A) 5
B) 10
C) 15
D) 20
Correct Answer: B
Solution :
Given: \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{8}{1}=8:1\] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{36}{35}=36:35\] \[\frac{{{D}_{1}}}{{{D}_{2}}}=\frac{4}{1}=4:1\]or \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{4}{1}\] \[\frac{{{\rho }_{1}}}{{{\rho }_{2}}}=\frac{1}{2}\] Since, density \[\rho =\frac{M}{V}\] \[\Rightarrow \] \[M=\rho V=\pi {{r}^{2}}l\rho \] \[\Rightarrow \] \[\frac{M}{l}=\pi {{r}^{2}}\rho \] The frequency of wire vibrating given as \[n=\frac{1}{2l}\sqrt{\frac{T}{M}}\] where, \[m=\frac{M}{l}\] Hence, \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{\frac{1}{2{{l}_{1}}}\sqrt{\frac{{{T}_{1}}}{{{m}_{12}}}}}{\frac{1}{2{{l}_{2}}}\sqrt{\frac{{{T}_{2}}}{{{m}_{2}}}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}}\times \frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\sqrt{\frac{\pi r_{2}^{2}{{\rho }_{2}}}{\pi r_{1}^{2}{{\rho }_{1}}}\times \frac{{{T}_{1}}}{{{T}_{2}}}}\] Now, putting the given values in Eq. (i) \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{35}{36}\sqrt{\frac{1}{16}\times \frac{2}{1}\times 8}=\frac{35}{36}\] \[{{n}_{1}}=\frac{35}{36}{{n}_{2}}\] (Given: \[{{n}_{2}}=360\,Hz\]) \[=\frac{35}{36}\times 360=350\,Hz\] Number of beats produced per second is \[{{n}_{2}}-{{n}_{1}}=360-350=10Hz\]
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