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A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied by a horizontal string which has a tension of 50 N . The coefficient of friction between the surfaces of contact is :

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1/4

3/4

2/3

1/2

answer is A.

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Detailed Solution

$\begin{array}{l} 15 \times 10 \times \frac{1}{\sqrt{2}} = \frac{50}{\sqrt{2}} + μ (\frac{150}{\sqrt{2}} + \frac{50}{\sqrt{2}}) \\ \frac{100}{\sqrt{2}} = (μ) (\frac{200}{\sqrt{2}}) \Rightarrow μ = \frac{1}{2} \end{array}$

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