In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle `theta` with the vertical.

Options

A. `theta=cos^(-1)``(1)/(3)`
B. `theta=60^(@)`
C. `theta=cos^(-1)``(2)/(3)`
D. `theta=0`

Explanation

To solve the problem, we will analyze the forces acting on the bob of the pendulum when it is at an angle θ with the vertical. We will use the principles of mechanics to derive the required angle.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Bob:**
When the bob is at an angle θ with the vertical, the forces acting on it are:
- The gravitational force (weight) acting downwards: \( W = mg \)
- The tension in the string (T) acting along the string towards the pivot.

2. **Resolve the Weight into Components:**
The weight can be resolved into two components:
- A component along the direction of the tension: \( W_{\parallel} = mg \cos \theta \)
- A component perpendicular to the tension: \( W_{\perpendicular} = mg \sin \theta \)

3. **Apply Newton's Second Law:**
The net force acting towards the center of the circular path (centripetal force) is provided by the tension in the string minus the component of weight acting along the string:
\[ T - mg \cos \theta = \frac{mv^2}{r} \]
where \( v \) is the speed of the bob at angle θ and \( r \) is the length of the string.

4. **Breaking Strength of the String:**
According to the problem, the breaking strength of the string is double the weight of the bob:
\[ T = 2mg \]

5. **Substituting the Tension into the Force Equation:**
Substitute \( T = 2mg \) into the equation from step 3:
\[ 2mg - mg \cos \theta = \frac{mv^2}{r} \]
Simplifying this gives:
\[ mg(2 - \cos \theta) = \frac{mv^2}{r} \]

6. **Finding the Velocity at Angle θ:**
The bob is released from rest when the string is horizontal. The height (h) from which it falls to angle θ is given by:
\[ h = r(1 - \cos \theta) \]
Using conservation of energy, the potential energy at the top is converted to kinetic energy at angle θ:
\[ mg h = \frac{1}{2} mv^2 \]
Substituting for h:
\[ mg(r(1 - \cos \theta)) = \frac{1}{2} mv^2 \]
Simplifying gives:
\[ v^2 = 2g r(1 - \cos \theta) \]

7. **Substituting Velocity into the Force Equation:**
Substitute \( v^2 \) back into the centripetal force equation:
\[ mg(2 - \cos \theta) = \frac{m(2gr(1 - \cos \theta))}{r} \]
Simplifying gives:
\[ mg(2 - \cos \theta) = 2mg(1 - \cos \theta) \]
Dividing by \( mg \) (assuming \( m \neq 0 \)):
\[ 2 - \cos \theta = 2(1 - \cos \theta) \]

8. **Solving for cos θ:**
Expanding and rearranging gives:
\[ 2 - \cos \theta = 2 - 2\cos \theta \]
Rearranging leads to:
\[ \cos \theta = \frac{2}{3} \]

9. **Finding the Angle θ:**
Finally, we find θ:
\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]

### Conclusion:
The angle θ at which the string breaks is:
\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]