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 θ with the vertical

The correct option is θ = cos⁻¹(2/3).

Let us analyze the motion of the simple pendulum step-by-step to find the angle at which the string breaks.

Step 1: Understand the given values and parameters
Let:
- The mass of the bob be $m$.
- The length of the string be $L$.
- The acceleration due to gravity be $g$.
- The weight of the bob be $W=mg$.
- The breaking strength (maximum tension capacity) of the string is given as double the weight of the bob, which can be written as:

$T_{\text{break}}=2mg$

Step 2: Apply the Principle of Conservation of Mechanical Energy
The bob is released from rest when the string is horizontal. At this initial position, the angle with the vertical is 90°.
When the string makes an angle $\theta$ with the vertical, the vertical distance $h$ descended by the bob from its initial horizontal level is:

$h=L\cos \theta$

By conservation of mechanical energy, the loss in gravitational potential energy as the bob falls is equal to the gain in its kinetic energy:

$mgh=\frac{1}{2}m{v}^2$

Substituting $h=L\cos \theta$ into the energy equation:

$mgL\cos \theta =\frac{1}{2}m{v}^2$

Simplifying to find $v^2$:

$v^2=2gL\cos \theta$

Step 3: Analyze the forces acting on the bob in the radial direction
At the angle $\theta$ with the vertical, the forces acting on the bob along the line of the string (radial direction) are:
1. Tension $T$ pulling inwards towards the center of rotation (support point).
2. The component of gravity, $mg\cos \theta$, acting radially outwards away from the support.

The net force directed towards the center provides the centripetal force required for circular motion:

$T-mg\cos \theta =\frac{m{v}^2}{L}$

Step 4: Express tension T as a function of θ
We substitute the value of $v^2$ obtained from the conservation of energy in Step 2 into the centripetal force equation:

$T-mg\cos \theta =\frac{m\left(2gL\cos \theta \right)}{L}$

$T-mg\cos \theta =2mg\cos \theta$

Rearranging the equation to solve for tension $T$:

$T=3mg\cos \theta$

Step 5: Calculate the angle θ at which the string breaks
The string breaks when the tension in the string equals its breaking strength $T_{\text{break}}=2mg$:

$3mg\cos \theta =2mg$

Dividing both sides of the equation by $mg$:

$3\cos \theta =2$

$\cos \theta =\frac{2}{3}$

Taking the inverse cosine:

$\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$

Thus, the string breaks when the angle it makes with the vertical is $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$.