JEE Main · 2008 · 27 Apr 2008 · Offline · Conservation of Momentum

The centre of mass is the mass-weighted mean position, $x_{cm}=\frac{{\int }_0^{L}x\lambda dx}{{\int }_0^L\lambda dx}$ with $\lambda =k{\left(\frac{x}{L}\right)}^n$. Both integrals are simple powers of $x$: the numerator yields a term in $x^{n+2}$ and the denominator a term in $x^{n+1}$, and after evaluating between $0$ and $L$ the constants and powers of $L$ combine to $x_{cm}=\frac{n+1}{n+2}L$. At $n=0$ the rod is uniform and $x_{cm}=\frac{L}{2}$; as $n$ grows the mass crowds towards the far end and $x_{cm}$ climbs through $\frac{2L}{3},\frac{3L}{4},\dots$, rising steadily and levelling off towards $L$ without ever reaching it. The curve that rises monotonically from $\frac{L}{2}$ and saturates near $L$ is the one that fits this dependence.