JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2025
- Medium
- MCQ
The center of mass of a thin rectangular plate (fig - x ) with sides of length \(a\) and \(b\), whose mass per unit area \((\sigma)\) varies as \(\sigma=\frac{\sigma_0 x}{a b}\) (where \(\sigma_0\) is a constant), would be \(\qquad\)

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