A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N . What is the maximum speed with which the ball can be moved

Options

A. `14 m//s`
B. `3m//s`
C. `3.92 m//s`
D. `5m//s`

Answer

To solve the problem, we need to find the maximum speed with which the ball can be moved without breaking the string. The tension in the string provides the centripetal force required to keep the ball moving in a circular path.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the ball, \( m = 0.25 \, \text{kg} \)
- Length of the string (radius of the circle), \( R = 1.96 \, \text{m} \)
- Maximum tension in the string, \( T_{\text{max}} = 25 \, \text{N} \)

2. **Understand the relationship between tension and centripetal force:**
The tension in the string provides the centripetal force required to keep the ball moving in a circular path. The formula for centripetal force is given by:
\[ F_c = \frac{m v^2}{R} \]
where \( v \) is the speed of the ball.

3. **Set the maximum tension equal to the centripetal force:**
Since the maximum tension is 25 N, we set it equal to the centripetal force:
\[ T_{\text{max}} = \frac{m v^2}{R} \]
Substituting the known values:
\[ 25 = \frac{0.25 v^2}{1.96} \]

4. **Rearrange the equation to solve for \( v^2 \):**
Multiply both sides by \( 1.96 \):
\[ 25 \times 1.96 = 0.25 v^2 \]
\[ 49 = 0.25 v^2 \]

5. **Isolate \( v^2 \):**
Divide both sides by \( 0.25 \):
\[ v^2 = \frac{49}{0.25} \]
\[ v^2 = 196 \]

6. **Take the square root to find \( v \):**
\[ v = \sqrt{196} = 14 \, \text{m/s} \]

### Conclusion:
The maximum speed with which the ball can be moved without breaking the string is \( 14 \, \text{m/s} \).