When a force $F_1$ is applied on a metallic wire, the length of the wire is $L_1$. If a force $F_2$ is applied on the same wire, the length of the wire is $L_2$. The original length of the wire $L$ is

• A

  $\frac{L_1 F_1+L_2 F_2}{F_1+F_2}$
• B

  $\frac{L_2-L_1}{F_1+F_2}$
• Correct answer
  C

  $\frac{F_1 L_2-F_2 L_1}{F_1-F_2}$
• D

  $\frac{F_1 L_1-F_2 L_2}{F_1-F_2}$

Solution

Change in length of wire of length $L$ is directly proportional to the external force applied
$$ \begin{array}{ll} & F_1 \propto\left(L_1-L\right) \\ \Rightarrow & F_1=K\left(L_1-L\right) \\ \text { Similarly, } & F_2=K\left(L_2-L\right) \end{array} $$ Dividing Eq. (i) by Eq. (ii), we get $$ \begin{aligned} \frac{F_1}{F_2} & =\frac{L_1-L}{L_2-L} \\ \Rightarrow F_1 L_2-F_1 L & =F_2 L_1-F_2 L \\ \Rightarrow \quad L & =\frac{F_1 L_2-F_2 L_1}{F_1-F_2} \end{aligned} $$

Hence, the correct answer is option: 'C'.