Q 11-06-072JEE MainJEE Main 2025 (28 Jan, Shift 1)Medium
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 $\sigma = \dfrac{\sigma_0x}{ab}$ (where $\sigma_0$ is a constant), would be
Answer: (A) $\left(\dfrac{2}{3}a, \dfrac{b}{2}\right)$
The density depends only on $x$, so by symmetry $y_{cm} = b/2$.
Take strips of width $dx$: $dm = \sigma b\,dx \propto x\,dx$.
$$x_{cm} = \frac{\int_0^a x\cdot x\,dx}{\int_0^a x\,dx} = \frac{a^3/3}{a^2/2} = \frac{2a}{3}$$
Centre of mass: $\left(\dfrac{2a}{3}, \dfrac{b}{2}\right)$
Solution by Sreeraj P, M.Sc Physics