A particle is released one by one from the top of two inclined rough surfaces of height h each. The angles of inclination of the two planes are 30^o and 60^o, respectively. All other factors (e.g., coefficient of friction, mass of block, etc.) are same in both the cases. Let K₁ and K₂ be the kinetic energies of the particle at the bottom of the plane in the two cases. Then

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$K_{1} = K_{2}$

$K_{1} > K_{2}$

$K_{1} < K_{2}$

Data insufficient

answer is C.

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Detailed Solution

Work done by friction:

$W = (μmgcos θ) S = (μmgcos θ) \frac{h}{\sin θ} = μmghcot θ$

$Now \cot θ_{1} = \cot 30^{\circ} = \sqrt{3}$

$\cot θ_{2} = \cot 60^{\circ} = \frac{1}{\sqrt{3}}$

i.e., kinetic energy $(KE = mgh - W)$ in first case will be less or K₁ < K₂.

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