Question

Copper of fixed volume $V$; is drawn into wire of length $l$. When this wire is subjected to a constant force ${}^{\prime }{F}^{\prime }$, the extension produced in the wire is $\Delta l$. Which of the following graphs is a straight line?

• A. $\Delta l$ versus $=\frac{1}{l}$
• B. $\Delta l$ versus $l^2$
• C. $\Delta l$ versus $=\frac{1}{l^2}$
• D. $\Delta l$ versus $l$

Answer 1

The correct option is B $\Delta l$ versus $l^2$

We know that, Young's modulus $\left(Y\right)$
$Y=\frac{Fl}{A\Delta l}...\left(1\right)$
Volume, $V=Al...\left(2\right)$
Eq.$\left(1\right)$ can be written as,
$\Rightarrow Y=\frac{F{l}^2}{Al\Delta l}$

From Eq.$\left(1\right)$ and Eq.$\left(2\right)$,
$\Delta l=\frac{F{l}^2}{VY}...\left(3\right)$
[Here, $F,V$ and $Y$ are constant, therefore $\frac{F}{VY}$ is constant]
$\Rightarrow \Delta l\propto l^2$

On comparing Eq.$\left(3\right)$ with the equation of a straight line $y=mx+c$,
$y=\Delta l,c=0$
Slope, $m=\frac{F}{VY}=\text{constant}$
& $x=l^2$
$\Rightarrow$ $\Delta l$ versus $l^2$ graph is of the form $y=mx$.

Hence, $\Delta l$ versus $l^2$ graph is a straight line passing through origin.