Question

A uniform rod of mass m, length L, area of cross-section A and Youngs modulus Y hangs from a rigid support. Its elongation under its own weight will be:

• A. zero
• B. mgL/2YA
• C. mgL/YA
• D. 2mgL/YA.

Answer 1

The correct option is B mgL/2YA

$\begin{array}{c}M=\text{Mass}\\ A=\text{Area of cross Sechon}\\ L=\text{Length of Rod}\end{array}$

$Y=\text{Yourgls modulus}$

$\text{mass of}\mathrm{leng}thL=M$

$\text{Mass of length}x=\left(\frac{M}{L}\right)x$

$\begin{array}{c}\text{Tension at point}B=\left(\frac{M}{L}x\right)g\\ Y=\frac{\text{Tension/Area}}{\frac{\Delta L}{L}}=\text{strain}\end{array}$

$\begin{array}{c}\therefore Y=\frac{\frac{I}{A}}{\frac{dc}{dx}}(dc=\begin{array}{c}\text{incressed}\\ \text{length}\end{array})\\ dc=\int \frac{T}{y_A}dx\end{array}$

$\begin{array}{c}\text{integrating botu sides}\\ \Delta L={\int }_0^L\frac{\left(\frac{M}{L}x\right)g}{Y_A}dx\end{array}$

$\Delta L=\frac{Mg}{AYL}{\int }_0^{L}xd\therefore \Delta L=\frac{MgL}{2YA}$

Hence option B is correct