Question

Uniform rod of mass $m$, length $l$, area of cross-section $A$ has Young's modulus $Y$. If it is hanged vertically, elongation under its own weight will be :

• A. $\frac{mgl}{2AY}$
• B. $\frac{2mgl}{AY}$
• C. $\frac{mgl}{AY}$
• D. $\frac{mgY}{Al}$

Answer 1

Answer: (C)

$\frac{mgl}{AY}$

$Y=\frac{Fl}{A\Delta l}\Rightarrow \Delta l=\frac{Fl}{YA}=\frac{mgl}{YA}$ (in this question whole mass is considered on centre of mass and hence for calculating max displacement, we are calculating he displacement of centre of mass only)