Q. An object of mass 10 kg is whirled round a horizontal circle of radius 4 m by a revolving string inclined 30${^\circ}\ $ to the vertical. If the uniform speed of the object is 5 $ms^{- 1}$, the tension in the string (approximately) is
A) 720 N
B) 960 N
C) 114 N
D) 125 N
Explanation
(d) In figure, sin$30{^\circ} = \frac{AB}{OA}$
[image:4_image87.jpeg]
or $OA = \frac{AB}{\sin{30{^\circ}}} = \frac{4}{1/2} = 8$m
$\frac{T}{AO} = \frac{F}{AB} = \frac{mg}{OB}$
$F = \frac{AO}{AB} \times F = \frac{AO}{AB\ }\frac{mv^{2}}{r} = \frac{8}{4} \times 10 \times \frac{5^{2}}{4} \approx 125\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.