A solid sphere of mass 'm' and radius 'r' is allowed to roll without slipping from the highest point of an inclined plane of lengh 'L' and makes an angle 30° with the horizontal. The speed of the particle at the bottom of the plane is . If the angle of in

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Solution

Q.
A solid sphere of mass 'm' and radius 'r' is allowed to roll without slipping from the highest point of an inclined plane of length 'L' and makes an angle 30° with the horizontal. The speed of the particle at the bottom of the plane is $v_{1}$ . If the angle of inclination is increased to 45° while keeping L constant. Then the new speed of the sphere at the bottom of the plane is $v_{2}$ . The ratio of $v_{1}^{2} : v_{2}^{2}$ is [2025]

1 $1 : \sqrt{2}$

2 1 : 3

3 1 : 2

4 $1 : \sqrt{3}$

Ans.
(1)

Loss in P.E. = gain in K.E.

$m g L \sin θ = \frac{1}{2} (\frac{7}{5}) m v^{2} \Rightarrow v^{2} \propto \sin θ$

$\Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{\sin 30 °}{\sin 45 °} = \frac{1}{\sqrt{2}}$

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