A body slides down a smooth inclined plane having angle ' $\theta$ ' and reaches the bottom with velocity…
A body slides down a smooth inclined plane having angle ' $\theta$ ' and reaches the bottom with velocity 'v'. If a body is a sphere then its linear velocity at the bottom of the plane is
$\sqrt{\frac{2}{7}} \mathrm{v}$
$\sqrt{\frac{3}{7}} \mathrm{v}$
$\sqrt{\frac{5}{7}} \mathrm{v}$
$\sqrt{\frac{9}{7}} \mathrm{v}$
Solution
The linear velocity of the body, $v=\sqrt{2 g h}$
The velocity of the sphere about its centre,
$v_{C M}=\sqrt{\frac{2 g h}{1+\frac{K^{2}}{R^{2}}}}=\frac{v}{\sqrt{1+\frac{K^{2}}{R^{2}}}} \ldots(\text { i })$
For uniform solid sphere,
$\frac{K^{2}}{R^{2}}=\frac{2}{5}$
Substituting value in Eq. (i), we get
$v_{C M}=\frac{v}{\sqrt{1+\left(\frac{2}{5}\right)}}=\sqrt{\frac{5}{7} v}$