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 30º with the horizontal. The speed of the particle at the bottom of the plane is v₁. If the angle of inclination is increased to 45º while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v₂. The ratio of v₁² : v₂² is

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1 : $\sqrt{2}$

1 ; 3

1 : $\sqrt{3}$

1 : 2

answer is A.

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Detailed Solution

using WET
W_g = k_f – k_i
Mg L sin $θ$ = k_f – k_i
K.E. in pure rolling $\frac{1}{2} {mV}_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}$
$\begin{array}{l} = \frac{1}{2} {mV}^{2} + \frac{1}{2} \times \frac{2}{5} {mR}^{2} \frac{V^{2}}{R^{2}} \\ \frac{7}{10} {mV}^{2} \\ mgLsin θ = \frac{7}{10} {mV}_{f}^{2} - 0 \\ V_{f}^{2} \propto \sin θ \\ {(\frac{V_{1}}{V_{2}})}^{2} = \frac{\sin θ_{1}}{\sin θ_{2}} = \frac{\sin 30^{\circ}}{\sin 45^{\circ}} = \frac{1}{\sqrt{2}} \end{array}$

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