A wire of 9.8×10^–3kgm^–1 passes over a friction-less 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 the 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. Chose the correct option

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m = 2kg

m = 7kg

m = 5kg

m = 20kg

answer is A.

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Detailed Solution

$\large v\, = \,\sqrt {\frac{T}{\mu }} \, = \,\sqrt {\frac{{mg\sin \theta }}{\mu }}$

$\large \therefore \,100\, = \,\sqrt {\frac{{m9.8\sin {{30}^0}}}{{9.8 \times {{10}^{ - 3}}}}} \,\therefore \,m\, = \,20kg$

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