Q 11-06-173JEE MainJEE Main 2020 (9 Jan, Shift 2)Medium
A rod of length $l$ has non-uniform linear mass density given by $\rho(x) = a + b\left(\dfrac{x}{l}\right)^2$, where $a$ and $b$ are constants and $0 \le x \le l$. The value of $x$ for the centre of mass of the rod is at:
Answer: (B) $\dfrac{3}{4}\left(\dfrac{2a+b}{3a+b}\right)l$
$$M = \int_0^l\left(a + \frac{bx^2}{l^2}\right)dx = al + \frac{bl}{3} = \frac{(3a+b)l}{3}$$
$$\int_0^l x\left(a + \frac{bx^2}{l^2}\right)dx = \frac{al^2}{2} + \frac{bl^2}{4} = \frac{(2a+b)l^2}{4}$$
$$x_{\text{cm}} = \frac{(2a+b)l^2/4}{(3a+b)l/3} = \frac{3}{4}\left(\frac{2a+b}{3a+b}\right)l$$
Solution by Sreeraj P, M.Sc Physics