The diameter of a wire stretched by a force of 144 N between two bridges distant 50 cm is 1 mm. It produces 4 beats with a tuning fork while vibrating in fundamental mode. On reducing the tension to 121 N, again 4 beats/sec are heard. The frequency of fork is

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195 Hz

85 Hz

92 Hz

75 Hz

answer is D.

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Detailed Solution

we know that $n \propto \sqrt{T}$
Let the frequency of fork be n. Then
$n_{1} - n = 4 or n_{1} = n + 4 and n - n_{2} = 4 or n_{2} = n - 4 \frac{n_{1}}{n_{2}} = \frac{n + 4}{n - 4} From eqs. (1) and (2) \frac{12}{11} = \frac{n + 4}{n - 4} n = 92 Hz$

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