Q. A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2/\pi$ revolutions per $second$ around the vertical axis through the fixed end as shown in figure , then tension in the string is
A) $ML$
B) $2\ ML$
C) $4\ ML$
D) $16\ ML$
Explanation
(d) $T\sin\theta = M\omega^{2}R$ …(i)
$T\sin\theta = M\omega^{2}L\sin\theta$ …(ii)
From (i) and (ii)
$T = M\omega^{2}L$
$= M\ 4\pi^{2}n^{2}L$
$= M\ 4\pi^{2}\left( \frac{2}{\pi} \right)^{2}L = 16\ ML$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.