Q. 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) $\frac{\rho^{2}{gL}^{2}}{2Y}$
B) $\frac{\rho{gL}^{2}}{2Y}$
C) $\frac{\rho^{2}g^{2}L^{2}}{2Y}$
D) $\frac{\rho gL}{2Y}$
Explanation
(d) The weight of the rod can be assumed to act at its mid-point.
Now, the mass of the rod is
$M = V\rho$
⟹ $M = AL\rho$
Here, $A = area\ of\ cross - sections,$
L= length of the rod.
Now, we know that the Young’s modulus
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$Y = \frac{\frac{MgL}{2}}{A \bullet l\ }\ \ \ \ (Here,\ L = \frac{L}{2},l = extension)$
⟹ $l = \frac{\frac{MgL}{2}}{AY}$
or $l = \frac{MgL}{2AY}$
On putting the value of M from Eq.(i), we get
$l = \frac{AL\rho \bullet gL}{2AY}$
or $l = \frac{\rho gL^{2}}{2Y}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.