A wire of linear mass density mu = 10^{-2} , | vedclass

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(C) The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density

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$m = 20 \, kg$

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(C) The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Given $v = 100 \, m \, s^{-1}$ and $\mu = 10^{-2} \, kg \, m^{-1}$.$100 = \sqrt{\frac{T}{10^{-2}}} \implies 100^2 = \frac{T}{10^{-2}} \implies T = 10000 	imes 10^{-2} = 100 \, N$.Since the system is in equilibrium,the tension $T$ in the string is equal to the weight of the hanging mass $M$,so $T = Mg = 100 \, N$. Taking $g = 10 \, m \, s^{-2}$,we get $M = \frac{100}{10} = 10 \, kg$.For the mass $m$ on the inclined plane,the component of gravity along the plane must balance the tension $T$,so $T = mg \sin(30^o)$.$100 = m 	imes 10 	imes \frac{1}{2} \implies 100 = 5m \implies m = 20 \, kg$.Thus,$\frac{m}{M} = \frac{20}{10} = 2$. Looking at the options,$m = 20 \, kg$ is correct.

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