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A wire of 10^-2 kgm^-1 passes over a frictionless light pulley fixed on the top of a frictionless inclined plane which makes an angle of 30° with the 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 in equilibrium and a transverse wave propagates along the wire with a velocity of 100 ms^-1. Then

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$\frac{m}{M} = 4$

M=5 kg

m=20 kg

$\frac{m}{M} = \frac{1}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} mgsin θ = Mg \Rightarrow \frac{m}{M} = 2 \\ V = \sqrt{\frac{T}{μ}} \\ \Rightarrow 100 = \sqrt{\frac{mgsin θ}{10^{- 2}}} \\ \Rightarrow 100 = m \times 10 \times \frac{1}{2} \Rightarrow m = 20 kg \end{array}$

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