A submarine A travelling at $18$ km/hr is being chased along the line of its velocity by another submarine B travelling at $27$ km/hr. B sends a sonar signal of $500$ Hz to detect A and receives a reflected sound of frequency $\nu$. The value of $\nu$ is close to (Speed of sound in water $1500\ \text{m s}^{-1}$)
Answer: (C) $502$ Hz
$v_A = 5$ m/s, $v_B = 7.5$ m/s, both moving in the same direction with B behind.
Frequency received by A (observer moving away, source approaching):
$$f_1 = 500\times\frac{1500 - 5}{1500 - 7.5}$$
A re-emits $f_1$ while moving away from B, and B moves towards it:
$$\nu = f_1\times\frac{1500 + 7.5}{1500 + 5} = 500\times\frac{1495}{1492.5}\times\frac{1507.5}{1505} \approx 501.7 \approx 502\ \text{Hz}$$
Solution by Sreeraj P, M.Sc Physics