A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is g = 10 m s 2

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TestHub · Accepted Answer

Given, height of inclined plane, h = 7m
G = $\frac{10 m}{s^{2}}$
By conservation of energy,
Potential energy lost by the solid sphere in rolling down the inclined plane = Kinetic energy gained by the sphere.
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} l ω^{2}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} . \frac{2}{5} m R^{2} {(\frac{v}{R})}^{2}$
$= \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2}$
$m g h = \frac{7}{10} m v^{2}$
$∴ v^{2} = \frac{10}{7} g h$
$\Rightarrow v = \sqrt{\frac{10}{7}} g h = \sqrt{\frac{10}{7} \times 10 \times 7} = \frac{\sqrt{100} m}{s}$

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