A metal wire of diameter 1.5 mm is held on two knife edges separated by a distance 50 cm the tension in the wire is 100 N the wire vibrating with its fundamental frequency and vibrating tuning fork together produces 5 beats per second. The tension in the wire is then reduced to 81 N, when the two are excited, beats are heard at the same rate. Calculate frequency of the fork.

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95 Hz

110 Hz

100 Hz

180 Hz

answer is B.

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Detailed Solution

Let N be the frequency of the tuning fork then the frequency of wire, when tension is 100 N will be (N+5) and the tension is 81 N it is (N-5) since each case 5 beats are heard per second.

$∴ N + 5 = \frac{1}{2 l} \sqrt{\frac{T_{1}}{μ}} = \frac{1}{2 \times 0.5} \sqrt{\frac{100}{μ}} = \frac{10}{\sqrt{μ}} \to (i) a n d N - 5 = \frac{1}{2 l} \sqrt{\frac{T_{2}}{μ}} = \frac{1}{2 \times 0.5} \sqrt{\frac{81}{μ}} = \frac{9}{\sqrt{μ}} \to (i i) S u b t r a c t i n g (i i) f r o m (i) w e h a v e 10 = \frac{1}{\sqrt{μ}} (o r) μ = 0.01 k g / m U \sin g t h i s v a l u e i n (i) (o r) (i i) N = 95 H z$