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 $ ρ $ and $ Y $ are the density and Young’s modulus of the copper respectively?

• A. $ \frac{\rho^{2}gL^{2}}{2Y} $
• B. $ \frac{\rho gL^{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 $l=\frac{\rho gL^{2}}{2Y}$