Question

A copper rod of length $L$ and radius $r$ is suspended from the ceiling by one of its ends. What will be elongation of the rod due to its own weight when $\rho$ and$Y$ are the density and Young's modulus of the copper respectively?

• A. $\frac{{\rho }^{2}g{L}^2}{2Y}$
• B. $\frac{\rho g{L}^2}{2Y}$
• C. $\frac{{\rho }^2{g}^2{L}^2}{2Y}$
• D. $\frac{\rho gL}{2Y}$

Answer 1

Answer: (B)

$\frac{\rho g{L}^2}{2Y}$

Weight of rod W=$A\times L\times \rho \times g$
The weight increases linearly from 0 to W with increase in length.
So an average value of force is to be taken. $F_{av}$=$\frac{W}{2}$
$$F_{av}=\frac{A\times L\times \rho \times g}{2}$$
Stress=$\frac{F}{A}$
Strain=$\frac{\delta L}{L}$
Now Y=$\frac{stress}{strain}$
Therefore elongation
$$\delta L=\frac{F\times L}{A\times Y}$$
$$\delta L=\frac{A\times L\times \rho \times g\times L}{2A\times Y}$$
$$\delta L=\frac{\rho g{L}^2}{2Y}$$