Question

An Indian rubber cord $L\text{m}$ long and area of cross-section $A{\text{m}}^2$ is suspended vertically. Density of rubber is $D{\text{kg/m}}^3$ and Young's modulus of rubber is $E{\text{N/m}}^3.$ If the wire extends by $l\text{m}$ under its own weight, then extension $l$ is

• A. $L^{2}Dg/E$
• B. $L^{2}Dg/2E$
• C. $L^{2}Dg/4E$
• D. $L$

Answer 1

The correct option is B $L^{2}Dg/2E$

A small differential element $dx$ at distance $x$ from the bottom of cord
Force acting on this element $F=\frac{M}{L}\times x\times g$
If extension in this element is $dl$
Then,
$dl=\frac{Fdx}{AE}=\frac{Mg}{LAE}xdx$
${\int }_0^{l}dl=\frac{Mg}{ALE}{\int }_0^{L}xdx=\frac{MgL}{2AE}$
$l=\frac{MgL}{2AE}$
If density is D then,
$D=\frac{M}{AL}$
So, $l=\frac{Dg{L}^2}{2E}$
Hence, option B is the correct answer.