A) 12
B) 0
C) 1
D) 6
Correct Answer: D
Solution :
Let \[{{\lambda }_{1}}=5.0\,m,\,v=330\,m/s\]and \[{{\lambda }_{2}}=5.5\,m\] The relation between frequency, wavelength and velocity is given by \[v\,=n\,\lambda \] \[\Rightarrow n=\frac{v}{\lambda }....(i)\] The frequency corresponding to wavelength \[{{\lambda }_{1}},\] \[{{n}_{1}}=\frac{v}{{{\lambda }_{1}}}=\frac{330}{5.0}=66\,Hz\] The frequency corresponding to wavelength \[{{\lambda }_{2}}\], \[{{n}_{2}}=\frac{v}{{{\lambda }_{2}}}=\frac{330}{5.5}=60\,Hz\] Hence, no. of bears per second \[={{n}_{1}}-{{n}_{2}}\] \[=66-60\] \[=6\]
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