Physics · System Of Particles

JEE Main 2025 — 28 January, Morning Shift — Question 65

The centre of mass of a thin rectangular plate (fig $x$ ) with sides of length $a$ and $b$ , whose mass per unit area $(\sigma)$ varies as $\sigma=\frac{\sigma_{0} \mathrm{x}}{\mathrm{ab}}$ (where $\sigma_{0}$ is a constant), would be $\qquad$

Option A:
$\left(\frac{2a}{3} , \frac{b}{2}\right)$
Correct
Option B:
$\left(\frac{2a}{3} , \frac{2b}{3}\right)$
Option C:
$\left(\frac{a}{2}, \frac{b}{2}\right)$
Option D:
$\left(\frac{a}{3} , \frac{b}{2}\right)$

Answer: A

Step-by-step solution

$\sigma$ is constant in $y$ -direction

So, $\mathrm{y}_{\mathrm{cm}}=\mathrm{b} / 2$

$x_{c m}=\frac{\int_{0}^{a} x d m}{\int_{0}^{a} d m}$

$=\frac{\int_{0}^{a} x \sigma_{x} d A}{\int_{0}^{a} \sigma_{x} d A}$

$=\frac{\int_{0}^{a} x \frac{\sigma_{0} x}{a b} b d x}{\int_{0}^{a} \frac{\sigma_{0} x}{a b} b d x}$

$x_{c m}=\frac{\int_{0}^{a} x^{2} d x}{\int_{0}^{a} x d x}$

$=\frac{\left(\frac{x^{3}}{3}\right)_{0}^{a}}{\left(\frac{x^{2}}{2}\right)_{0}^{a}}=\frac{a^{3} / 3}{a^{2} / 2}$

$=\frac{2 \mathrm{a}}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: System Of Particles
Topic: Position of Center of Mass