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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° and 60°, respectively. All other factors (e.g., coefficient of friction, mass of block, etc.) are same in both the cases. Let $K_{1}$ and $K_{2}$ 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: $W = (μ m g c o s θ) S = (μ m g c o s θ)$ $\frac{h}{\sin θ}$ . $N o w c o t θ_{1} = c o t 30 °$ , $c o t θ_{2} = c o t 60 °$ .i.e., kinetic energy (KE = mgh - W) in first case will be less or $K_{1}$ < $K_{2}$ .

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