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Q.
A spherical shell and a solid cylinder of same radius rolls down an inclined plane. The ratio of their accelerations will be:-
NTA AbhyasNTA Abhyas 2022
Solution:
Acceleration of a purely rolling object on an inclined plane is :-
$a=\frac{g sin \theta }{\left(1 + \frac{K^{2}}{R^{2}}\right)}$
for spherical shell, $\frac{K^{2}}{R^{2}}=\frac{2}{3}$
for solid cylinder, $\frac{K^{2}}{R^{2}}=\frac{1}{2}$
so, $\frac{a_{\text{shell }}}{a_{\text{cylinder }}}=\frac{\frac{g sin \theta }{\left(1 + \frac{2}{3}\right)}}{\frac{g sin \theta }{\left(1 + \frac{1}{2}\right)}}=\frac{9}{10}$