A) 8
B) 5
C) 10
D) 6
Correct Answer: C
Solution :
Frequency of sonometer wire is given by \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}=\frac{1}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}\rho }}\] \[{{n}_{1}}=\frac{1}{2{{l}_{1}}}\sqrt{\frac{{{T}_{1}}}{\pi r_{1}^{2}{{\rho }_{1}}}}\] \[{{n}_{2}}=\frac{1}{2{{l}_{2}}}\sqrt{\frac{{{T}_{2}}}{\pi r_{2}^{2}{{\rho }_{2}}}}\] ... (ii) \[\therefore \] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}\times \frac{r_{2}^{2}}{r_{1}^{2}}\times \frac{{{\rho }_{2}}}{{{\rho }_{1}}}}\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{35}{36}\sqrt{\frac{8}{1}\times \frac{1}{16}\times \frac{2}{1}}\] \[\because \] \[{{n}_{1}}<{{n}_{2}}\]and \[{{n}_{2}}=360\] Therefore, \[n=360\times \frac{35}{36}\] \[{{n}_{1}}=350\,Hz\] So, number of beats produced \[={{n}_{1}}-{{n}_{2}}\] \[=360-350=10\]
You need to login to perform this action.
You will be redirected in
3 sec
