Physics · Simple Harmonic Motion

JEE Main 2025 — 23 January, Morning Shift — Question 50

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 $\mathrm{y} \pi \times 10^{-2} \mathrm{~s}$ , where the value of y is\ (Acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ , density of water $=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ )

Option A:
2
Correct
Option B:
6
Option C:
4
Option D:
1

Answer: A

Step-by-step solution

$a^{2} \times \rho g=m a_{\text {net }}$ \ $\frac{L^{2} \rho g}{m} x=a_{\text {net }}$ \ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~L}^{2} \rho \mathrm{~g}}}$ \ where $\mathrm{m}=10 \mathrm{~g}, \mathrm{~L}=10 \mathrm{~cm}, \rho=1000 \mathrm{~kg} / \mathrm{m}^{3}$

Answer key and solution verified before publishing.

Practise Simple Harmonic Motion

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Simple Pendulum and Angular SHM