Q. The maximum and minimum tension in the string whirling in a circle of radius $2.5\ m$ with constant velocity are in the ratio $5:3$ the the velocity is
A) $\sqrt{98}\ m/s$
B) $7\ m/s$
C) $\sqrt{490}\ m/s$
D) $\sqrt{4.9}$
Explanation
(a) In this problem it is assumed that particle although moving in a vertical loop but its speed remain constant
Tension at lowest point $T_{\max} = \frac{mv^{2}}{r} + mg$
Tension at highest point $T_{\min} = \frac{mv^{2}}{r} - mg$
$\frac{T_{\max}}{T_{\min}} = \frac{\frac{mv^{2}}{r} + mg}{\frac{mv^{2}}{r} - mg} = \frac{5}{3}$
By solving we get, $v = \sqrt{4gr} = \sqrt{4 \times 9.8 \times 2.5} = \sqrt{98}\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.