Question

A stone is tied at the end of a rope of length 1 m and whirled in a vertical circle. The ratio of velocity at the highest point to the lowest point will be

• A. \(1:\sqrt{5}\)
• B. \(\sqrt{3}:1\)
• C. \(\sqrt{3}:\sqrt{5}\)
• D. \(\sqrt{5}:1\)

Hint:

In vertical circular motion, use energy conservation between top and bottom points.

Answer 1

The Correct Option is A

Step 1: Apply conservation of energy.
Let velocity at the top be \(v_t\) and at the bottom be \(v_b\). The height difference between top and bottom is \(2R\).
Step 2: Energy relation.
\[ \frac{1}{2} m v_b^2 = \frac{1}{2} m v_t^2 + mg(2R) \]
Step 3: Minimum condition at top.
For a taut string at the top:
\[ v_t^2 = gR \]
Step 4: Substitute and solve.
\[ v_b^2 = gR + 4gR = 5gR \]
Step 5: Ratio of velocities.
\[ v_t : v_b = \sqrt{gR} : \sqrt{5gR} = 1:\sqrt{5} \]
Step 6: Conclusion.
The required ratio is \(1:\sqrt{5}\).