A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is `(g = 10 m//s^(2))`

Options

A. zero
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Explanation

To solve the problem, we need to analyze the forces acting on the plumb bob suspended from the roof of the car that is moving in a circular path. Here are the steps to find the angle made by the rod with the vertical:

### Step 1: Identify the forces acting on the plumb bob
The two main forces acting on the plumb bob are:
1. The gravitational force (weight) acting downwards, \( F_g = mg \)
2. The centripetal force required to keep the car moving in a circular path, which causes the bob to deflect from the vertical.

### Step 2: Calculate the centripetal acceleration
The centripetal acceleration \( a_c \) can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
Where:
- \( v = 10 \, \text{m/s} \) (the speed of the car)
- \( r = 10 \, \text{m} \) (the radius of the circular track)

Substituting the values:
\[ a_c = \frac{(10 \, \text{m/s})^2}{10 \, \text{m}} = \frac{100 \, \text{m}^2/\text{s}^2}{10 \, \text{m}} = 10 \, \text{m/s}^2 \]

### Step 3: Relate the forces to find the angle
The angle \( \theta \) made by the rod with the vertical can be found using the tangent function:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{a_c}{g} \]
Where:
- \( a_c = 10 \, \text{m/s}^2 \) (centripetal acceleration)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ \tan \theta = \frac{10 \, \text{m/s}^2}{10 \, \text{m/s}^2} = 1 \]

### Step 4: Calculate the angle \( \theta \)
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Conclusion
The angle made by the rod with the vertical is \( 45^\circ \).