Q. Find the extension produced in a copper of length 2 m and diameter 3 mm, when a force of 30 N is applied. Young’s modulus for copper = $1.1\ \times 10^{11}Nm^{- 2}$
A) 0.2 mm
B) 0.04 mm
C) 0.08 mm
D) 0.68 mm
Explanation
(c) $\mathrm{\Delta}l = \frac{4Fl}{\pi D^{2}Y}$
= $\frac{4 \times 30 \times 2 \times 7}{22\ \times \left( 3 \times 10^{- 3} \right)^{2} \times 1.1\ \times 10^{11}}$
= $7.7\ \times 10^{- 5}m = 0.077mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.