10 \times 0.1}{10^3 \times 10}} \\ & =2 \pi \times 10^{-2}\end{aligned}$
$\begin{aligned} & g \rho a^2 x=\sigma a^3 A \\ & A=\frac{\rho}{\sigma} \frac{g}{a} x \\ & T=2 \pi \sqrt{\frac{\sigma a}{\rho g}} \end{aligned}$ Now, $\sigma=\frac{10 \times 10^{-3}}{10^{-3}}=10$ $\begin{aligned} \Rightarrow \quad T & =2 \pi \sqrt{\frac{10 \times 0.1}{10^3 \times 10}} \\ & =2 \pi \times 10^{-2}\end{aligned}$
Hence, the correct answer is option: 'B'.