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 of finding the ratio of tension at the top and bottom of a stone rotated in a vertical circle, we can follow these steps:

### Step 1: Identify the given parameters
- The speed at the bottommost point is given as \( v = \sqrt{8gR} \), where \( R \) is the radius of the circle and \( g \) is the acceleration due to gravity.

### Step 2: Analyze forces at the bottommost point
At the bottommost point, the forces acting on the stone are:
- Weight of the stone (\( mg \)) acting downwards.
- Tension in the string (\( T_b \)) acting upwards.
- Centripetal force requirement, which is provided by the net force acting towards the center.

Using Newton's second law, we can write the equation for the bottommost point:
\[ T_b - mg = \frac{mv^2}{R} \]
Substituting \( v = \sqrt{8gR} \):
\[ T_b - mg = \frac{m(\sqrt{8gR})^2}{R} \]
\[ T_b - mg = \frac{m(8gR)}{R} \]
\[ T_b - mg = 8mg \]
\[ T_b = 9mg \]

### Step 3: Analyze forces at the topmost point
At the topmost point, the forces acting on the stone are:
- Weight of the stone (\( mg \)) acting downwards.
- Tension in the string (\( T_t \)) acting downwards.
- Centripetal force requirement.

Using Newton's second law, we can write the equation for the topmost point:
\[ T_t + mg = \frac{mv'^2}{R} \]
Where \( v' \) is the speed at the topmost point. We need to find \( v' \) using the conservation of energy.

### Step 4: Apply conservation of energy
The total mechanical energy at the bottommost point must equal the total mechanical energy at the topmost point:
\[ \frac{1}{2}mv^2 + mg(0) = \frac{1}{2}mv'^2 + mg(2R) \]
Substituting \( v = \sqrt{8gR} \):
\[ \frac{1}{2}m(8gR) = \frac{1}{2}mv'^2 + mg(2R) \]
Cancelling \( m \) from both sides:
\[ 4gR = \frac{1}{2}v'^2 + 2gR \]
Rearranging gives:
\[ 4gR - 2gR = \frac{1}{2}v'^2 \]
\[ 2gR = \frac{1}{2}v'^2 \]
Multiplying both sides by 2:
\[ 4gR = v'^2 \]
Thus, \( v' = \sqrt{4gR} \).

### Step 5: Substitute \( v' \) into the tension equation at the topmost point
Now substituting \( v' \) into the tension equation at the topmost point:
\[ T_t + mg = \frac{m(\sqrt{4gR})^2}{R} \]
\[ T_t + mg = \frac{m(4gR)}{R} \]
\[ T_t + mg = 4mg \]
\[ T_t = 4mg - mg \]
\[ T_t = 3mg \]

### Step 6: Find the ratio of tensions
Now we can find the ratio of tension at the topmost point to that at the bottommost point:
\[ \text{Ratio} = \frac{T_t}{T_b} = \frac{3mg}{9mg} = \frac{3}{9} = \frac{1}{3} \]

### Final Answer
The ratio of tension at the top and bottom is \( \frac{1}{3} \).

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