Question

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, original length of the wire is

• A. $ \frac{l_{1}F_{1}+l_{2}\,f_{2}}{F_{1}+F_{2}} $
• B. $ \frac{l_{1}-l_{2}}{F_{2}-F_{1}} $
• C. $ \frac{l_{1}F_{2}-l_{2}\,F_{1}}{F_{2}-F_{1}} $
• D. $ \frac{l_{1}F_{1}-l_{2}\,F_{2}}{F_{2}-F_{1}} $

Answer 1

Answer: (C)

As, $F \alpha \Delta l$ where, $\Delta 1=$ change in length Now, suppose the origin length of the metal wire is $l$ then,
$F_{1} \propto\left(l_{1}-l\right)$
Similarly, $F_{2} \propto\left(1_{2}-1\right)$
The ratio, $\frac{F_{1}}{F_{2}}=\frac{l_{1}-1}{l_{2}-1}$
$\Rightarrow F_{1} l_{2}-F_{1} l=F_{2} l_{1}-F_{2} l$
$\Rightarrow \left(F_{2}-F_{1}\right) I=F_{2} l_{1}-F_{1} l_{2}$
$\Rightarrow l=\frac{F_{2} I_{1}-F_{1} I_{2}}{F_{2}-F_{1}}$