Question

A stone of mass \(1\text{ kg}\) tied to a light inextensible string of length \(L = \frac{5}{3}\text{ m}\) is rotating in a circular path of radius \(L\) in a vertical plane. If the ratio of maximum tension in the string to the minimum tension in the string is 3 , the speed of the stone at the highest point of the circle is ( \(g =\) acceleration due to gravity)

• A. \(\sqrt{gL}\)
• B. \(\sqrt{2gL}\)
• C. \(\sqrt{4gL}\)
• D. \(\sqrt{8gL}\)

Hint:

Use energy + tension formulas together for vertical circular motion.

Answer 1

The Correct Option is A

Concept:
In vertical circular motion:

• Maximum tension occurs at lowest point
• Minimum tension occurs at highest point

\[ T_{\text{bottom}} = \frac{mv_b^2}{L} + mg \] \[ T_{\text{top}} = \frac{mv_t^2}{L} - mg \] Step 1: Given ratio. \[ \frac{T_{\text{max}}}{T_{\text{min}}} = 3 \Rightarrow \frac{T_{\text{bottom}}}{T_{\text{top}}} = 3 \]
Step 2: Apply energy conservation. Between top and bottom: \[ \frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + 2mgL \] \[ v_b^2 = v_t^2 + 4gL \]
Step 3: Substitute in tension ratio. \[ \frac{\frac{m(v_t^2 + 4gL)}{L} + mg}{\frac{mv_t^2}{L} - mg} = 3 \] Cancel \(m\): \[ \frac{\frac{v_t^2}{L} + 4g + g}{\frac{v_t^2}{L} - g} = 3 \] \[ \frac{\frac{v_t^2}{L} + 5g}{\frac{v_t^2}{L} - g} = 3 \]
Step 4: Solve equation. Let \(x = \frac{v_t^2}{L}\) \[ \frac{x + 5g}{x - g} = 3 \] \[ x + 5g = 3x - 3g \] \[ 2x = 8g \Rightarrow x = 4g \] \[ \frac{v_t^2}{L} = g \Rightarrow v_t^2 = gL \]
Step 5: Conclusion. \[ v_t = \sqrt{gL} \]