A string breaks if its tension exceeds 10 newtons . A stone of mass 250 gm tied to this string of length 10 cm is rotated in a horizontal circle. The maximum angular velocity of rotation can be

Options

A. `20 rad//s`
B. `40 rad//s`
C. `100 rad//s`
D. `200 rad//s`

Answer

To find the maximum angular velocity of the stone tied to the string before the string breaks, we can follow these steps:

### Step 1: Identify the given values
- Mass of the stone, \( m = 250 \, \text{g} = 0.25 \, \text{kg} \)
- Maximum tension in the string, \( T_{\text{max}} = 10 \, \text{N} \)
- Length of the string (which is the radius of the circular motion), \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Step 2: Write the formula for centripetal force
The centripetal force required to keep an object moving in a circle is given by the equation:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the linear velocity.

### Step 3: Relate linear velocity to angular velocity
The linear velocity \( v \) can be expressed in terms of angular velocity \( \omega \) as:
\[ v = r \omega \]
Substituting this into the centripetal force equation gives:
\[ F_c = \frac{m(r \omega)^2}{r} = m r \omega^2 \]

### Step 4: Set the centripetal force equal to the maximum tension
Since the maximum tension in the string provides the maximum centripetal force, we can set:
\[ T_{\text{max}} = m r \omega^2 \]
Substituting the known values:
\[ 10 \, \text{N} = 0.25 \, \text{kg} \times 0.1 \, \text{m} \times \omega^2 \]

### Step 5: Solve for \( \omega^2 \)
Rearranging the equation gives:
\[ \omega^2 = \frac{10 \, \text{N}}{0.25 \, \text{kg} \times 0.1 \, \text{m}} = \frac{10}{0.025} = 400 \]

### Step 6: Calculate \( \omega \)
Taking the square root of both sides:
\[ \omega = \sqrt{400} = 20 \, \text{rad/s} \]

### Conclusion
The maximum angular velocity of rotation before the string breaks is:
\[ \omega = 20 \, \text{rad/s} \]