Question

Two stones of masses \(m\) and \(3m\) are whirled in horizontal circles, the heavier one in a radius \(\frac{r}{3}\) and the lighter one in radius \(r\). When both the stones experience same centripetal forces, the tangential speed of lighter stone is '\(n\)' times that of the value of heavier stone. The value of '\(n\)' is

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Equate m v^2 / r for both stones and solve for the speed ratio.

Answer 1

The Correct Option is C

Step 1: Ratio Method:
Equal forces give $\dfrac{m_1v_1^2}{r_1}=\dfrac{m_2v_2^2}{r_2}$, so $\left(\dfrac{v_1}{v_2}\right)^2=\dfrac{m_2}{m_1}\cdot\dfrac{r_1}{r_2}$.

Step 2: Numbers:
$\dfrac{m_2}{m_1}=3$ and $\dfrac{r_1}{r_2}=\dfrac{r}{r/3}=3$. So $\left(\dfrac{v_1}{v_2}\right)^2=9$.

Step 3: Answer:
$\dfrac{v_1}{v_2}=3$, option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } 3} \]