A stone is rotated in a vertical circle. Speed at bottommost point is `sqrt(8gR)`, where R is the radius of circle. The ratio of tension at the top and the bottom is

Options

A. `1:2`
B. `1:3`
C. `2:3`
D. `1:4`

Explanation

To solve the problem, we need to find the ratio of the tension in the string at the topmost point (T_B) to the tension at the bottommost point (T_A) of the vertical circle in which the stone is rotating.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Speed at the bottommost point (A): \( v_A = \sqrt{8gR} \)
- Radius of the circle: \( R \)

2. **Calculate Tension at the Bottommost Point (T_A):**
- At the bottommost point, the forces acting on the stone are:
- Tension \( T_A \) acting upwards
- Weight \( mg \) acting downwards
- Centripetal force requirement: \( \frac{mv_A^2}{R} \) acting downwards
- The equation of motion at point A can be written as:
\[ T_A - mg = \frac{mv_A^2}{R} \]
- Substitute \( v_A \):
\[ T_A - mg = \frac{m(\sqrt{8gR})^2}{R} \]
\[ T_A - mg = \frac{m \cdot 8gR}{R} \]
\[ T_A - mg = 8mg \]
- Therefore, we can solve for \( T_A \):
\[ T_A = mg + 8mg = 9mg \]

3. **Calculate Speed at the Topmost Point (B):**
- To find the speed at the topmost point (B), we can use the conservation of mechanical energy between points A and B.
- At point A (bottom):
- Kinetic Energy (KE_A) = \( \frac{1}{2} m v_A^2 = \frac{1}{2} m (8gR) = 4mgR \)
- Potential Energy (PE_A) = 0 (taking this as the reference level)
- At point B (top):
- Kinetic Energy (KE_B) = \( \frac{1}{2} m u^2 \)
- Potential Energy (PE_B) = \( mg(2R) = 2mgR \)
- By conservation of energy:
\[ KE_A + PE_A = KE_B + PE_B \]
\[ 4mgR + 0 = \frac{1}{2} m u^2 + 2mgR \]
- Rearranging gives:
\[ 4mgR = \frac{1}{2} m u^2 + 2mgR \]
\[ 2mgR = \frac{1}{2} m u^2 \]
\[ 4gR = u^2 \]

4. **Calculate Tension at the Topmost Point (T_B):**
- At the topmost point (B), the forces acting on the stone are:
- Tension \( T_B \) acting downwards
- Weight \( mg \) acting downwards
- Centripetal force requirement: \( \frac{mu^2}{R} \) acting downwards
- The equation of motion at point B can be written as:
\[ T_B + mg = \frac{mu^2}{R} \]
- Substitute \( u^2 \):
\[ T_B + mg = \frac{m(4gR)}{R} \]
\[ T_B + mg = 4mg \]
- Therefore, we can solve for \( T_B \):
\[ T_B = 4mg - mg = 3mg \]

5. **Calculate the Ratio of Tensions:**
- Now, we can find the ratio of tensions:
\[ \frac{T_B}{T_A} = \frac{3mg}{9mg} = \frac{1}{3} \]

### Final Answer:
The ratio of tension at the topmost point to the tension at the bottommost point is \( \frac{1}{3} \).