Q. Two bars $A\ and\ B$ of circular cross-section and of same volume and made of the same material are subjected to tension. If the diameter of A is half that of B and if the force applied to both the rods is the same and it is in the elastic limit, the ratio of extension of A to that of B will be
A) 16
B) 8
C) 4
D) 7
Explanation
(a) $Y = \frac{FV}{A^{2}\mathrm{\Delta}l}\ $
$\mathrm{\Delta}l\ \propto \frac{1}{A^{2}}\ \ or\ \ \mathrm{\Delta}l\ \propto \frac{1}{D^{4}}$
$\therefore\ \frac{\mathrm{\Delta}l_{A}}{\mathrm{\Delta}l_{B}} = \frac{D_{B}^{4}}{D_{A}^{4}} = \ \frac{1^{4}}{\left( \frac{1}{2} \right)^{4}} = 16$
$Y = \frac{F}{A}\ \times \frac{l}{\mathrm{\Delta}l}\ $
Now, $V = Al\ \ or\ l\ \frac{V}{A}\ \therefore\ \ Y = \frac{FV}{A^{2}\mathrm{\Delta}l}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.