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**Hint**To solve this question, we need to use the formula for the fundamental frequency in a stretched string. We have to apply this formula for the three segments and for the original string. Then, on manipulating the four equations thus formed, we will get the answer.

**Formula Used**The formula used to solve this question is

$\nu = \dfrac{1}{{2l}}\sqrt {\dfrac{F}{\mu }} $, where $\nu $ is the fundamental frequency of a string of length $l$ having $\mu $ mass per unit length, which is stretched by a tension $F$

**Complete step-by-step solution**

Let the original length of the string be $l$

Since the string is divided into three segments of lengths,${l_1}$, ${l_2}$ and ${l_3}$, so we get the length $l$ as

$l = {l_1} + {l_2} + {l_3}$ (i)

We know that for a stretched string of length $l$, the fundamental frequency is given by

$\nu = \dfrac{1}{{2l}}\sqrt {\dfrac{F}{\mu }} $

Or, $\nu = \dfrac{{\sqrt F }}{{2\sqrt \mu }} \times \dfrac{1}{l}$

Since the mass per unit length, $\mu $ is a property of the string material, so it is a constant.

Also, for the same tension $F$ in the original string and it’s each segment, $F$ is a constant.

So, replacing all the constants in the above equations with $k$, we get the fundamental frequency

$\nu = \dfrac{k}{l}$

From here, the length is given as

\[l = \dfrac{k}{\nu }\] (ii)

This is the expression of the length of the original string with its fundamental frequency.

Now, applying equation (i) for the first segment, we get its length ${l_1}$ as

\[{l_1} = \dfrac{k}{{{\nu _1}}}\] (iii)

Similarly, we get the lengths ${l_1}$ and ${l_1}$ of the second and third segments as

\[{l_2} = \dfrac{k}{{{\nu _2}}}\] (iv)

\[{l_3} = \dfrac{k}{{{\nu _3}}}\] (v)

On adding the equations (iii) (iv) and (v), we get

\[{l_1} + {l_2} + {l_3} = \dfrac{k}{{{\nu _1}}} + \dfrac{k}{{{\nu _2}}} + \dfrac{k}{{{\nu _3}}}\]

From (i)

\[l = \dfrac{k}{{{\nu _1}}} + \dfrac{k}{{{\nu _2}}} + \dfrac{k}{{{\nu _3}}}\]

Substituting $l$ from (ii), we get

\[\dfrac{k}{\nu } = \dfrac{k}{{{\nu _1}}} + \dfrac{k}{{{\nu _2}}} + \dfrac{k}{{{\nu _3}}}\]

Taking $k$as common on the RHS

\[\dfrac{k}{\nu } = k\left( {\dfrac{1}{{{\nu _1}}} + \dfrac{1}{{{\nu _2}}} + \dfrac{1}{{{\nu _3}}}} \right)\]

Dividing both sides by $k$, we get

\[\dfrac{1}{\nu } = \dfrac{1}{{{\nu _1}}} + \dfrac{1}{{{\nu _2}}} + \dfrac{1}{{{\nu _3}}}\]

This is the required expression for the original frequency of the string.

**Hence, the correct answer is option C.**

**Note**In case if we don’t remember the exact formula for the fundamental frequency of a stretched string, then also we can attempt this question correctly. The simple method used for this purpose is the dimensional analysis. As in the question, the information about the frequencies and the lengths of the strings are given, so we must obtain the relation between the frequency and the length. We can easily do this by performing dimensional analysis, and then carry out the rest of the calculations discussed in the solution.

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