The velocity of a body moving in a vertical circle of radius r is `sqrt(7gr)` at the lowest point of the circle. What is the ratio of maximum and minimum tension?

Options

A. `4:1`
B. `sqrt7:1`
C. `3:1`
D. `2:1`

Answer

To solve the problem, we need to find the ratio of maximum and minimum tension in a body moving in a vertical circle of radius \( r \) with a given velocity at the lowest point. The velocity at the lowest point is given as \( v = \sqrt{7gr} \).

### Step-by-Step Solution:

1. **Identify the Points of Interest**:
- The lowest point (Point 2) where the tension is maximum.
- The highest point (Point 1) where the tension is minimum.

2. **Calculate the Velocities**:
- At the lowest point (Point 2), the velocity \( v_2 = \sqrt{7gr} \).
- At the highest point (Point 1), we need to find \( v_1 \) using conservation of energy.

3. **Apply Conservation of Energy**:
- The total mechanical energy at the lowest point (Point 2) is the sum of kinetic and potential energy:
\[ KE_2 + PE_2 = KE_1 + PE_1 \]
- At Point 2 (lowest point):
- Kinetic Energy \( KE_2 = \frac{1}{2} mv_2^2 = \frac{1}{2} m (7gr) = \frac{7}{2} mgr \)
- Potential Energy \( PE_2 = 0 \) (taking this point as reference).
- At Point 1 (highest point):
- Kinetic Energy \( KE_1 = \frac{1}{2} mv_1^2 \)
- Potential Energy \( PE_1 = mg(2r) = 2mgr \)

Therefore, the equation becomes:
\[ 0 + \frac{7}{2} mgr = \frac{1}{2} mv_1^2 + 2mgr \]

4. **Solve for \( v_1^2 \)**:
- Rearranging gives:
\[ \frac{7}{2} mgr - 2mgr = \frac{1}{2} mv_1^2 \]
- Simplifying:
\[ \frac{7}{2} mgr - \frac{4}{2} mgr = \frac{1}{2} mv_1^2 \]
\[ \frac{3}{2} mgr = \frac{1}{2} mv_1^2 \]
- Cancel \( m \) and multiply by 2:
\[ 3gr = v_1^2 \implies v_1 = \sqrt{3gr} \]

5. **Calculate Tension at Both Points**:
- **At the highest point (Point 1)**:
\[ T_{min} = mg + \frac{mv_1^2}{r} = mg + \frac{m(3gr)}{r} = mg + 3mg = 4mg \]
- **At the lowest point (Point 2)**:
\[ T_{max} = \frac{mv_2^2}{r} - mg = \frac{m(7gr)}{r} - mg = 7mg - mg = 6mg \]

6. **Find the Ratio of Tensions**:
\[ \text{Ratio} = \frac{T_{max}}{T_{min}} = \frac{6mg}{4mg} = \frac{6}{4} = \frac{3}{2} \]

### Final Answer:
The ratio of maximum tension to minimum tension is \( \frac{3}{2} \).