Q.

A wire having a linear density of  0.05 g/cm is stretched between two rigid supports with a tension of 4.5×102N it is observed that the wire resonance at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. Determine the length of the wire.

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a

2.14 m

b

3.2 m

c

4 m

d

8 m

answer is A.

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Detailed Solution

Expression for Frequency when both ends are tied f = n2LTμ  Now,  420=n2LTμ→(i) 490=n+12LTμ→(ii) Divide equation    i÷iin=6Putting in equation (i)420=62L4.5×1020.05×10−310−2⇒ L=2.14 m
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