Q. A cube of aluminium of sides 0.1 m is subjected to a sharing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be
A) 0.02
B) 0.1
C) 0.005
D) 0.002
Explanation
(d) Shearing strain = $\frac{0.02\ \times 10^{- 2}}{0.1\ } = 0.002$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.