The maximum and minimum tension in the string whirling in a circle of radius 2.5 m with constant velocity are in the ratio `5:3` then its velocity is

Options

A. `sqrt(98)ms^(-1)`
B. `7 ms^(-1)`
C. `sqrt490 ms^(-1)`
D. `sqrt4.9 ms^(-1)`

Answer

To solve the problem, we need to find the velocity of an object whirling in a vertical circle with a given radius and a specific ratio of maximum to minimum tension in the string.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- When the object is at the **top** of the circle, the forces acting on it are:
- Tension (T) acting downwards.
- Weight (mg) acting downwards.
- Centripetal force required for circular motion (mv²/r) acting upwards.
- The equation for the top position is:
\[ T + mg = \frac{mv^2}{r} \quad \text{(1)} \]

- When the object is at the **bottom** of the circle, the forces acting on it are:
- Tension (T) acting upwards.
- Weight (mg) acting downwards.
- Centripetal force required for circular motion (mv²/r) acting upwards.
- The equation for the bottom position is:
\[ T = mg + \frac{mv^2}{r} \quad \text{(2)} \]

2. **Expressing Tension**:
- From equation (1):
\[ T_{\text{min}} = \frac{mv^2}{r} - mg \]
- From equation (2):
\[ T_{\text{max}} = mg + \frac{mv^2}{r} \]

3. **Setting Up the Ratio**:
- We are given that the ratio of maximum tension to minimum tension is \( \frac{T_{\text{max}}}{T_{\text{min}}} = \frac{5}{3} \).
- Substituting the expressions for \( T_{\text{max}} \) and \( T_{\text{min}} \):
\[ \frac{mg + \frac{mv^2}{r}}{\frac{mv^2}{r} - mg} = \frac{5}{3} \]

4. **Cross-Multiplying**:
- Cross-multiplying gives us:
\[ 3\left(mg + \frac{mv^2}{r}\right) = 5\left(\frac{mv^2}{r} - mg\right) \]

5. **Simplifying the Equation**:
- Expanding both sides:
\[ 3mg + \frac{3mv^2}{r} = \frac{5mv^2}{r} - 5mg \]
- Rearranging terms:
\[ 3mg + 5mg = \frac{5mv^2}{r} - \frac{3mv^2}{r} \]
\[ 8mg = \frac{2mv^2}{r} \]

6. **Cancelling m and Rearranging**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 8g = \frac{2v^2}{r} \]
- Rearranging gives:
\[ v^2 = 4gr \]

7. **Substituting the Values**:
- Given \( r = 2.5 \, \text{m} \) and using \( g = 10 \, \text{m/s}^2 \):
\[ v^2 = 4 \times 10 \times 2.5 \]
\[ v^2 = 100 \]

8. **Finding the Velocity**:
- Taking the square root:
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answer:
The velocity of the object is \( 10 \, \text{m/s} \).