Q. If the shear modulus of a wire material is 5.9$\ \times 10^{11}\ dyne\ cm^{- 2}$ then the potential energy of a wire of $4 \times 10^{3}\ cm$ in diameter and 5 cm long twisted through an angle of 10’ , is
A) $1.253\ \times 10^{- 12}\ J$
B) $2.00\ \times 10^{- 12}\ J$
C) $1.00\ \times 10^{- 12}\ J$
D) $0.8\ \times 10^{- 12}\ J$
Explanation
(a) To twist the wire through the angle $d\theta,\ $it is necessary to do the work
$dW = \ \tau d\theta$
And $\theta = 10^{'} = \frac{10}{60}\ \times \frac{\pi}{180} = \frac{\pi}{1080}\ rad$
$W = \ \int_{0}^{\theta}{\tau\ d\theta = \ \ \int_{0}^{\theta}{\frac{\eta\pi r^{4}\theta d\theta}{2l} = \ \frac{\eta\pi r^{4}\theta}{4l}}}$
$W = \ \frac{5.9\ \times 10^{11} \times 10^{- 5} \times \ \pi\left( 2 \times 10^{- 5} \right)^{4}\pi^{2}\ }{10^{- 4} \times 4 \times 5 \times 10^{- 2} \times (1080)^{2}}$
$W = 1.253\ \times 10^{- 12}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.