JEE Main 2020 (9 Jan, Shift 2) Physics Q5: Center of Mass

Question

A rod of length $L$ has non-uniform linear mass density given by $\rho(x) = a + b\left(\frac{x}{L}\right)^2$, 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 :

Accepted Answer

Option 1: $\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$

Answer

$\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$

Answer

$\frac{3}{2}\left(\frac{2a+b}{3a+b}\right)L$

Answer

$\frac{4}{3}\left(\frac{a+b}{2a+3b}\right)L$

Answer

$\frac{3}{2}\left(\frac{a+b}{2a+b}\right)L$

JEE Main 9 January 2020, Shift 2, Physics Q5

Same idea in other shifts