Q. A uniform plank of Young’s modulus $Y$ is moved over a smooth horizontal surface by a constant force $F$. The area of cross section of the plank is $A$. The compressive strain on the plank in the direction of the force is
A) $F/AY$
B) $2F/AY$
C) $\frac{1}{2}(F/AY)$
D) $3F/AY$
Explanation
(a) $Y = \frac{F/A}{Strain} \Rightarrow strain = \frac{F}{AY}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.