Q 11-06-218JEE MainJEE Main 2019 (8 Apr, Shift 2)Easy
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights $h_{sph}$ and $h_{cyl}$ on the incline. The ratio $\dfrac{h_{sph}}{h_{cyl}}$ is given by
Answer: (C) $\dfrac{14}{15}$
For rolling without slipping, total kinetic energy is $\frac12mv^2\left(1 + \frac{k^2}{R^2}\right)$, and all of it becomes $mgh$:
$$h = \frac{v^2}{2g}\left(1 + \frac{k^2}{R^2}\right)$$
Sphere: $1 + \frac25 = \frac75$. Cylinder: $1 + \frac12 = \frac32$.
$$\frac{h_{sph}}{h_{cyl}} = \frac{7/5}{3/2} = \frac{14}{15}$$
Solution by Sreeraj P, M.Sc Physics