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**Hint:**If one source emits sound waves of frequency ${{f}_{1}}$ and other source emits sound waves of frequency ${{f}_{2}}$. Then the beat frequency between the two waves is given as $B=\left| {{f}_{1}}-{{f}_{2}} \right|$.

Use this formula in both the case and try to find the frequency of the unknown source.

**Formula used:**

$B=\left| {{f}_{1}}-{{f}_{2}} \right|$

**Complete step by step answer:**

When sound waves of different frequencies from two different sources superimpose, we get something called beat frequency. The beat frequency is nothing but the magnitude of the difference between the frequencies of the two sources.

Suppose there are two sources of sound. One emits sound waves of frequency ${{f}_{1}}$ and another source emits sound waves of frequency ${{f}_{2}}$. Then the beat frequency between the two waves is given as $B=\left| {{f}_{1}}-{{f}_{2}} \right|$.

One emits sound waves of frequency ${{f}_{1}}$ and another source emits sound waves of frequency ${{f}_{2}}$.

Then the beat frequency between the two waves is given as $B=\left| {{f}_{1}}-{{f}_{2}}\right|$.

Let the frequency of the unknown source be f.

It is given that when this source is sounded with a source with a frequency of 250 Hz, the beat frequency is 4 beats/s.

Therefore, $\Rightarrow 4=\left| f-250 \right|$

$\Rightarrow f-250=\pm 4$

$\Rightarrow f=250\pm 4$.

This means that the frequency of the unknown source can be either 254 Hz or 246 Hz.

Now, the second harmonic of the unknown source will be equal to 2f. It is said that the frequency

given 5 beats/s when sounded with a source of frequency 513 Hz.

Then,

$\Rightarrow 5=\left| 2f-513 \right|$

$\Rightarrow 2f-513=\pm 5$

$\Rightarrow 2f=513\pm 5$

This means that the second harmonic of the unknown source can be either 518 Hz or 508 Hz.

If $2f=518$, then $f=259Hz$.

If $2f=508$, then $f=254Hz$.

However, f can be either 254 Hz or 246 Hz.

This means that the frequency of unknown source is 254 Hz.

**Hence, the correct option is A.**

**Note:**This is one of the methods to find the frequency of an unknown source of sound. Another way to find the unknown frequency with the help of beats is by waxing of the unknown source. When wax is applied, its mass increases, which lowers its frequency.

Note that $\left| {{f}_{1}}-{{f}_{2}} \right|$ is a modulus function. It gives us only the positive numbers and zero. Suppose, we have a function $y=|x|$. Then the value of y is always positive.

If x is positive or equal to zero, then the value of x is equal to x.

i.e. for $x\ge 0$, $y=x$.

If x is negative, then the value of y is negative of x.

i.e. for x < 0, $y=-x$.

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