Q. The area of cross section of a steel wire $(Y = 2.0 \times 10^{11}N/m^{2})$ is $0.1cm^{2}$. The force required to double its length will be
A) $2 \times 10^{12}N$
B) $2 \times 10^{11}N$
C) $2 \times 10^{10}N$
D) $2 \times 10^{6}N$
Explanation
(d) When the length of wire is doubled then $l = L$ and strain = 1 $\therefore Y = strain = \frac{F}{A}$
$\therefore Force = Y \times A = 2 \times 10^{11} \times 0.1 \times 10^{- 4} = 2 \times 10^{6}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.