Question

A uniform cyclinder rod of length $L$, cross -sectional area $A$ and youngs modules $Y$ is acted upon by the forces shown in fig. The elongation of the rod is

• A. $\frac{3FL}{5AY}$
• B. $\frac{2FL}{5AY}$
• C. $\frac{3FL}{8AY}$
• D. $\frac{8FL}{3AY}$

Answer 1

The correct option is D $\frac{8FL}{3AY}$

$\begin{array}{c}\text{Using}\frac{F}{A}=Y\frac{\Delta l}{\ell }\\ \text{(1)}\Delta l_1=\frac{F_1{l}_1}{AY}\\ \Rightarrow \Delta l_1=\frac{3F\left(2L/3\right)}{AY}\\ \Rightarrow \Delta l_1=\frac{2FL}{AY}\end{array}$

$\begin{array}{c}\text{(2)}\Delta l_2=\frac{F_2{l}_2}{AY}\\ \Rightarrow \Delta l_2=\frac{2F\left(L/3\right)}{AY}\\ \Rightarrow \Delta l_2=\frac{2FL}{3AY}\end{array}$

$\begin{array}{cc}\text{Total elongation}& =\frac{2FL}{AY}+\frac{2FL}{3AY}=\frac{8FL}{3AY}\end{array}$