Question

A uniform cylindrical rod of length $L.$ cross-sectional area $A$ and Young's modulus $Y$ is acted is acted upon by the forces as shown in the figure. The elongation in the rod, is

• A. $\frac{4FL}{AY}$
• B. $\frac{15FL}{4AY}$
• C. $\frac{15FL}{2AY}$
• D. $\frac{FL}{3AY}$

Answer 1

The correct option is B $\frac{15FL}{4AY}$

$\begin{array}{c}\text{By equilibrium.}\\ \text{(because net force is o on rod)}\\ T^{\prime }=4F\text{.}\end{array}$

$\begin{array}{cc}\Delta l_1& =\text{elongation for}\frac{L}{4}\text{length}\\ & =\frac{Fl}{AY}=\frac{3Fd}{AY}\times \frac{L}{4}=\frac{3}{4}\frac{Fl}{AY}\end{array}$

$\begin{array}{cc}\Delta l_2& =\text{elongation for}34a\text{length}\\ & =\frac{T^{\prime }{l}^{\prime }}{AY}=\frac{4F}{AY}\times \frac{3L}{4}=\frac{3FL}{AY}\end{array}$

$\text{Total elongation}=\Delta l_1+\Delta l_2=\frac{15}{4}\frac{FL}{AY}$