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

The Correct Option is B

To solve this problem, we need to understand how the extension \(\Delta l\) of the wire is related to its length \(l\), given a constant force \(F\) and a fixed volume \(V\) of copper.

1. The extension of a wire when a force is applied can be determined using Hooke's Law for elasticity, expressed as: \[\Delta l = \frac{Fl}{AY}\], where \(A\) is the cross-sectional area and \(Y\) is Young's modulus of the material.
2. Since the volume \(V\) of the wire is constant, we have: \[V = Al\]. This implies that the cross-sectional area \(A\) can be written as \[A = \frac{V}{l}\].
3. Substituting the expression for \(A\) into the formula for \(\Delta l\), we get: \[\Delta l = \frac{Fl}{\left(\frac{V}{l}\right)Y} = \frac{F l^2}{V Y}\].
4. This equation shows that the extension \(\Delta l\) is directly proportional to \(l^2\), assuming constant \(F\), \(V\), and \(Y\).
5. Given this proportionality, the graph of \(\Delta l\) versus \(l^2\) will be a straight line.

Therefore, the correct answer to the problem is that the graph of \(\Delta l\) versus \(l^2\) is a straight line. The other options do not satisfy this linear relationship based on the derived formula.

Conclusion: The correct graph that forms a straight line is \(\Delta l \, \, \, \, versus \, \, \, \, l^2\).