Question

A uniform rod of length L, cross-sectional area A and Young's Y is stretched by applying two force $F_1$ and $F_2$ as shown in figure. The elongation in the rod will be $(F_1>F_2)$

• A. $\frac{(F_1+F_2)L}{AY}$
• B. $\frac{(F_1-F_2)L}{2AY}$
• C. $\frac{(F_1+F_2)L}{2AY}$
• D. $\frac{F_{1}L}{2AY}$

Answer 1

The correct option is C $\frac{(F_1+F_2)L}{2AY}$

$\begin{array}{c}\text{Abplying}NLM-2\text{law}\\ \begin{array}{cc}{F}_1-T& =\left(PAx\right).a\\ T-F_2& =\left(PA\right(L-x\left)\right]a\end{array}\end{array}$

$\begin{array}{cc}& \frac{F_1-T}{T-F_2}=\frac{x}{L-x}\\ T& =F_1-\frac{x}{L}(F_1-F_2)\end{array}$

$\text{By Hookes law,}\Delta l=\frac{Fl}{AY}$

$\text{Elongation}\left(\Delta l\right)={\int }_0^{\infty }\frac{Tdx}{Ay}={\int }_0^l\frac{[F_1-\frac{x}{L}\left(F_1{f}_2\right)]}{Ay}dx=\frac{(F_1+F_2)L}{2AY}$