A) 303 ms-1
B) 332 ms-1
C) 323.2 ms-1
D) 300 ms-1
Correct Answer: C
Solution :
Key Idea: Number of beats per second = difference of the frequencies of sound sources. Let \[l\] be the length of pipe and v is the velocity of sound, then the frequency of note emitted from the pipe is \[n=\frac{v}{4l}\] Number of beats in 1 s \[=\frac{16}{20}=\frac{4}{5}\] For a closed organ pipe \[x={{n}_{1}}-{{n}_{2}}=\frac{v}{4}\left( \frac{1}{{{l}_{1}}}-\frac{1}{{{l}_{2}}} \right)\] \[\frac{4}{5}=\frac{v}{4}\left( \frac{1}{1}-\frac{1}{1.01} \right)=\frac{0.01v}{4\times 1\times 1.01}\] \[\Rightarrow \]\[0.01\times 5v=16\times 1.01\] \[\therefore \] \[v=\frac{16\times 1.01}{0.05}=323.2\,m/s\]You need to login to perform this action.
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