Q. Two particles of equal mass are connected to a rope $AB$ of negligible mass such that one is at end $A$ and other dividing the length of rope in the ratio $1\ :2$ from $B$. The rope is rotated about end $B$ in a horizontal plane. Ratio of tensions in the smaller part to the other is (ignore effect of gravity)
A) $4\ :3$
B) $1\ :4$
C) $1\ :2$
D) $1\ :3$
Explanation
(a) Tensions in the respective parts are shown in figure
[image:4_image168.emf]
Let $\omega$ be angular velocity, then
$T_{1} - T_{2} = m\omega^{2} \times r\ $ …(i)
and $T_{2} = m\omega^{2}(r + 2r)$
$T_{2} = 3m\omega^{2}r$ …(ii)
From equation (i) and (ii)
$T_{1} = 4m\omega^{2}r \Rightarrow \frac{T_{1}}{T_{2}} = \frac{4}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.