A) 30 Hz
B) 40 Hz
C) 50 Hz
D) 60 Hz
Correct Answer: A
Solution :
Given \[l=1m,\]\[d=9\times {{10}^{3}}kg/{{m}^{3}}\] \[\Delta l=0.36mm=3.6\times {{10}^{-4}}m\] \[Y=9\times {{10}^{10}}N/{{m}^{2}}\] Frequency \[(v)=\frac{1}{2l}\sqrt{\frac{T}{m}}=\frac{1}{2l}\sqrt{\frac{T}{A\Delta l}}\] But Young's modulus \[Y=\frac{Tl}{A\Delta l}\] \[\therefore \] \[T=\frac{YA\Delta l}{l}\] \[\therefore \] \[n=\frac{1}{2l}\sqrt{\frac{YA\Delta l}{l\times Ad}}=\frac{1}{2l}\sqrt{\frac{Y\Delta l}{ld}}\] \[=\frac{1}{2\times 1}\sqrt{\frac{9\times {{10}^{10}}\times 3.6\times {{10}^{-4}}}{1\times 9\times {{10}^{3}}}}\] \[=30Hz\]
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