A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be , (Take `g = 10m/s^2`)

Options

A. `pi/6`
B. `pi/2`
C. `pi/4`
D. `pi/3`

Answer

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 in a circular path, and the bob will make an angle θ with the vertical due to the centripetal force required to keep it moving in a circle.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The forces acting on the bob are:
- The tension (T) in the string, which can be resolved into two components:
- T cos(θ) acting vertically upward (balancing the weight of the bob).
- T sin(θ) acting horizontally (providing the centripetal force).
- The weight of the bob (mg) acting downward.

2. **Write the Equations**:
- For vertical equilibrium (since the bob is not moving vertically):
\[ T \cos(θ) = mg \quad \text{(1)} \]
- For horizontal motion (providing the centripetal force):
\[ T \sin(θ) = \frac{mv^2}{R} \quad \text{(2)} \]

3. **Divide the Equations**:
- Dividing equation (2) by equation (1):
\[ \frac{T \sin(θ)}{T \cos(θ)} = \frac{\frac{mv^2}{R}}{mg} \]
- This simplifies to:
\[ \tan(θ) = \frac{v^2}{Rg} \quad \text{(3)} \]

4. **Substitute the Values**:
- Given:
- \( v = 20 \, \text{m/s} \)
- \( R = 40 \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)
- Substitute these values into equation (3):
\[ \tan(θ) = \frac{(20)^2}{40 \cdot 10} = \frac{400}{400} = 1 \]

5. **Find the Angle**:
- Now, we find θ:
\[ θ = \tan^{-1}(1) = \frac{\pi}{4} \, \text{radians} \]

### Final Answer:
The angle made by the string with the vertical is \( \frac{\pi}{4} \) radians.

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