Question

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. $\frac{L_{2}-L_{1}}{F_{1}+F_{2}}$
• C. $\frac{F_{1} L_{2}-F_{2} L_{1}}{F_{1}-F_{2}}$
• D. $\frac{F_{1} L_{1}-F_{2} L_{2}}{F_{1}-F_{2}}$

Answer 1

Answer: (C)

Change in length of wire of length $L$ is directly proportional to the external force applied

$ F_{1} \propto\left(L_{1}-L\right) $
$\Rightarrow F_{1}=K\left(L_{1}-L\right)\,.......(i)$
Similarly, $F_{2}=K\left(L_{2}-L\right)\,......(ii)$
Dividing Eq. (i) by Eq. (ii), we get
$\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 $
$\Rightarrow L =\frac{F_{1} L_{2}-F_{2} L_{1}}{F_{1}-F_{2}}$