Question

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

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

Answer 1

Answer: (B)

$\frac{\mu gL^{2}}{2YA}$

Answer 2

: 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

$T(x) = \mu gx$ ….(i)

Let dl be increase in length of the element. Then

$Y = \frac{T(x)/A}{dl/dx}$

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

Total extension produced in the wire is

$l = \int_{0}^{L}{\frac{\mu gx}{YA}dx = \frac{\mu g}{YA}}\left\lbrack \frac{x^{2}}{2} \right\rbrack_{0}^{L} = \frac{\mu gL^{2}}{2YA}$