Q. The upper end of a wire 1 m log and 2 mm radius is clamped. The lower end is twisted through and angle of $45{^\circ}.$ The angle of shear is
A) $0.09{^\circ}$
B) 0.9${^\circ}$
C) 9${^\circ}$
D) 90${^\circ}$
Explanation
(a) $strain = \frac{r}{l}\phi\ \frac{2\ \times 10^{- 3}}{1} \times 45{^\circ} = 0.9$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.