A) 200
B) 210
C) 205
D) 215
Correct Answer: C
Solution :
The frequency of sonometer wire is given by \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] where\[l\]is length, T the tension and m the mass per unit length. When the length of wire is 0.20 m. \[{{n}_{1}}=\frac{1}{2\times 0.2}\sqrt{\frac{T}{m}}\] When length is 0.21 m, then \[{{n}_{2}}=\frac{1}{0.21\times 2}\sqrt{\frac{T}{m}}\] Let frequency of tuning fork be N, \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{N+5}{N-5}\] \[\Rightarrow \] \[\frac{0.21}{0.20}=\frac{N+5}{N-5}\] \[\Rightarrow \] \[0.21\,N-1.05=0.2\,N+1\] \[\Rightarrow \] \[0.01\,N=2.05\] \[\Rightarrow \] \[N=\frac{2.05}{0.01}=205\,Hz\]
You need to login to perform this action.
You will be redirected in
3 sec
