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. If the mass $M$ is hung at the free end of the wire, then the extension produced in the wire is:

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

Answer 1

The correct option is C $\frac{\mu g{L}^2+2MgL}{2YA}$

Consider a small element of length $dx$ at a distance $x$ from the load as shown in the wire
Tension in the wire at a distance $x$ from the lower end is
$T\left(x\right)=\mu gx+Mg....\left(i\right)$
Let $dl$ be increase in length of the element
Then $Y=\frac{T\left(x\right)/A}{dl/dx}$
$dl=\frac{T\left(x\right)dx}{YA}=\left(\frac{\mu gx+Mg}{YA}\right)dx$ [Using (i)]
$\therefore$ Total extension produced in the wire is
$l=l={\int }_0^L\left(\frac{\mu gx+Mg}{YA}\right)dx=\frac{1}{YA}{[\frac{{\mu gx}^2}{2}+Mgx]}_0^L=\frac{1}{YA}[\frac{{\mu gL}^2}{2}+MgL]=\frac{{\mu gL}^2+2MgL}{2YA}$