D

To solve the problem, we can follow these steps:
Step 1: Consider the forces acting on the particle. When the particle is at an angle \( \theta \) with the vertical, the tension \( T \) in the string and the weight \( mg \) act on it. The acceleration of the particle is inclined to the string at \( 45^\circ \). This means we have to set up the equations of motion accordingly.
Step 2: Resolve the tension \( T \) into components:
- Vertical component: \( T \cos(\theta) \)
- Horizontal component: \( T \sin(\theta) \)
Step 3: The net vertical force is: \( T \cos(\theta) = mg \)
The net horizontal force, due to acceleration \( a \), is: \( T \sin(\theta) = ma \)
Step 4: Given that the acceleration of the particle is inclined to the string by \( 45^\circ \), we can write \( a = g \tan(\theta) \).
Step 5: Substitute this back: \( T \sin(\theta) = mg \tan(\theta) \).
Step 6: From the vertical forces: \( T = \frac{mg}{\cos(\theta)} \)
Step 7: Set the expressions for tension from both equations equal to each other: \( \frac{mg}{\cos(\theta)} \sin(\theta) = mg \tan(\theta) \).
Canceling \( mg \) and simplifying leads us to find that \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).
Step 8: Analyzing at \( 45^\circ \): The angle \( \theta \) when the tension's horizontal component equals the acceleration's component gives \( \theta + 45^\circ = 90^\circ \), thus \( \theta = 45^\circ \). Therefore the angle made by the string with the vertical is finally \( 45^\circ \). Hence, the answer is option D.