A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of a . What will be the angle of inclination with vertical

Options

A. `tan^(-1)(a//g)`
B. `tan^(-1)(g//a)`
C. `cos^(-1)(a//g)`
D. `cos^(-1)(g//a)`

Answer

To find the angle of inclination of a plumb line suspended from a ceiling of a car moving with a horizontal acceleration \( A \), we can analyze the forces acting on the plumb line.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the plumb line acts vertically downward with a force \( W = mg \), where \( m \) is the mass of the plumb line and \( g \) is the acceleration due to gravity.
- The tension \( T \) in the string acts along the string, directed towards the point of suspension.

2. **Establish the Coordinate System**:
- Let the angle of inclination of the plumb line with the vertical be \( \theta \).
- The horizontal acceleration \( A \) of the car will create an inertial force acting on the plumb line in the opposite direction (to the left if the car accelerates to the right).

3. **Draw a Free Body Diagram**:
- The vertical component of the tension \( T \) must balance the weight:
\[ T \cos \theta = mg \]
- The horizontal component of the tension must balance the inertial force due to the car's acceleration:
\[ T \sin \theta = ma \]

4. **Divide the Equations**:
- Dividing the second equation by the first gives:
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{ma}{mg} \]
- This simplifies to:
\[ \tan \theta = \frac{a}{g} \]

5. **Find the Angle**:
- To find the angle \( \theta \), take the arctangent of both sides:
\[ \theta = \tan^{-1}\left(\frac{a}{g}\right) \]

### Final Answer:
The angle of inclination of the plumb line with the vertical is given by:
\[ \theta = \tan^{-1}\left(\frac{a}{g}\right) \]