Q. A particle has velocity $\sqrt{3rg}$ at the highest pint in vertical circle. Find the ratio of tensions at the highest and lowest point
A) 1 : 6
B) 1 : 4
C) 1 : 3
D) 1 : 2
Explanation
(b) $T_{top} = \frac{mv^{2}}{r} - mg = 2mg$
$\frac{T_{top}}{T_{bottom}} = \frac{2mg}{2mg + 6mg} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.