Q. A copper wire of negligible mass, 1 m length and cross-sectional area $10^{- 6}$ is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rad $s^{- 1}.\ $If the elongation in the wire is $10^{- 3}m$, then the Young’s modulus is
A) $4\ \times 10^{11}Nm^{- 2}$
B) $6\ \times 10^{11}Nm^{- 2}$
C) $8\ \times 10^{11}Nm^{- 2}$
D) $10\ \times 10^{11}Nm^{- 2}$
Explanation
(a) $Y = \frac{Fl}{A\mathrm{\Delta}l} = \ \frac{\left( ml\omega^{2} \right)l}{A\ \mathrm{\Delta}l}\ or\ Y = \frac{ml^{2}\omega^{2}}{A\mathrm{\Delta}l}$
Or $Y = \ \frac{1 \times 1 \times 1 \times 20 \times 20}{10^{- 6} \times 10^{- 3}} = 4\ \times 10^{11}\ Nm^{- 2}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.