Question

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

• A. $\Delta l$ versus $1/l$
• B. $\Delta l$ versus $l^2$
• C. $\Delta l$ versus $1/l^2$
• D. $\Delta l$ versus $l$

Answer 1

Answer: (B)

$\Delta l$ versus $l^2$

Answer 2

The Young's modulus ($Y$) of a material is defined as the ratio of stress to strain: $Y = \frac{\text{Stress}}{\text{Strain}}$

For a wire subjected to a tensile force $F$, the stress is $\frac{F}{A}$, where $A$ is the cross-sectional area. The strain is $\frac{\Delta l}{l}$, where $\Delta l$ is the extension and $l$ is the original length. So, $Y = \frac{F/A}{\Delta l/l} = \frac{Fl}{A \Delta l}$.

Rearranging this equation to find the extension $\Delta l$: $\Delta l = \frac{Fl}{A Y}$

We are given that the wire is made from a fixed volume $V$ of copper. The volume of a wire is given by $V = A \times l$. From this, we can express the cross-sectional area $A$ in terms of $V$ and $l$: $A = \frac{V}{l}$

Substitute this expression for $A$ into the equation for $\Delta l$: $\Delta l = \frac{Fl}{(V/l)Y} = \frac{Fl^2}{VY}$

Since the force $F$, the volume $V$, and the Young's modulus $Y$ of copper are constant, we can write this relationship as: $\Delta l = K l^2$ where $K = \frac{F}{VY}$ is a constant.

This equation shows that the extension $\Delta l$ is directly proportional to the square of the length $l$. A graph of $\Delta l$ versus $l^2$ will be a straight line passing through the origin, with the slope equal to the constant $K$.

Let's examine the given options:

1. $\Delta l$ versus $1/l$: If we plot $\Delta l$ against $1/l$, let $x = 1/l$. Then $l = 1/x$. The relationship becomes $\Delta l = K (1/x)^2 = K/x^2$. This is not a linear relationship.
2. $\Delta l$ versus $l^2$: If we plot $\Delta l$ against $l^2$, let $x = l^2$. The relationship becomes $\Delta l = Kx$. This is of the form $y = mx$, which represents a straight line passing through the origin with slope $m=K$.
3. $\Delta l$ versus $1/l^2$: If we plot $\Delta l$ against $1/l^2$, let $x = 1/l^2$. Then $l^2 = 1/x$. The relationship becomes $\Delta l = K (1/x) = K/x$. This is not a linear relationship.
4. $\Delta l$ versus $l$: If we plot $\Delta l$ against $l$, let $x = l$. The relationship becomes $\Delta l = Kx^2$. This is a quadratic relationship, representing a parabola, not a straight line.

Therefore, the graph of $\Delta l$ versus $l^2$ is a straight line.