A car is moving on a plane inclined at `30^(@)` to the horizontal with an acceleration of `10m//s^(2)` parallel to the plane upward. A bob is suspended by a string from the roof. The angle in degrees which the string makes with the vertical is: (Assume that the bob does not move relative to car) `[g=10m//s^(2)]`

Options

A. `20^@`
B. `30^@`
C. `45^@`
D. `60^@`

Explanation

To solve the problem, we need to analyze the forces acting on the bob suspended from the roof of the car. The car is moving on an inclined plane, and we have to find the angle the string makes with the vertical.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Bob:**
- The forces acting on the bob are:
- The tension (T) in the string acting along the string.
- The gravitational force (mg) acting vertically downward.
- The pseudo force (ma) acting horizontally opposite to the direction of acceleration of the car.

2. **Set Up the Coordinate System:**
- Since the car is inclined at an angle of \( \theta = 30^\circ \) to the horizontal, we can resolve the forces into components along the inclined plane and perpendicular to it.
- Let \( \alpha \) be the angle the string makes with the vertical.

3. **Resolve Forces:**
- The tension can be resolved into two components:
- \( T \cos \alpha \) acting vertically upward.
- \( T \sin \alpha \) acting horizontally along the incline.
- The gravitational force can be resolved into two components as well:
- The component of gravity acting parallel to the incline: \( mg \sin \theta \).
- The component of gravity acting perpendicular to the incline: \( mg \cos \theta \).

4. **Apply Newton's Second Law:**
- In the direction parallel to the incline (horizontal):
\[ T \sin \alpha = ma \cos \theta \]
- In the direction perpendicular to the incline (vertical):
\[ T \cos \alpha = mg + ma \sin \theta \]

5. **Substitute Known Values:**
- Given:
- \( a = 10 \, \text{m/s}^2 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( \theta = 30^\circ \)
- Substitute \( \theta \) into the equations:
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 30^\circ = \frac{1}{2} \)

6. **Solve for \( \tan \alpha \):**
- From the equations:
\[ \tan \alpha = \frac{a \cos \theta}{a \sin \theta + g} \]
- Substitute the values:
\[ \tan \alpha = \frac{10 \cdot \frac{\sqrt{3}}{2}}{10 \cdot \frac{1}{2} + 10} = \frac{5\sqrt{3}}{5 + 10} = \frac{5\sqrt{3}}{15} = \frac{\sqrt{3}}{3} \]
- Thus, \( \tan \alpha = \frac{1}{\sqrt{3}} \).

7. **Find the Angle \( \alpha \):**
- Since \( \tan \alpha = \frac{1}{\sqrt{3}} \), we know that:
\[ \alpha = 30^\circ \]

### Final Answer:
The angle which the string makes with the vertical is \( 30^\circ \).