Question

A wire of length $L$ has a linear mass density $\mu$ and area of cross-section $A$ and Young's modulus $Y$ is suspended vertically from a rigid support. The extension produced in the wire due to its own weight is:

• A. $\frac{\mu g{L}^2}{YA}$
• B. $\frac{\mu g{L}^2}{2YA}$
• C. $\frac{2\mu g{L}^2}{YA}$
• D. $\frac{2\mu g{L}^2}{3YA}$

Answer 1

The correct option is B $\frac{\mu g{L}^2}{2YA}$

Consider a small element of length $dx$ at a distance $x$ from the free end of wire as shown in the figure. Tension in the wire at distance $x$ from the lower end is
$T\left(x\right)=\mu gx....\left(i\right)$
Let $dl$ be increase in length of the element. Then
$T=\frac{T\left(x\right)/A}{dl/dx}$

$dl=\frac{T\left(x\right)dx}{YA}=\frac{\mu gxdx}{YA}$ [Using (i)]

Total extension produced in the wire is
$l={\int }_0^L\frac{\mu gx}{YA}=\frac{\mu g}{YA}{\left[\frac{x^2}{2}\right]}_0^L=\frac{\mu g{L}^2}{2YA}$