A stone of mass 2 kg is tied to a string of length 0.5 m. If the breaking tension of the string is 900 N, then the maximum angular velocity, the stone can have in uniform circular motion is

Options

A. 30 `"rad s"^(-1)`
B. `20 "rad s"^(-1)`
C. `10 "rad s"^(-1)`
D. `25 "rad s"^(-1)`

Explanation

To solve the problem, we need to determine the maximum angular velocity (\( \omega \)) at which a stone of mass 2 kg can rotate in a circular motion without breaking the string tied to it. The breaking tension of the string is given as 900 N, and the length of the string is 0.5 m.

### Step-by-Step Solution:

1. **Identify the Forces Involved**:
In uniform circular motion, the tension in the string provides the necessary centripetal force to keep the stone moving in a circle. The centripetal force (\( F_c \)) required for circular motion is given by:
\[ F_c = m \cdot a_c \]
where \( a_c \) is the centripetal acceleration.

2. **Centripetal Acceleration**:
The centripetal acceleration can be expressed in terms of angular velocity (\( \omega \)):
\[ a_c = r \cdot \omega^2 \]
where \( r \) is the radius of the circular path (length of the string).

3. **Setting Up the Equation**:
The tension in the string at the maximum angular velocity is equal to the breaking tension. Therefore, we can set up the equation:
\[ T = m \cdot r \cdot \omega^2 \]
where \( T \) is the breaking tension (900 N), \( m \) is the mass of the stone (2 kg), and \( r \) is the length of the string (0.5 m).

4. **Substituting the Values**:
Substitute the known values into the equation:
\[ 900 = 2 \cdot 0.5 \cdot \omega^2 \]

5. **Simplifying the Equation**:
Simplifying the right side:
\[ 900 = 1 \cdot \omega^2 \]
Therefore, we have:
\[ \omega^2 = 900 \]

6. **Calculating Angular Velocity**:
Taking the square root of both sides to find \( \omega \):
\[ \omega = \sqrt{900} = 30 \, \text{radians/second} \]

### Final Answer:
The maximum angular velocity (\( \omega \)) that the stone can have in uniform circular motion is **30 radians/second**.