JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2025
- Hard
- MCQ
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 \(=10^3 \mathrm{~kg} / \mathrm{m}^3\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
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, \(\sigma=\frac{10 \times 10^{-3}}{10^{-3}}=10\)
\(\begin{aligned} \Rightarrow \quad T & =2 \pi \sqrt{\frac{10 \times 0.1}{10^3 \times 10}} \\ & =2 \pi \times 10^{-2}\end{aligned}\)
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