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Question

Two long strings A and B each having linear mass density 1.2×102 kg/m are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x=0. Wave pulses are produced on the string at the left ends at t=0 on string A and t=20 ms on string B. When will the pulse on B overtake that on A for the first time?

A
0.5 s
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B
0.25 s
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C
0.1 s
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D
0.4 s
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Solution

The correct option is C 0.1 s
Given that
TA=4.8 N TB=7.5 N
μA=1.2×1012 kg/m μB=1.2×102 kg/m
So
vA=TAμA=4.81.2×102=20 m/s
vB=TBμB=7.51.2×102=25 m/s
The pulse produced on A and B meets when, xA=xB

Let t be the time at which both pulses meet
So,
xA=xB20t=25(t0.02)
5t=0.5=0.1 s

Thus, option (c) is the correct answer.

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