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

Options

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

Explanation

To find the maximum speed with which the ball can be moved without breaking the string, we can follow these steps:

### Step 1: Understand the forces acting on the ball
When the ball is moving in a horizontal circle, the only force providing the necessary centripetal force is the tension in the string. The tension (T) in the string must equal the centripetal force required to keep the ball moving in a circle.

### Step 2: Write the formula for centripetal force
The centripetal force (Fc) required to keep an object of mass \( m \) moving at speed \( v \) in a circle of radius \( r \) is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m = 0.25 \, \text{kg} \) (mass of the ball)
- \( r = 1.96 \, \text{m} \) (length of the string)

### Step 3: Set the tension equal to the centripetal force
Since the tension in the string must not exceed \( 25 \, \text{N} \), we can set up the inequality:
\[ T \leq 25 \, \text{N} \]
Thus, we have:
\[ \frac{mv^2}{r} \leq 25 \]

### Step 4: Substitute the known values into the inequality
Substituting \( m = 0.25 \, \text{kg} \) and \( r = 1.96 \, \text{m} \) into the equation gives:
\[ \frac{0.25v^2}{1.96} \leq 25 \]

### Step 5: Solve for \( v^2 \)
To isolate \( v^2 \), multiply both sides by \( 1.96 \):
\[ 0.25v^2 \leq 25 \times 1.96 \]
Calculating the right side:
\[ 0.25v^2 \leq 49 \]
Now, divide both sides by \( 0.25 \):
\[ v^2 \leq \frac{49}{0.25} \]
Calculating the division:
\[ v^2 \leq 196 \]

### Step 6: Take the square root to find \( v \)
Taking the square root of both sides gives:
\[ v \leq \sqrt{196} \]
Calculating the square root:
\[ v \leq 14 \, \text{m/s} \]

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