Question

A copper rod of length \[l\] is suspended from the ceiling by one of its ends. Find the elongation \[\Delta l\] of the rod due to its own weight.
A. \[\Delta l = \dfrac{1}{2}\dfrac{{\rho g{l^2}}}{E}\]
B. \[\Delta l = \dfrac{1}{3}\dfrac{{\rho g{l^2}}}{E}\]
C. \[\Delta l = \dfrac{1}{4}\dfrac{{\rho g{l^2}}}{E}\]
D. \[\Delta l = \dfrac{1}{5}\dfrac{{\rho g{l^2}}}{E}\]

Answer 1

Hint:We are given then the length of the rod. And as we know the relation of the elongation, force, area, and Young’s modulus . This relation is also known as Young’s relation. Assume the quantities which are not given. These assumed quantities will further get eliminated.

Complete step-by-step solution:
We know that weight is the product of mass and gravitational length.
Let W be weight:
\[W = v\rho g\]
\[v = Area \times length\]
So \[v = AL\rho g\]
Now \[{F_{av}}\] be the average force acting on the rod.
As density of rod is uniform thus \[{F_{av}}\] is force which is at the middle of the rod
Thus \[{F_{av}} = \dfrac{W}{2}\]
\[{F_{av}} = \dfrac{{AL\rho g}}{2}\]
Now young’s relation is:
\[Y = \dfrac{{stress}}{{stain}} = \dfrac{{\Delta l}}{l}\]
Or \[\Delta l = \dfrac{{FL}}{{AY}}\]
Putting the value of force in it:
\[\Delta l = \dfrac{{AL\rho gL}}{{2AY}}\]
\[\Delta l = \dfrac{{\rho g{L^2}}}{{2Y}}\]
Thus, option A is correct

Note:- In solid mechanics, Young’s modulus is defined as the ratio of the longitudinal stress over longitudinal strain, in the range of elasticity the Hooke's law holds (stress is directly proportional to strain). It is a measure of stiffness of elastic material.
A solid material will undergo elastic deformation of \[\Delta l\] when a small load is applied to it in compression or extension. Elastic deformation is reversible (the material returns to its original shape after the load is removed).