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 then its velocity is
A) $\sqrt{98}\ ms^{- 1}$
B) $7\ ms^{- 1}$
C) $\sqrt{490}\ ms^{- 1}$
D) $\sqrt{4.9}\ ms^{- 1}$
Explanation
(a) In this problem it is assumed that particle although moving in a vertical loop but it speed remain constant
Tension at lowest point $T_{\max} = \frac{mv^{2}}{r} + m$g
Tension at highest point $T_{\min} = \frac{mv^{2}}{r} - mg$
$\frac{T_{\max}}{T_{\min}} = \frac{\frac{mv^{2}}{r} + mg}{\frac{mv^{2}}{r} - mg} = \frac{5}{3}$
$\therefore\ v = \sqrt{4gr} = \sqrt{4 \times 9.8 \times 2.5} = \sqrt{98}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.