Q125 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Physics
Consider a particle is moving with a minimum speed $v$ at highest point of vertical circle of radius $R$. If the radius of the circle doubled the corresponding minimum speed will be
- A$V$
- B$\frac{v}{\sqrt{2}}$
- C$\sqrt{3} v$
- Correct answerD$\sqrt{2} v$
Solution
When a particle moves along a vertical circle, then it has its minimum speed at the highest point of the vertical circle and in this case it just
completes the loop. The minimum speed is
$\begin{aligned} v_{\min } & =\sqrt{g R} \Rightarrow v_{\min } \propto \sqrt{R} \\ \text { So, } \quad & \frac{\left(v_{\min }\right)_1}{\left(v_{\min }\right)_2}=\sqrt{\frac{R_1}{R_2}} \Rightarrow \frac{v}{\left(v_{\min }\right)_2}=\sqrt{\frac{R}{2 R}} \\ \Rightarrow \quad & \frac{v}{\left(v_{\min }\right)_2}=\frac{1}{\sqrt{2}} \Rightarrow\left(v_{\min }\right)_2=\sqrt{2} v\end{aligned}$
Hence, the correct answer is option: 'D'.