A submarine A travelling at 18 km hr$$^{-1}$$ is being chased along the line of its velocity by another submarine B travelling at 27 km hr$$^{-1}$$. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency v. The value of v is close to (Speed of sound in water 1500 m s$$^{-1}$$)

We first change the speeds of the two submarines from km hr-1 to m s-1. The relation is

$$1\;\text{km hr}^{-1}= \frac{1000\ \text{m}}{3600\ \text{s}}=\frac{5}{18}\ \text{m s}^{-1}.$$

So

$$u_A = 18\;\text{km hr}^{-1}\times\frac{5}{18}=5\ \text{m s}^{-1},\qquad u_B = 27\;\text{km hr}^{-1}\times\frac{5}{18}=7.5\ \text{m s}^{-1}.$$

The speed of sound in water is given as

$$v = 1500\ \text{m s}^{-1}.$$

Submarine B emits a sonar signal of frequency

$$f_0 = 500\ \text{Hz}.$$

The sound first travels from B (source) to A (receiver). For a moving source and a moving observer on the same line, the classical Doppler formula is stated as

$$f' = f\,\frac{v+v_o}{v-v_s},$$

where

• $$v_o$$ is the component of the observer’s velocity toward the source (take it positive when toward),
• $$v_s$$ is the component of the source’s velocity toward the observer (take it positive when toward).

During the first trip (B → A):

• B is moving toward A, so $$v_s = +u_B = +7.5\ \text{m s}^{-1}.$$
• A is moving away from B, so $$v_o = -u_A = -5\ \text{m s}^{-1}.$$

Hence the frequency heard by A is

$$\begin{aligned} f_1 &= f_0\,\frac{v+v_o}{v-v_s} \\ &= 500\;\text{Hz}\,\frac{1500-5}{1500-7.5} \\ &= 500\;\text{Hz}\,\frac{1495}{1492.5}. \end{aligned}$$

Evaluating the fraction step by step,

$$\frac{1495}{1492.5}=1+\frac{2.5}{1492.5}\approx1+0.001676=1.001676,$$

so

$$f_1 \approx 500\times1.001676 = 500.838\ \text{Hz}.$$

This sound is reflected from A. On the return trip (A → B) A now behaves like a moving source emitting at frequency $$f_1$$, while B is the observer.

During the second trip:

• A is moving away from B, hence its velocity toward B is negative: $$v_s'=-u_A=-5\ \text{m s}^{-1}.$$
• B is moving toward A, so $$v_o'=+u_B=+7.5\ \text{m s}^{-1}.$$

Applying the same formula again, the frequency finally received by B is

$$\begin{aligned} f_2 &= f_1\,\frac{v+v_o'}{v-v_s'} \\ &= 500.838\;\text{Hz}\,\frac{1500+7.5}{1500+5} \\ &= 500.838\;\text{Hz}\,\frac{1507.5}{1505}. \end{aligned}$$

Now

$$\frac{1507.5}{1505}=1+\frac{2.5}{1505}\approx1+0.001662=1.001662,$$

and therefore

$$f_2 \approx 500.838\times1.001662 = 501.672\ \text{Hz}.$$

This is very close to 502 Hz, the nearest value among the choices.

Hence, the correct answer is Option C.