A source of sound A emitting waves of frequency 1800 Hz is falling towards ground with a terminal speed v. The observer B on the ground directly beneath the source receives waves of frequency 2150 Hz. The source A receives waves, reflected from ground of frequency nearly: (Speed of sound = 343 m/s)

First, we are given that a source of sound A emits waves at a frequency of 1800 Hz and is falling towards the ground with a terminal speed $$v$$. An observer B on the ground directly beneath the source receives waves at a frequency of 2150 Hz. The speed of sound is 343 m/s. We need to find the frequency of the waves that source A receives after they are reflected from the ground.

To solve this, we use the Doppler effect formula for sound. The general formula when the source and observer are moving relative to each other is:

$$f' = f \left( \frac{v_{\text{sound}} \pm v_{\text{observer}}}{v_{\text{sound}} \mp v_{\text{source}}} \right)$$

The signs depend on the directions of motion. For the first part, source A is moving towards the stationary observer B on the ground. Since the source is moving towards the observer, the observed frequency increases. The formula becomes:

$$f_B = f \left( \frac{v_{\text{sound}}}{v_{\text{sound}} - v_{\text{source}}} \right)$$

Here, $$f = 1800$$ Hz, $$f_B = 2150$$ Hz, and $$v_{\text{sound}} = 343$$ m/s. Plugging in the values:

$$2150 = 1800 \left( \frac{343}{343 - v} \right)$$

Now, solve for $$v$$. Rearrange the equation:

$$\frac{2150}{1800} = \frac{343}{343 - v}$$

Simplify the left side:

$$\frac{2150}{1800} = \frac{2150 \div 50}{1800 \div 50} = \frac{43}{36}$$

So:

$$\frac{43}{36} = \frac{343}{343 - v}$$

Cross-multiply:

$$43 \times (343 - v) = 36 \times 343$$

Calculate $$36 \times 343$$:

$$343 \times 30 = 10290, \quad 343 \times 6 = 2058, \quad 10290 + 2058 = 12348$$

Calculate $$43 \times 343$$:

$$343 \times 40 = 13720, \quad 343 \times 3 = 1029, \quad 13720 + 1029 = 14749$$

So:

$$14749 - 43v = 12348$$

Rearrange to solve for $$v$$:

$$14749 - 12348 = 43v$$

$$2401 = 43v$$

$$v = \frac{2401}{43}$$

Simplify:

$$2401 \div 43 = 55.8372, \quad \text{but we keep it as a fraction:} \quad v = \frac{2401}{43} \text{m/s}$$

Note that $$2401 = 343 \times 7$$ because $$343 \times 7 = 2401$$. So:

$$v = \frac{343 \times 7}{43} \text{m/s}$$

Now, we need to find the frequency that source A receives from the waves reflected from the ground. The ground reflects the sound waves, and since the ground is stationary, the frequency of the reflected wave is the same as the frequency received by the ground. Observer B on the ground receives 2150 Hz, so the reflected wave has a frequency of 2150 Hz and is emitted by the stationary ground.

Source A is moving towards the ground (downwards) while the reflected wave is traveling upwards. So, for the reflected wave:

• The source (ground) is stationary, so $$v_{\text{source}} = 0$$.
• The observer (source A) is moving towards the source (ground), so it is moving towards the sound wave.

The Doppler effect formula for a stationary source and an observer moving towards the source is:

$$f'' = f_{\text{reflected}} \left( \frac{v_{\text{sound}} + v_{\text{observer}}}{v_{\text{sound}}} \right)$$

Here, $$f_{\text{reflected}} = 2150$$ Hz, $$v_{\text{sound}} = 343$$ m/s, and $$v_{\text{observer}} = v = \frac{2401}{43}$$ m/s. Plugging in:

$$f'' = 2150 \left( \frac{343 + v}{343} \right) = 2150 \left(1 + \frac{v}{343}\right)$$

Substitute $$v = \frac{343 \times 7}{43}$$:

$$f'' = 2150 \left(1 + \frac{\frac{343 \times 7}{43}}{343}\right) = 2150 \left(1 + \frac{7}{43}\right)$$

Simplify inside the parentheses:

$$1 + \frac{7}{43} = \frac{43}{43} + \frac{7}{43} = \frac{50}{43}$$

So:

$$f'' = 2150 \times \frac{50}{43}$$

Note that $$2150 = 43 \times 50$$ because $$43 \times 50 = 2150$$. Therefore:

$$f'' = 43 \times 50 \times \frac{50}{43} = 50 \times 50 = 2500$$

Thus, the frequency that source A receives from the reflected waves is 2500 Hz.

Hence, the correct answer is Option B.