JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Rotational Mechanics

Look at the piece of rod beyond a distance $x$ from the axis; the tension $T(x)$ at that cross-section must supply the centripetal force for all of that outer piece. For a uniform rod of linear mass density $\lambda$, $T(x)={\int }_x^{\ell }\lambda {\omega }^{2}rdr=\frac{\lambda {\omega }^2}{2}({\ell }^2-x^2)$. This is a downward-opening parabola in $x$: it is largest at the axis, $T(0)=\frac{\lambda {\omega }^2{\ell }^2}{2}$, and falls smoothly to $T(\ell )=0$ at the free end.