Q. A tensile force of $2 \times 10^{3}\ N$ doubles the length of a rubber band of cross-sectional area $2 \times 10^{- 4}m^{2}.$ The Young’s modulus of elasticity of the rubber band is
A) $4\ \times 10^{7}Nm^{- 2}$
B) $2\ \times 10^{2}\ Nm^{- 2}$
C) $10^{7}\ Nm^{- 2}$
D) $0.5\ \times 10^{7}\ Nm^{- 2}$
Explanation
(c) Strain = $\frac{\mathrm{\Delta}l}{l} = \frac{l}{l} = 1$
$\therefore\ Y = stress = \ \frac{2\ \times 10^{3}N}{2 \times 10^{- 4}m^{2}} = 10^{7}Nm^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.