A toy-car, blowing its horn, is moving with a steady speed of 5 m s$$^{-1}$$, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m s$$^{-1}$$, the frequency of the horn of the toy car is close to

Let the true frequency of the horn of the toy-car be $$f\;{\rm Hz}$$. The speed of sound in air is given to be $$v = 340\;{\rm m\,s^{-1}}$$ and the speed of the toy-car is $$u = 5\;{\rm m\,s^{-1}}$$.

The observer receives sound by two different paths:

1. Direct sound from the horn.   The toy-car (source) is moving towards the observer.   For a stationary observer and a moving source, the Doppler-shift formula is

$$f_{\text{observed}} = \frac{v}{v - u_s}\,f,$$

where $$u_s$$ is the speed of the source towards the observer. Here $$u_s = u = 5\;{\rm m\,s^{-1}}$$, so the frequency heard directly is

$$f_1 = \frac{v}{v - u}\,f = \frac{340}{340 - 5}\,f = \frac{340}{335}\,f.$$

2. Reflected sound from the wall (echo).   First, the sound reaches the wall. The car is moving away from the wall, so for the wall (which is a stationary observer) the received frequency is

$$f_{\text{wall}} = \frac{v}{v + u}\,f = \frac{340}{340 + 5}\,f = \frac{340}{345}\,f.$$

The wall is at rest, so it reflects this sound without any further Doppler shift; the wall now behaves as a stationary source emitting frequency $$f_{\text{wall}}$$. Because the final observer is also stationary, the frequency finally heard from the echo remains

$$f_2 = f_{\text{wall}} = \frac{340}{345}\,f.$$

The observer hears beats produced by the superposition of the direct sound ($$f_1$$) and the reflected sound ($$f_2$$). The beat frequency equals the absolute difference of the two frequencies, and this is given as 5 beats per second:

$$|\,f_1 - f_2\,| = 5.$$

Explicitly substituting $$f_1$$ and $$f_2$$ we get

$$\left|\,\frac{340}{335}\,f - \frac{340}{345}\,f\,\right| = 5.$$

Pulling out the common factor $$340f$$ gives

$$340f\left|\,\frac{1}{335} - \frac{1}{345}\,\right| = 5.$$

The difference inside the modulus is

$$\frac{1}{335} - \frac{1}{345} = \frac{345 - 335}{335 \times 345} = \frac{10}{335 \times 345}.$$

Therefore,

$$340f \times \frac{10}{335 \times 345} = 5.$$

We now solve for $$f$$ step by step:

$$f = 5 \times \frac{335 \times 345}{340 \times 10}.$$

Simplifying the numerator and denominator,

$$f = 5 \times \frac{335 \times 345}{3400}.$$ $$f = \frac{5}{3400} \times 335 \times 345.$$ $$f = \frac{5 \times 335 \times 345}{3400}.$$

Take out the common factor 5 from 3400:

$$f = \frac{335 \times 345}{680}.$$

Now divide 335 by 5 to make the arithmetic easier:

$$335 = 5 \times 67,$$ so substitute:

$$f = \frac{(5 \times 67) \times 345}{680} = \frac{5 \times 67 \times 345}{5 \times 136}$$ $$f = \frac{67 \times 345}{136}.$$

Divide numerator and denominator by 17:

$$\frac{345}{17} = 20.294\ldots \quad\text{and}\quad \frac{136}{17} = 8.$$ So

$$f \approx \frac{67 \times 20.294}{8} = \frac{1359.7}{8} \approx 169.96.$$

Rounding to the nearest whole number gives

$$f \approx 170\;{\rm Hz}.$$

Among the given options, this matches Option D.

Hence, the correct answer is Option D.