A bat moving at 10 m s$$^{-1}$$ towards a wall sends a sound signal of 8000 Hz towards it. On reflection, it hears a sound of frequency $$f$$. The value of $$f$$ in Hz is close to (speed of sound = 320 m s$$^{-1}$$)

A bat is moving towards a wall at 10 m/s and emits a sound of frequency 8000 Hz. The speed of sound is 320 m/s. We need to find the frequency $$f$$ that the bat hears after the sound reflects off the wall. This involves the Doppler effect in two steps: first, the sound travels to the wall, and then the reflected sound travels back to the bat.

In the first step, the bat (source) is moving towards the stationary wall (observer). The frequency $$f_1$$ heard by the wall is given by the Doppler effect formula for a source moving towards a stationary observer:

$$ f_1 = \left( \frac{v}{v - v_b} \right) f_0 $$

where $$v = 320$$ m/s is the speed of sound, $$v_b = 10$$ m/s is the speed of the bat, and $$f_0 = 8000$$ Hz is the emitted frequency. Substituting the values:

$$ f_1 = \left( \frac{320}{320 - 10} \right) \times 8000 = \left( \frac{320}{310} \right) \times 8000 $$

Simplify the fraction $$\frac{320}{310}$$ by dividing numerator and denominator by 10:

$$ \frac{320}{310} = \frac{32}{31} $$

So,

$$ f_1 = \frac{32}{31} \times 8000 $$

Calculate $$8000 \times 32$$:

$$ 8000 \times 32 = 256000 $$

Thus,

$$ f_1 = \frac{256000}{31} $$

In the second step, the wall reflects the sound and acts as a stationary source emitting frequency $$f_1$$. The bat is now moving towards this stationary source. The frequency $$f$$ heard by the bat (observer moving towards a stationary source) is given by:

$$ f = \left( \frac{v + v_b}{v} \right) f_1 $$

Substituting the values:

$$ f = \left( \frac{320 + 10}{320} \right) f_1 = \left( \frac{330}{320} \right) f_1 $$

Simplify $$\frac{330}{320}$$ by dividing numerator and denominator by 10:

$$ \frac{330}{320} = \frac{33}{32} $$

So,

$$ f = \frac{33}{32} \times f_1 $$

Substitute $$f_1 = \frac{256000}{31}$$:

$$ f = \frac{33}{32} \times \frac{256000}{31} $$

Notice that 32 in the denominator and 256000 in the numerator can be simplified. Since $$256000 = 8000 \times 32$$, we can write:

$$ f = \frac{33}{32} \times \frac{8000 \times 32}{31} = \frac{33}{31} \times 8000 $$

The 32 cancels out:

$$ f = \frac{33}{31} \times 8000 $$

Now, compute $$33 \times 8000$$:

$$ 33 \times 8000 = 264000 $$

So,

$$ f = \frac{264000}{31} $$

Divide 264000 by 31:

$$ 264000 \div 31 $$

31 multiplied by 8516 gives:

$$ 31 \times 8516 = 31 \times (8000 + 516) = (31 \times 8000) + (31 \times 516) = 248000 + 15996 = 263996 $$

Subtract from 264000:

$$ 264000 - 263996 = 4 $$

So,

$$ f = 8516 + \frac{4}{31} \approx 8516.129 \text{ Hz} $$

The value is close to 8516 Hz. Comparing with the options:

A. 8258

B. 8424

C. 8000

D. 8516

Hence, the correct answer is Option D.