A wire of linear mass density mu = 9.8 times | vedclass

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

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$20$

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(A) The velocity of a transverse wave on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the string and $\mu$ is the linear mass density.For the system in equilibrium,the tension $T$ in the string is equal to the weight of the hanging mass $M$,so $T = Mg$.Also,for the mass $m$ on the inclined plane to be in equilibrium,the component of its weight along the plane must balance the tension: $T = mg \sin 30^{\circ}$.Equating the two expressions for tension: $Mg = mg \sin 30^{\circ} \implies M = m \sin 30^{\circ} = \frac{m}{2} \implies m = 2M$.Given $v = 100 \, m \, s^{-1}$ and $\mu = 9.8 	imes 10^{-3} \, kg \, m^{-1}$,we have:$100 = \sqrt{\frac{Mg}{9.8 	imes 10^{-3}}}$$100^2 = \frac{M(9.8)}{9.8 	imes 10^{-3}}$$10000 = M 	imes 1000$$M = 10 \, kg$.Since $m = 2M$,we get $m = 2 	imes 10 = 20 \, kg$.

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