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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2025 · 28 January · Morning Shift · Physics · Center of Mass, Momentum and Collision

The center of mass of a thin rectangular plate (fig - x ) with...

The center of mass of a thin rectangular plate (fig - x ) with sides of length aa and bb, whose mass per unit area (σ)(\sigma) varies as σ=σ0xab\sigma=\frac{\sigma_0 x}{a b} (where σ0\sigma_0 is a constant), would be \qquad The center of mass of a thin rectangular plate (fig - x ) with sides of length a and b , whose mass per unit area (σ)

Options

  1. A(23a, b2)\left(\frac{2}{3} a, \frac{\mathrm{~b}}{2}\right)
  2. B(a2, b2)\left(\frac{a}{2}, \frac{\mathrm{~b}}{2}\right)
  3. C(13a, b2)\left(\frac{1}{3} a, \frac{\mathrm{~b}}{2}\right)
  4. D(23a,23b)\left(\frac{2}{3} a, \frac{2}{3} b\right)

Correct answer

A. (23a, b2)\left(\frac{2}{3} a, \frac{\mathrm{~b}}{2}\right)

Step-by-step solution

\begin{aligned} & d m=\sigma d A \\ & \begin{aligned} d & =\sigma(d x)(d y)=\frac{\sigma_0 x}{a b}(d x)(d y) \\ x_{c o m} & =\frac{\int x d m}{\int d m}=\frac{\int x \frac{\left(\sigma_0 x\right)}{a b}(d x)(d y)}{\int_0 \frac{\sigma_0}{a b}(d x)(d y)} \\ & =\frac{\int_0^a x^2 d x \int_0^b d y}{\int_0^b x d x \int_0^b d y}=\frac{2 a}{3} \\ y_{c o m} & =\frac{\int y d m}{\int d m}=\frac{\int y\left(\frac{\sigma_0 x}{a b}\right)(d x)(d y)}{\int \frac{\sigma_0 x}{a b}(d x)(d y)}\end{aligned}\end{aligned}=\frac{\int_0^a x d x \int_0^b y d y}{\int_0^a x d x \int_0^b d y}=\frac{b}{2} i.e., \vec{r}_{\mathrm{com}}\equiv \left(\frac{2 a}{3},\frac{b}{2}\right)
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