A wire PQ has length 4.8 m and mass 0.06 kg | vedclass

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(C) Mass per unit length for wire $PQ$ is $m_{PQ} = \frac{0.06}{4.8} = \frac{1}{80} \ kg/m$. Mass per unit length for wire $QR$ is $m_{QR} = \frac{0.2

Vedclass · Accepted Answer

$0.14$

Vedclass · Answer

(C) Mass per unit length for wire $PQ$ is $m_{PQ} = \frac{0.06}{4.8} = \frac{1}{80} \ kg/m$. Mass per unit length for wire $QR$ is $m_{QR} = \frac{0.2}{2.56} = \frac{5}{64} \ kg/m$. The velocity of the wave in wire $PQ$ is $v_{PQ} = \sqrt{\frac{T}{m_{PQ}}} = \sqrt{\frac{80}{1/80}} = \sqrt{6400} = 80 \ m/s$. The time taken to travel through wire $PQ$ is $t_1 = \frac{L_{PQ}}{v_{PQ}} = \frac{4.8}{80} = 0.06 \ s$. The velocity of the wave in wire $QR$ is $v_{QR} = \sqrt{\frac{T}{m_{QR}}} = \sqrt{\frac{80}{5/64}} = \sqrt{\frac{80 	imes 64}{5}} = \sqrt{16 	imes 64} = 4 	imes 8 = 32 \ m/s$. The time taken to travel through wire $QR$ is $t_2 = \frac{L_{QR}}{v_{QR}} = \frac{2.56}{32} = 0.08 \ s$. The total time taken is $t = t_1 + t_2 = 0.06 + 0.08 = 0.14 \ s$.

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