Two blocks are connected over a mass less pulley as shown in the figure. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2, Block A slides down the incline at constant speed. The mass of block B in Kg

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3.0

3.5

3.3

2.5

answer is B.

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Detailed Solution

$\begin{array}{l} m g \sin θ = μ m g \cos θ + M g \\ M = m \sin θ - μ m \cos θ \\ = m (\sin θ - μ \cos θ) \\ = 10 (\sin 30^{\circ} - 0.2 \cos 30^{\circ}) \\ = 10 (0.5 - 0.2 \times \frac{\sqrt{3}}{2}) \\ = 5 - 1.732 = 3.268 \\ = 3.3 Kg \end{array}$

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