A) 400 m
B) 300 Hz
C) 200 Hz
D) 100 Hz
Correct Answer: C
Solution :
As stretched string of 1m long is plucked at a distance of 25 cm from end, so there will be 2 loops. Frequency, \[n=\frac{p}{2l}\sqrt{\frac{T}{m}}\] Given, \[p=2,l=1\,m,T=20N,\] \[m=0.5\times {{10}^{-3}}kg/m\] \[\therefore \] \[n=\frac{2}{2\times 1}\sqrt{\frac{20}{0.5\times {{10}^{-3}}}}\] \[\Rightarrow \] \[n=200\,Hz\]You need to login to perform this action.
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