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6.System of Particles and Rotational Motion
hard
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 radio $\frac{{{h_{sph}}}}{{{h_{cyl}}}}$ is given by
A
$1$
B
$\frac{4}{5}$
C
$\frac{2}{{\sqrt 5 }}$
D
$\frac{14}{15}$
(JEE MAIN-2019)
Solution
For solid sphere
$\frac{1}{2}m{v^2} + \frac{1}{2} \cdot \frac{1}{2}m{R^2} \cdot \frac{{{V^2}}}{{{R^2}}} = mg{h_{sph}}$
For solid cylinder
$\frac{1}{2}m{V^2} + \frac{1}{2} \cdot \frac{1}{2}m{R^2} \cdot \frac{{{V^2}}}{{{R^2}}} = mg{h_{cyl}}$
$ \Rightarrow \frac{{{h_{sph}}}}{{{h_{cyl}}}} = \frac{{7/5}}{{3/2}} = \frac{{14}}{{15}}$
Standard 11
Physics
