Question

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

• A. $20m/s$
• B. $10\sqrt{3}m/s$
• C. $10m/s$
• D. $5\sqrt{2}m/s$

Answer 1

The correct option is C $10m/s$

When the stone is revolved in a vertical circle, tension is maximum at bottom and minimum at top. So we have, Tension at bottom as $T_B$ and Tension at top as $T_T$.

We know, $\frac{T_B}{T_T}=4$

$\Rightarrow \frac{\frac{m{v}_B^2}{L}+mg}{\frac{m{v}_T^2}{L}-mg}=4$

$\Rightarrow \frac{m{v}_B^2}{L}+mg=4(\frac{m{v}_T^2}{L}-mg)$

$\Rightarrow \frac{m{v}_B^2}{L}=\frac{4m{v}_T^2}{L}-5mg$

$\Rightarrow m{v}_B^2=4m{v}_T^2-5mgL$ ......$\left(I\right)$

Now using the law of conservation of energy, at top and bottom of the vertical circle, we have:

$\frac{1}{2}m{v}_T^2+2mgL=\frac{1}{2}m{v}_B^2$

$\Rightarrow m{v_B}^2=m{v_T}^2+4mgL$ ......$\left(II\right)$

from $\left(I\right)$ and $\left(II\right)$, we have

$4m{v}_T^2-5mgL=m{v_T}^2+4mgL\Rightarrow 3m{v}_T^2=9mgL\Rightarrow v_T=\sqrt{3gL}\Rightarrow v_T=\sqrt{3\times 10\times \left(\frac{10}{3}\right)}=10m/s$