Question

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

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

Answer 1

Answer: (B)

Consider a small element oflength $dx$ 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
$T\left(x\right)$ $=\mu gx$ $\quad\ldots\left(i\right)$

Let $dl$ be increase in length of the element. Then
$Y$ $=\frac{T\left(x\right) /A}{dl / dx}$
$dl$ $=\frac{T\left(x\right)dx}{YA}$ $=\frac{\mu gx \, dx}{YA}$ $\quad\left[Using\left(i\right)\right]$
Total extension produced in the wire is
$l$ $=\int\limits_{o}^{L} \frac{\mu gx}{YA} dx$ $=\frac{\mu g}{YA} \left[\frac{x^{2}}{2}\right]_{o}^{L}$ $=\frac{\mu gL^{2}}{2 YA}$