Question

A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity $\sqrt{3gr}$ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is

• A. 4 : 3
• B. 5 : 4
• C. 1 : 4
• D. 3 : 2
• E. 1 : 2

Answer 1

Answer: (C)

1 : 4

Tension at highest point
$T_H=\frac{m{v}^2}{r}-mg=3mg-mg=2mg$
and tension at lowest point
$T_L=\frac{m{v}^2}{r}+mg$
Here, $V_L^2=3gr+2g.2r=7gr$
So, $T_L=7mg+mg=8mg$
Hence, $\frac{T_H}{T_L}=\frac{2mg}{8mg}=\frac{1}{4}$