A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of inclination `30^(@)` with horizontal. Acceleration of train up the plane is `a=g//2` . The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is.

Options

A. `30^(@)`
B. `tan^(-1)(2//(sqrt(3))`
C. `tan^(-1)(sqrt(3)//2)`
D. `tan^(-1)(2)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the plumb bob and use the equilibrium conditions to find the angle the string makes with the normal to the ceiling of the train compartment.

### Step 1: Understand the setup
- The train is moving up an inclined plane with an inclination of \(30^\circ\) and has an acceleration \(a = \frac{g}{2}\).
- The plumb bob is hanging from the ceiling of the train compartment, which is also inclined at \(30^\circ\).

### Step 2: Identify the forces acting on the bob
In the frame of the train compartment:
1. The weight of the bob acts downward: \(W = mg\).
2. The tension \(T\) in the string acts along the string.
3. A pseudo force acts on the bob due to the acceleration of the train, directed down the incline: \(F_{pseudo} = ma = m\left(\frac{g}{2}\right)\).

### Step 3: Resolve forces into components
- The weight of the bob can be resolved into two components:
- Perpendicular to the incline: \(W_{\perp} = mg \cos(30^\circ)\)
- Parallel to the incline: \(W_{\parallel} = mg \sin(30^\circ)\)

- The pseudo force can also be resolved into components:
- Perpendicular to the incline: \(F_{pseudo \perp} = \frac{mg}{2} \cos(30^\circ)\)
- Parallel to the incline: \(F_{pseudo \parallel} = \frac{mg}{2} \sin(30^\circ)\)

### Step 4: Write the equilibrium equations
In equilibrium, the sum of forces in both the perpendicular and parallel directions must be zero.

1. **Vertical (perpendicular to incline) equilibrium**:
\[ T \cos(\theta) = mg \cos(30^\circ) + \frac{mg}{2} \cos(30^\circ) \]
Simplifying gives:
\[ T \cos(\theta) = \frac{3mg}{2} \cos(30^\circ) \]

2. **Horizontal (parallel to incline) equilibrium**:
\[ T \sin(\theta) = mg \sin(30^\circ) + \frac{mg}{2} \sin(30^\circ) \]
Simplifying gives:
\[ T \sin(\theta) = \frac{3mg}{2} \sin(30^\circ) \]

### Step 5: Divide the equations to eliminate T
Dividing the vertical equation by the horizontal equation:
\[ \frac{T \cos(\theta)}{T \sin(\theta)} = \frac{\frac{3mg}{2} \cos(30^\circ)}{\frac{3mg}{2} \sin(30^\circ)} \]
This simplifies to:
\[ \cot(\theta) = \frac{\cos(30^\circ)}{\sin(30^\circ)} \]
Using the values:
\[ \cot(\theta) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]
Thus:
\[ \theta = 30^\circ \]

### Step 6: Find the angle with respect to the normal
The angle that the string makes with the normal to the ceiling is:
\[ \phi = 90^\circ - \theta = 90^\circ - 30^\circ = 60^\circ \]

### Final Answer
The angle which the string supporting the bob makes with the normal to the ceiling in equilibrium is \(60^\circ\).

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