A car is moving with constant speed of 10m/s on a horizontal circular path of radius `10sqrt(3m)` . A bob of mass m suspended through a light string from the roof of the car. What is the angle made by the string with the vertical if the bob is stationary with respect to the car? (g = 10 m/`s^(2)`)

Options

A. `tan^(-1) (sqrt(3))`
B. `tan^(-1) ((1)/(sqrt(2)))`
C. `30^(@)`
D. `45^(@)`

Explanation

To solve the problem, we need to analyze the forces acting on the bob suspended from the roof of the car as it moves in a circular path. The bob is stationary with respect to the car, which means we can consider the forces in a non-inertial frame of reference (the car).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Bob:**
- The weight of the bob, \( mg \), acts vertically downward.
- A pseudo force \( F_p \) acts horizontally outward due to the circular motion of the car. This pseudo force is given by \( F_p = \frac{mv^2}{r} \), where \( v \) is the speed of the car and \( r \) is the radius of the circular path.

2. **Given Values:**
- Speed of the car, \( v = 10 \, \text{m/s} \)
- Radius of the circular path, \( r = 10\sqrt{3} \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

3. **Calculate the Pseudo Force:**
\[ F_p = \frac{mv^2}{r} = \frac{m(10)^2}{10\sqrt{3}} = \frac{100m}{10\sqrt{3}} = \frac{10m}{\sqrt{3}} \]

4. **Draw the Free Body Diagram (FBD):**
- The weight \( mg \) acts downwards.
- The pseudo force \( F_p \) acts horizontally outward.
- The string makes an angle \( \theta \) with the vertical.

5. **Set Up the Equation Using Tan Function:**
- From the FBD, we can relate the forces using the tangent of the angle \( \theta \):
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{F_p}{mg} \]
\[ \tan \theta = \frac{\frac{10m}{\sqrt{3}}}{mg} = \frac{10}{g\sqrt{3}} = \frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}} \]

6. **Find the Angle \( \theta \):**
- We know that \( \tan \theta = \frac{1}{\sqrt{3}} \).
- The angle \( \theta \) corresponding to \( \tan \theta = \frac{1}{\sqrt{3}} \) is \( 30^\circ \).

### Final Answer:
The angle made by the string with the vertical is \( \theta = 30^\circ \).

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