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Question

# A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. The wavelength of the fourth harmonic of the fundamental will be

A
80 cm
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B
60 cm
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C
40 cm
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D
20 cm
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Solution

## The correct option is D 40 cmm= mass per unit length=4×10−380×10−2=0.005Kg/mgiven T=50NL=80cm=0.8m∴v=√Tm=√500.005=100m/secFundamental frequency f0=12L√Tm=12×0.8√500.005=625Hz∴f4 Frequency of fourth harmonic 4f0=4×62.5=250HzAs we knowv4=f4λ4∴λ4=v4f4=100250=0.4m=40cm

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