Q. A stone of mass $1\ kg$ tied to a light inextensible string of length $L = \frac{10}{3}m$ is whirling in a circular path of radius $L$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if $g$ is taken to be $10m/\sec^{2}\ ,$ the speed of the stone at the highest point of the circle is
A) $20m/sec$
B) $10\sqrt{3}m/sec$
C) $5\sqrt{2}m/sec$
D) $10m/sec$
Explanation
(d) Since the maximum tension $T_{B}$ in the string moving in the vertical circle is at the bottom and minimum tension $T_{T}$ is at the top
$\therefore T_{B} = \frac{mv_{B}^{2}}{L} + mg$ and $T_{T} = \frac{mv_{T}^{2}}{L} - mg$
$\therefore\frac{T_{B}}{T_{T}} = \frac{\frac{mv_{B}^{2}}{L} + mg}{\frac{mv_{T}^{2}}{L} - mg} = \frac{4}{1}\ or\ \frac{v_{B}^{2} + gL}{v_{T}^{2} - gL} = \frac{4}{1}$
Or $v_{B}^{2} + gL = 4v_{T}^{2} - 4gL$ but $v_{B}^{2} = v_{T}^{2} + 4gL$
$\therefore v_{T}^{2} + 4gL + gL = 4v_{T}^{2} - 4gL \Rightarrow 3v_{T}^{2} = 9gL$
$\therefore v_{T}^{2} = 3 \times g \times L = 3 \times 10 \times \frac{10}{3}\ or\ v_{T} = 10\ m/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.