A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ($$\theta$$) of thread deflection in the extreme position will be :

At the extreme position, the velocity is zero. The acceleration is purely tangential (gravity component): $$a_{ext} = g\sin\theta$$.

At the lowest position, the acceleration is centripetal: $$a_{low} = \frac{v^2}{L}$$.

Using energy conservation from extreme to lowest position:

$$\frac{1}{2}mv^2 = mgL(1 - \cos\theta)$$

$$v^2 = 2gL(1 - \cos\theta)$$

So $$a_{low} = \frac{2gL(1 - \cos\theta)}{L} = 2g(1 - \cos\theta)$$.

Setting $$a_{ext} = a_{low}$$:

$$g\sin\theta = 2g(1 - \cos\theta)$$

$$\sin\theta = 2(1 - \cos\theta)$$

Using half-angle: $$2\sin(\theta/2)\cos(\theta/2) = 2 \cdot 2\sin^2(\theta/2)$$

$$\cos(\theta/2) = 2\sin(\theta/2)$$

$$\tan(\theta/2) = \frac{1}{2}$$

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

The answer corresponds to Option (2).