A) 1: 2
B) 1 : 3
C) 1 : 4
D) 1 : 6
Correct Answer: B
Solution :
First overtone of string\[A\]= Second overtone of string B. \[\Rightarrow \] Second harmonic of \[A\] = Third harmonic of B \[\Rightarrow \] \[{{n}_{2}}={{n}_{3}}\]\[\Rightarrow \] \[{{\left[ 2({{n}_{1}}) \right]}_{A}}={{\left[ 3({{n}_{1}}) \right]}_{B}}\] (\[\because \] \[{{n}_{1}}=\frac{1}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}\rho }}\]) Þ\[2\,\left[ \frac{1}{2{{l}_{A}}{{r}_{A}}}\sqrt{\frac{T}{\pi \rho }} \right]=3\,\left[ \frac{1}{2{{l}_{B}}{{r}_{B}}}\sqrt{\frac{T}{\pi \rho }} \right]\] \[\frac{{{l}_{A}}}{{{l}_{B}}}=\frac{2}{3}\frac{{{r}_{B}}}{{{r}_{A}}}\Rightarrow \frac{{{l}_{A}}}{{{l}_{B}}}=\frac{2}{3}\times \frac{{{r}_{B}}}{(2{{r}_{B}})}=\frac{1}{3}\]You need to login to perform this action.
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