Question

A stone of mass $1\text{kg}$ tied to a light inextensible string of length $L=\frac{10}{3}\text{m}$ 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 is $4$ and if $g$ is taken to be $10{\text{m/s}}^2$, the speed of the stone at the highest point of the circle is

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

Answer 1

The correct option is D $10\text{m/s}$

In case of circular motion in a vertical plane, the tension is maximum at the lowest point $A$ but minimum at the highest point $C$.
For circular motion at lowest point A,
$T_A-Mg=\frac{M{V}_A^2}{L}$
$\Rightarrow T_A=Mg+\frac{M{V}_A^2}{L}=T_{max}$

For circular motion at the highest point C,
$T_c+Mg=\frac{M{V}_C^2}{L}$
$\Rightarrow T_c=\frac{M{V}_C^2}{L}-Mg=T_{min}$

Given that
$\Rightarrow \frac{T_{max}}{T_{min}}=\frac{4}{1}=\frac{\frac{M{V}_A^2}{L}+Mg}{\frac{M{V}_C^2}{L}-Mg}\Rightarrow \frac{V_A^2+gL}{V_C^2-gL}=\frac{4}{1}$

But from work energy theorem, $V_A^2=V_C^2+2g\left(2L\right)=V_C^2+4gL$
$\Rightarrow \frac{V_C^2+5gL}{V_C^2-gL}=\frac{4}{1}\Rightarrow 3V_C^2=9gL$
$\Rightarrow V_C^2=3gL=3\times 10\times \frac{10}{3}=100$
$\therefore V_C=10\text{m/s}$
Hence, the correct answer is option (d).