A) 5
B) 8
C) 6
D) 10
Correct Answer: D
Solution :
Frequency in a stretched string is given by \[n=\frac{1}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}\rho }}=\frac{1}{l}\sqrt{\frac{T}{\pi {{d}^{2}}\rho }}\] (d = Diameter of string) Þ \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}\times {{\left( \frac{{{d}_{2}}}{{{d}_{1}}} \right)}^{2}}\times \left( \frac{{{\rho }_{2}}}{{{\rho }_{1}}} \right)}\] \[=\frac{35}{36}\sqrt{\frac{8}{1}\times {{\left( \frac{1}{4} \right)}^{2}}\times \frac{2}{1}}=\frac{35}{36}\]\[\Rightarrow {{n}_{2}}=\frac{36}{35}\times 360=370\] Hence beat frequency = \[{{n}_{2}}-{{n}_{1}}=10\]
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