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)

The weight of the rod can be assumed to act at its mid-point
Now, the mass of the rod is
$M=V_{\rho }$
$\Rightarrow M=AL\rho \dots \left(i\right)$
Here, A = area of cross-section
L = length of the rod
Now, we know that the Young’s modulus
$Y=\frac{MgL}{\frac{2}{A\cdot l}}$ [Here, $L=\frac{L}{2},l$ =extension]
$\Rightarrow l=\frac{2}{AY}$
or $l=\frac{Mgl}{2AY}$
On putting the value of $M$ from Eq.$(i)$,we get
$l=\frac{AL\rho \cdot gL}{2AY}$
or $l=\frac{\rho g{L}^2}{2Y}$