Question

A rod of length $L$ kept on a smooth horizontal surface is pulled along its length by a force $F$. The area of cross-sectional is $A$ and Young's modulus is $Y$. The extension in the rod is

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

Answer 1

Answer: (A)

$\frac{FL}{AY}$

Answer 2

Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$. Rearranging for the extension $\Delta L$, we get $\Delta L = \frac{FL}{AY}$.