Q. A chain of 125 links is 1.25 m long and has mass of 2 kg with the ends fastened together. It is set for rotating at 50 $revs^{- 1}$, centripetal force on each links is
A) 3.14 N
B) 0.314 N
C) 314 N
D) None of these
Explanation
(c) $F = m{r\omega}^{2}$
$= \frac{2}{125} \times \frac{1.25}{2\pi} \times (100\pi)^{2}\ \ \ (\therefore\omega = 2\pi f = 2 \times 50 \times \pi)$
$= 100\pi N = 314N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.