Q. A bob of mass 10 $kg\ $ is attached to wire 0.3 $m$ long. Its breaking stress is $4.8 \times 10^{7}N/m^{2}$. The area of cross section of the wire is $10^{- 6}m^{2}$. The maximum angular velocity with which it can be rotated in a horizontal circle
A) 8 $rad/sec$
B) 4$\ rad/sec$
C) 2$\ rad/sec$
D) 1$\ rad/sec$
Explanation
(b) Centripetal force = breaking force
$\Rightarrow m\omega^{2}r = breaking\ stress\ \times cross\ sectional\ area$
$\Rightarrow m\omega^{2}r = p \times A\ \Rightarrow \omega = \ \sqrt{\frac{p \times A}{mr}} = \ \sqrt{\frac{4.8 \times 10^{7} \times 10^{- 6}}{10 \times 0.3}}$
$\therefore\ \omega = 4\ rad/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.