A) 404 m/s
B) 4.04 m/s
C) 414 m/s
D) 400 m/s
Correct Answer: A
Solution :
Let\[{{n}_{1}}\]and\[{{n}_{2}}\]be the frequencies of the two notes of wavelengths\[{{\lambda }_{1}}\]and\[{{\lambda }_{2}}\]. Here, \[{{\lambda }_{1}}=1m,\,\,{{\lambda }_{2}}=1.01m\] If\[v\]is velocity of sound, then \[{{n}_{1}}=\frac{v}{1}\]and\[{{n}_{2}}=\frac{v}{1.01}\] If the number of beats produced per second is\[b\], then \[b={{n}_{1}}-{{n}_{2}}\] \[4=\frac{v}{1}-\frac{v}{1.01}\] \[\therefore \] \[v=\frac{4.04}{0.01}=404\,\,m/s\]
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