Q. Consider an iron rod of length 1 m and cross-section 1 $cm^{2}$ with a Young’s modulus of $10^{12}$ dyne $cm^{- 2}$. We wish to calculate the force with which the two ends must be pulled to produce an elongation of 1mm. It is equal to
A) $10^{9}$ dyne
B) $10^{8}$ dyne
C) $10^{6}$ dyne
D) $10^{17}$ dyne
Explanation
(a) $L = 1\ m = 100\ cm$
$A = 1\ cm^{2}$
$Y = 10^{12}\ dyne\ cm^{- 2}$
$\ l = 1 \times 10^{- 1}\ cm$
Force, $F = \frac{AYl}{L} = \frac{1 \times 10^{12} \times 10^{- 1}}{100}$
$\ = 10^{9}\ dyne$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.