An underwater sonar source operating at a frequency of $60 KHz$ directs its beam towards the surface. If the velocity of sound in air is $330\, m/s,$ the wavelength and frequency of waves in air are:
A$5.5 mm, 60 KHz$
B$330 m, 60 KHz$
C$5.5 mm, 20 KHz$
D$5.5 mm, 80 KHz$
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A$5.5 mm, 60 KHz$
a (a) Frequency of waves remains same, i.e. $60 Hz$
and wavelength $\lambda = \frac{v}{n} = \frac{{330}}{{60 \times {{10}^3}}}= 5.5 \,mm.$
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