Q. A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing modulus of cube is
A) $2 \times 10^{9}{Nm}^{- 2}$
B) $4 \times 10^{9}{Nm}^{- 2}$
C) $8 \times 10^{9}{Nm}^{- 2}$
D) $16 \times 10^{9}{Nm}^{- 2}$
Explanation
(a) Shearing modulus of cube
$ɳ = \frac{FL}{Al}$
$= \frac{8 \times 10^{3} \times 40 \times 10^{- 3}}{{(40 \times 10}^{- 3})^{2} \times (0.1 \times 10^{- 3})} = 2 \times 10^{9}{Nm}^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.