Question

A wire of mass $M$ and length $L$ is hanging from point $A$. Find the increase in its length due to its own weight. The Young's modulus of elasticity of the wire is $Y$ and area of cross-section of the wire is $A$.

• A. $\frac{mgL}{2AY}$
• B. $\frac{mgL}{3AY}$
• C. $\frac{mgL}{AY}$
• D. $\frac{2mgL}{AY}$

Answer 1

The correct option is A $\frac{mgL}{2AY}$

Consider a small section of the wire located at point $C$, at a distance $x$ from the end $B$. The weight of the wire, of length $x$ is,
$W=\left(\frac{mg}{L}\right)x$

The elongation in the section $dx$ will be,
$dl=\left(\frac{W}{AY}\right)dx=\left(\frac{mg}{LAY}\right)x\times dx$

Total elongation in the wire can be obtained by integrating this expression for $x=0$ to $x=L$

$\therefore \Delta l={\int }_{x=0}^{x=L}dl$

$=\left(\frac{mg}{LAY}\right){\int }_0^{L}xdx$

After integrating, we get, $\Delta l=\frac{mgL}{2AY}$

Hence, $\left(A\right)$ is the correct answer.

Why this question? In this case, different parts of the wire does not elongate to the same extent. The element closer to the support elongates more as the stress is higher in comparison with the elements closer to the free end of the wire.