For a particle rotating in a vertical circle with uniform speed, the maximum and minimum tension in the string are in the ratio 5 : 3. If the radius of vertical circle is 2 m, the speed of revolving body is (`g=10m//s^(2)`)

Options

A. `sqrt(5)m//s`
B. `4sqrt(5)m//s`
C. 5 m/s
D. 10 m/s

Answer

To solve the problem, we need to analyze the forces acting on a particle rotating in a vertical circle and use the given ratio of maximum and minimum tension in the string.

### Step-by-Step Solution:

1. **Identify the Forces**:
- At the top of the vertical circle (point A), the forces acting on the particle are:
- Tension (T_A) acting downwards
- Weight (mg) acting downwards
- Centripetal force (F_c) required to keep the particle in circular motion, which is provided by the tension and weight.
- At the bottom of the vertical circle (point B), the forces are:
- Tension (T_B) acting upwards
- Weight (mg) acting downwards
- Centripetal force (F_c) is provided by the tension minus the weight.

2. **Write the Equations for Tension**:
- At the top (point A):
\[ T_A + mg = F_c \quad \text{(1)} \]
- At the bottom (point B):
\[ T_B - mg = F_c \quad \text{(2)} \]

3. **Express Centripetal Force**:
- Centripetal force is given by:
\[ F_c = \frac{mv^2}{r} \]
- Where \( v \) is the speed of the particle and \( r \) is the radius of the circle.

4. **Substituting Centripetal Force into Tension Equations**:
- From equation (1):
\[ T_A = F_c - mg = \frac{mv^2}{r} - mg \]
- From equation (2):
\[ T_B = F_c + mg = \frac{mv^2}{r} + mg \]

5. **Set Up the Ratio of Tensions**:
- Given the ratio of maximum tension (T_B) to minimum tension (T_A) is \( \frac{5}{3} \):
\[ \frac{T_B}{T_A} = \frac{\frac{mv^2}{r} + mg}{\frac{mv^2}{r} - mg} = \frac{5}{3} \]

6. **Cross Multiply to Solve for v**:
- Cross multiplying gives:
\[ 3\left(\frac{mv^2}{r} + mg\right) = 5\left(\frac{mv^2}{r} - mg\right) \]
- Expanding both sides:
\[ 3\frac{mv^2}{r} + 3mg = 5\frac{mv^2}{r} - 5mg \]
- Rearranging terms:
\[ 3mg + 5mg = 5\frac{mv^2}{r} - 3\frac{mv^2}{r} \]
\[ 8mg = 2\frac{mv^2}{r} \]

7. **Simplifying the Equation**:
- Cancel \( m \) from both sides:
\[ 8g = 2\frac{v^2}{r} \]
- Rearranging gives:
\[ v^2 = 4gr \]

8. **Substituting Values**:
- Given \( g = 10 \, \text{m/s}^2 \) and \( r = 2 \, \text{m} \):
\[ v^2 = 4 \times 10 \times 2 = 80 \]
- Therefore, taking the square root:
\[ v = \sqrt{80} = 4\sqrt{5} \, \text{m/s} \]

### Final Answer:
The speed of the revolving body is \( 4\sqrt{5} \, \text{m/s} \).