The velocity of a body revolving in a vertical circle of radius r at the lowest point `sqrt(7gr)`. The ratio of maximum tensions in the string is

Options

A. `8 : 1`
B. `4 : 1`
C. `sqrt(7) : 1`
D. `1 : sqrt(7)`

Answer

To solve the problem, we need to find the ratio of the maximum tensions in the string when the body is at the lowest point and the highest point of the vertical circle.

### Step-by-Step Solution:

1. **Identify the given values:**
- The velocity of the body at the lowest point, \( v = \sqrt{7gr} \).
- The radius of the vertical circle, \( r \).

2. **Calculate the tension at the lowest point (T):**
- At the lowest point, the tension in the string can be calculated using the formula:
\[ T = mg + \frac{mv^2}{r} \]
- Here, \( v^2 = 7gr \), so substituting this into the equation gives:
\[ T = mg + \frac{m(7gr)}{r} \]
- Simplifying this:
\[ T = mg + 7mg = 8mg \]

3. **Calculate the velocity at the highest point (v'):**
- Using conservation of energy between the lowest and highest points:
\[ \frac{1}{2} mv^2 = mgh + \frac{1}{2} mv'^2 \]
- The height \( h \) when the body is at the highest point is \( 2r \), so:
\[ \frac{1}{2} m(7gr) = mg(2r) + \frac{1}{2} mv'^2 \]
- Rearranging gives:
\[ \frac{1}{2} mv'^2 = \frac{1}{2} m(7gr) - 2mgr \]
- Simplifying:
\[ \frac{1}{2} mv'^2 = \frac{1}{2} m(7gr - 4gr) = \frac{3}{2} mgr \]
- Thus, \( v'^2 = 3gr \).

4. **Calculate the tension at the highest point (T'):**
- At the highest point, the tension can be calculated using:
\[ T' = \frac{mv'^2}{r} - mg \]
- Substituting \( v'^2 = 3gr \):
\[ T' = \frac{m(3gr)}{r} - mg = 3mg - mg = 2mg \]

5. **Calculate the ratio of tensions:**
- The ratio of the tension at the lowest point (T) to the tension at the highest point (T'):
\[ \text{Ratio} = \frac{T}{T'} = \frac{8mg}{2mg} = 4 \]

### Final Answer:
The ratio of the maximum tensions in the string is \( 4:1 \).

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