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Question

A steel wire is rigidly fixed at both ends. Its length, mass and cross-sectional area are 1 m, 0.1 kg and 106 m2respectively. Tension in the wire is produced by lowering the temperature by 20C. If transverse waves are set up by plucking the wire at 0.25 m from one end, and assuming that the wire vibrates with minimum number of loops possible for such a case. The frequency of vibration (in Hz) is found to be 11×K. Find the value of K.(Given α=1.21×105 C1,Y=2×1011N/m2)


A
2
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B
4
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C
11
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D
22
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Solution

The correct option is A 2
As the temperature decreases length of wire tends to decrease by
ΔL=αLΔT
But since both ends of the wire are fixed it can't decrease its length, so stress is developed in the wire,
strain =ΔLL=αΔT
Stress =YΔLL=Ta=YΔT
T=ayαΔTT=106×2×1011×1.21×105×20 =48.4 N


from the figure
λ4=0.25 mλ=1 m
So, standing wave is in second harmonic.

For frequncy f=vλ=TM/L1λ
f=48.40.1/1×11=22
so k=2

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