A small ball is suspended by a string from the ceiling of a car. As the car accelerates at a rate ‘a’ the string makes an angle '/theta' with the vertical in equilibrium. Then the tension in the string is :

Options

A. `m sqrt(a^(2) + g^(2))`
B. `m a sqrt(a^(2) + g^(2))`
C. `m g sin theta`
D. `m g cos theta`

Explanation

To find the tension in the string when a small ball is suspended from the ceiling of a car that is accelerating at a rate 'a', and the string makes an angle θ with the vertical, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Ball**:
- The ball experiences two main forces:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The tension in the string acting at an angle θ from the vertical: \( T \)

2. **Consider the Pseudo Force**:
- Since the car is accelerating, we can consider a pseudo force acting on the ball in the opposite direction of the car's acceleration. This pseudo force is given by:
\[ F_{pseudo} = ma \]

3. **Draw the Free Body Diagram**:
- In the frame of reference of the car, the ball is in equilibrium. The tension \( T \) can be resolved into two components:
- A vertical component: \( T \cos \theta \)
- A horizontal component: \( T \sin \theta \)

4. **Set Up the Equilibrium Conditions**:
- Since the ball is in equilibrium, the sum of forces in both the vertical and horizontal directions must be zero.
- In the vertical direction:
\[ T \cos \theta = mg \] (1)
- In the horizontal direction:
\[ T \sin \theta = ma \] (2)

5. **Divide Equation (2) by Equation (1)**:
- To eliminate \( T \), we can divide the horizontal force equation by the vertical force equation:
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{ma}{mg} \]
- This simplifies to:
\[ \tan \theta = \frac{a}{g} \]

6. **Express Tension in Terms of a and g**:
- From equation (1), we can express \( T \):
\[ T = \frac{mg}{\cos \theta} \tag{3} \]
- To find \( T \) in terms of \( a \) and \( g \), we can use the relationship derived from the tangent:
\[ \cos \theta = \frac{g}{\sqrt{a^2 + g^2}} \quad \text{(using the identity } \sin^2 \theta + \cos^2 \theta = 1\text{)} \]
- Substituting this into equation (3):
\[ T = \frac{mg}{\frac{g}{\sqrt{a^2 + g^2}}} = mg \cdot \frac{\sqrt{a^2 + g^2}}{g} = m \sqrt{a^2 + g^2} \]

7. **Final Result**:
- Therefore, the tension in the string is:
\[ T = m \sqrt{a^2 + g^2} \]