A) 40 Hz
B) 35 Hz
C) 30 Hz
D) 25 Hz
Correct Answer: B
Solution :
For wire if M = mass, r = density, A = Area of cross section V = volume, l = length, Dl = change in length Then mass per unit length \[m=\frac{M}{l}=\frac{Al\rho }{l}=A\rho \] And Young?s modules of elasticity \[y=\frac{T/A}{\Delta l/l}\] Þ \[T=\frac{Y\Delta lA}{l}\]. Hence lowest frequency of vibration \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\]\[=\frac{1}{2l}\sqrt{\frac{y\left( \frac{\Delta l}{l} \right)A}{A\rho }}=\frac{1}{2l}\sqrt{\frac{y\Delta l}{l\rho }}\] Þ \[n=\frac{1}{2\times 1}\sqrt{\frac{9\times {{10}^{10}}\times 4.9\times {{10}^{-4}}}{1\times 9\times {{10}^{3}}}}=35Hz\]
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