Question

A stone of mass $1kg$ tied to a light inextensible string of length $L=\frac{5}{3}m$, is whirling in a vertical circle of radius $L$. If the ratio of the maximum and minimum tension in the string is $4$ and $g=10m/s^2,$ the speed of stone at the highest point of the circle is

• A. $5m/s$
• B. $5\sqrt{3}m/s$
• C. $5\sqrt{2}m/s$
• D. $10m/s$

Answer 1

Answer: (C)

In the vertical circular motion, the tension in the string is maximum at $A$ and minimum at $B$.

$T_A=\frac{m{v}_A^2}{L}+mg=T_{\max }$
$T_B=\frac{m{v}_B^2}{L}-mg=T_{\min }$
$\frac{T_{\max }}{T_{\min }}=4=\frac{\frac{m{v}_A^2}{L}+mg}{\frac{m{v}_B^2}{L}-mg}$
or $\frac{v_A^2+gL}{v_B^2-gL}=\frac{4}{1}\dots$(i)
But $v_A^2=v_B^2+2g(2L)=v_B^2+4gL$
(from the law of conservation of energy)
From (i) $\frac{v_B^2+5gL}{v_B^2-gL}=\frac{4}{1}$
On solving, we get
$v_B=\sqrt{3gL}=\sqrt{3\times 10\times \frac{5}{3}}$
$=\sqrt{50}=5\sqrt{2}m/s$