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A wire of $10^{- 2} k g m^{- 1}$ passes over a frictionless light pulley fixed on the top of a frictionless inclined plane which makes an angle $30 °$ with horizontal. Masses m and M are tied at two ends of wire such that m rests on the plane and M hangs freely vertically downwards. The entire system is equilibrium and the wave velocity along wire is $100 m s^{- 1}$ , then the value of $m =______k g$

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answer is 20.

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Detailed Solution

$m g \sin θ = M g \Rightarrow \frac{m}{M} = 2$

$V = \sqrt{\frac{T}{μ}} \Rightarrow 100 = \sqrt{\frac{m g \sin θ}{10^{- 2}}} 100 = m \times 10 \times \frac{1}{2} m = 20 k g$

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