A block of mass 10kg slides down from the top of an inclined plane of length 3m. The first 1m of the plane is smooth and next 2m is rough. The block is released from rest from top of the plane and again comes to rest at the bottom of the plane. If the angle of inclination of plane is $30^{\circ}$ , the coefficient of friction between plane and block on rough portion is

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$\frac{1}{\sqrt{3}}$

$\frac{\sqrt{3}}{4}$

$\frac{\sqrt{3}}{2}$

$\frac{2}{\sqrt{3}}$

answer is C.

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Detailed Solution

$\begin{array}{l} W = ΔK . E \\ W_{g} + W_{f} = 0 \\ m g h - f s = 0 \\ m g \sin θ \times l - μ mgcos θ (l - l_{r}) = 0 \\ m g \sin 30^{\circ} \times 3 = μ mgcos 30^{\circ} \times (3 - 1) \\ \frac{1}{2} \times 3 = μ \frac{\sqrt{3}}{2} \times 2 \\ μ = \frac{\sqrt{3}}{2} \end{array}$