Question

A bar of mass $m$ and length $l$ is hanging from point $\text{A}$ as shown in figure. 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 wires is $A$.

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

Answer 1

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

Consider a small section $dx$ of the bar at a distance $x$ from $\text{B}$. The weight of the bar for a length $x$ is,


$W=\left(\frac{mg}{l}\right)x$

Elongation in section $dx$ will be,

$dl=\left(\frac{W}{AY}\right)dx=\left(\frac{mg}{lAY}\right)xdx$

Total elongation in the bar,

$\therefore \Delta l={\int }_{x=0}^{x=l}dl=\left(\frac{mg}{AYl}\right){\int }_0^{l}xdx$

$=\frac{mg}{AYl}{\left[\frac{x^2}{2}\right]}_0^l=\frac{mgl}{2AY}$

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

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