A wire of $9.8 \times 10^{-3} \mathrm{kg} \mathrm{m}^{-1}$ 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 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
Text Solution
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$v = \sqrt{\frac{T}{\mu}}$
For equilibrium $Mg = mg \sin 30 = T$
⇒ ⇒ $\mathbf{M} = \frac{m}{2}$ ⇒ ⇒ $100 = \sqrt{\frac{Mg}{9.8 \times 10^{-3}}} = \sqrt{\frac{M(9.8)}{9.8 \times 10^{-3}}}$
⇒ ⇒ $100 = \sqrt{M(1000)}$ ⇒ ⇒ $M = 10 \mathrm{kg}$ and m = 20kg
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