A pendulum is hanging from the ceiling of a railway carriage making an angle of `30^(@)` with the vertical direction . The acceleration of the carriage is `:`

Options

A. `( g )/( sqrt( 3))`
B. `( 2g )/( sqrt( 3))`
C. ` g sqrt( 3)`
D. `( sqrt( 3)g)/( 2)`

Answer

To solve the problem, we will analyze the forces acting on the pendulum bob and apply the concepts of equilibrium in a non-inertial frame (the accelerating carriage).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Pendulum Bob:**
- The forces acting on the pendulum bob are:
- The gravitational force (weight) acting downward, \( mg \).
- The tension \( T \) in the string acting along the string.
- A pseudo-force \( F_{pseudo} \) acting horizontally in the opposite direction of the acceleration of the carriage, which has a magnitude of \( ma \), where \( a \) is the acceleration of the carriage.

2. **Set Up the Coordinate System:**
- Let’s consider the vertical direction as downward and the horizontal direction as the direction of the carriage's acceleration. The pendulum makes an angle of \( 30^\circ \) with the vertical.

3. **Resolve the Tension into Components:**
- The tension \( T \) can be resolved into two components:
- Vertical component: \( T \cos(30^\circ) \)
- Horizontal component: \( T \sin(30^\circ) \)

4. **Apply Equilibrium Conditions:**
- In the vertical direction, the forces must balance:
\[ T \cos(30^\circ) = mg \quad \text{(1)} \]
- In the horizontal direction, the net force is equal to the pseudo-force:
\[ T \sin(30^\circ) = ma \quad \text{(2)} \]

5. **Divide Equation (2) by Equation (1):**
- By dividing the two equations, we eliminate \( T \):
\[ \frac{T \sin(30^\circ)}{T \cos(30^\circ)} = \frac{ma}{mg} \]
- This simplifies to:
\[ \tan(30^\circ) = \frac{a}{g} \]

6. **Solve for the Acceleration \( a \):**
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{a}{g} \]
- Rearranging gives:
\[ a = g \cdot \frac{1}{\sqrt{3}} = \frac{g}{\sqrt{3}} \]

### Final Answer:
The acceleration of the carriage is:
\[ a = \frac{g}{\sqrt{3}} \]