Q136 / 150 · MHT CET 2019 (Shift 1) · Physics

A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is $(g = \frac{10 m}{s^{2}})$

A
$\frac{\sqrt{70} m}{s}$
B
$\frac{\sqrt{\frac{140}{3}} m}{s}$
C
$\frac{\sqrt{\frac{280}{3}} m}{s}$
Correct answer
D
$\frac{\sqrt{100} m}{s}$

Solution

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}

Hence, the correct answer is option: 'D'.

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