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In the figure shown an observer $O_{1}$ floats (static) on water surface with ears in air while another observer $O_{2}$ is moving upwards with constant velocity $v_{1} = v / 5$ in water.
The source moves down with constant velocity $v_{s} = v / 5$ and emits sound of frequency $f$ .
The velocity of sound in air is $v$ and that in water is $4 v$ . For the situation shown in figure

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The wavelength of the sound received by $O_{1}$ is $4 v / 5 f$

The wavelength of the sound received by $O_{1}$ is $v / f$

The frequency of the sound received by $O_{2}$ is $21 f / 16$

The wavelength of the sound received by $O_{2}$ is $16 v / 5 f$

answer is A, C, D.

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Detailed Solution

For observer $O_{1}$ ,

$λ_{1} = \frac{v - v_{s}}{f} = \frac{v - v / 5}{f} = \frac{4 v}{5 f}$

For $O_{2}$ , there is change of medium, hence at the surface of water, keeping frequency unchanged

$f_{w} = f \frac{(1 + \frac{v / 5}{4 v})}{(1 - \frac{v / 5}{v})} = \frac{21}{16} f$

$F o r O_{2}, λ_{w} = \frac{v e l o c i t y o f w a v e r e l a t i v e t o o b s e r v e r}{f_{w}} = \frac{4 v + \frac{v}{5}}{f_{w}} = \frac{21 v}{5} \frac{16 v}{21 f} = \frac{16 v}{5 f}$

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