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Question

A wire of density $$9\times10\space kg\space m^{-3}$$ stretched between two clamps $$1\space m$$ apart is subjected to an extension of $$4.9\times10^{-4}\space m$$. If Young's modulus of the wire is $$9\times10^{-4}\space m$$. The lowest frequency of transverse vibrations in the wire is


A
35 Hz
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B
70 Hz
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C
105 Hz
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D
140 Hz
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Solution

The correct option is A $$35\space Hz$$
Fundamental frequency $$\nu=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}$$.
Now, Young's modulus $$Y=\dfrac{T/A}{\Delta l/l}=\dfrac{Tl}{A\Delta l}\Rightarrow T= \dfrac{YA\Delta l}{l}$$. 
So, $$\nu=\dfrac{1}{2l}\sqrt{\dfrac{\dfrac{YA\Delta l}{l}}{A\rho}}=\dfrac{1}{2l}\sqrt{\dfrac{YA\Delta l}{lA\rho}}=\dfrac{1}{2}\sqrt{\dfrac{9\times10^{10}\times 4.9\times 10^{-4}}{9\times10^3}}=35$$ Hz.
Ans: A

Physics

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