A) 104.8cm
B) 95.4cm
C) 104cm
D) 96cm
Correct Answer: B
Solution :
When I length of closed organ pipe and \[\lambda \] be the wavelength, then frequency \[n=\frac{v}{4l}\] Also, number of beats = difference of frequencies of sound-source \[{{n}_{2}}-{{n}_{1}}=\frac{v}{4{{l}_{2}}}-\frac{v}{4{{l}_{1}}}\] \[{{n}_{2}}-{{n}_{1}}=\frac{v}{4}\left( \frac{1}{{{l}_{2}}}-\frac{1}{{{l}_{1}}} \right)\] \[\Rightarrow \] \[4=\frac{330}{4}\left( \frac{1}{{{l}_{2}}}-\frac{1}{1} \right)\] \[\Rightarrow \] \[\frac{1}{{{l}_{2}}}=\frac{4\times 4}{330}+1\] \[\Rightarrow \] \[\frac{1}{{{l}_{2}}}=\frac{346}{330}\] \[\Rightarrow \] \[{{l}_{2}}=\frac{330}{346}=0.954\,\,m=95.4\,\,cm\]You need to login to perform this action.
You will be redirected in
3 sec