1. Interpret the figure and condition

A bob of mass $M$ is attached to a string of length $L$. At the lowest point $A$, it is given a horizontal speed $V$.

“Just sufficient to make it reach point $B$” means the bob barely reaches the topmost point $B$ with zero speed.

So, using conservation of mechanical energy between bottom $A$ and top $B$:

$\frac12 MV^2 = Mg(2L)$

Hence,

$V^2 = 4gL$

2. Speed at an intermediate angle $\theta$

Let $\theta$ be the angle made by the string with the downward vertical at some intermediate position.

At this angle, the bob has risen by height

$h = L(1-\cos\theta)$

Applying conservation of energy between $A$ and this position:

$\frac12 MV^2 = \frac12 Mv^2 + MgL(1-\cos\theta)$

We are given that at angle $\theta$, the speed becomes half of that at $A$:

$v = \frac{V}{2}$

Substitute into energy equation:

$\frac12 MV^2 = \frac12 M\left(\frac{V}{2}\right)^2 + MgL(1-\cos\theta)$

$\frac12 MV^2 = \frac18 MV^2 + MgL(1-\cos\theta)$

$MgL(1-\cos\theta) = \frac38 MV^2$

Now use $V^2 = 4gL$:

$MgL(1-\cos\theta) = \frac38 M(4gL)$

$MgL(1-\cos\theta) = \frac32 MgL$

So,

$1-\cos\theta = \frac32$

$-\cos\theta = \frac12$

$\cos\theta = -\frac12$

Thus,

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

3. Check the interval

$\frac{2\pi}{3} = 120^\circ$

This lies in the interval

$\frac{\pi}{2} < \theta < \frac{3\pi}{4}$

since

$90^\circ < 120^\circ < 135^\circ$

Therefore, the correct option is:

$\boxed{\text{C}}$

4. Comparison with stored answer

Stored correct answer: $\text{D}$

But our derivation gives $\theta = \dfrac{2\pi}{3}$, which belongs to option $\text{C}$, not $\text{D}$.

So the stored answer appears to be incorrect.