The length of a metal wire is $l_{1}$ when the tension in it is $F_{1}$ and $l_{2}$ when the tension is $F_{2} .$ Then natural length of the wire is

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$\frac{l_{2} - l_{1}}{F_{2} - F_{1}}$

$\frac{l_{1} F_{1} - l_{2} F_{2}}{F_{2} - F_{1}}$

$\frac{l_{1} F_{1} + l_{2} F_{2}}{F_{1} + F_{2}}$

$\frac{l_{1} F_{2} - l_{2} F_{1}}{F_{2} - F_{1}}$

answer is C.

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Detailed Solution

Let ‘l’ be the natural length of the wire.
For the force $F_{1}$ if elongation is $e_{1}, e_{1} = \frac{F_{1} l}{A Y} \Rightarrow Y = \frac{F_{1} l}{A e_{1}}$
Similarly for the force $F_{2},$ if elongation is $e_{2}$
$e_{2} = \frac{F_{2} l}{A Y} \Rightarrow Y = \frac{F_{2} l}{A e_{2}}$
$l_{1} = l + e_{1}$ and $l_{2} = l + e_{2}$
$Y = \frac{F_{1} l}{A (l_{1} - l)} = \frac{F_{2} l}{A (l_{2} - l)}$ then $F_{1} (l_{2} - l) = F_{2} (l_{1} - l)$
And $F_{1} l_{2} - F_{1} l = F_{2} l_{1} - F_{2} l$
$(F_{2} - F_{1}) l = l_{1} F_{2} - l_{2} F_{1}$ or $l = \frac{l_{1} F_{2} - l_{2} F_{1}}{(F_{2} - F_{1})}$
So, the correct answer is 3