A car is moving along a circular track of radius `10sqrt(3)`m with a constant speed of 36kmph. A plumb bob is suspended from the roof of the car by a light rigid rod of length Im. The angle made by the rod with the track is (`g= 10ms^(-2)` )

Options

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

Answer

To solve the problem step by step, we will analyze the forces acting on the plumb bob suspended from the car and find the angle made by the rod with the track.

### Step 1: Understand the scenario
A car is moving in a circular path with a radius of \( r = 10\sqrt{3} \) m at a constant speed of \( v = 36 \) km/h. A plumb bob is suspended from the roof of the car by a light rigid rod of length \( L = 1 \) m. We need to find the angle \( \theta \) made by the rod with the vertical.

### Step 2: Convert speed from km/h to m/s
To work with SI units, we need to convert the speed from kilometers per hour to meters per second.
\[ v = 36 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 10 \text{ m/s} \]

### Step 3: Identify the forces acting on the bob
The forces acting on the bob are:
1. The gravitational force \( mg \) acting downwards.
2. The tension \( T \) in the rod acting along the rod.

### Step 4: Analyze the circular motion
Since the car is moving in a circular path, the bob will experience a centripetal force directed towards the center of the circular path. This force is provided by the horizontal component of the tension in the rod.

### Step 5: Set up the equations of motion
For the vertical direction:
\[ T \cos \theta = mg \quad \text{(1)} \]
For the horizontal direction (centripetal force):
\[ T \sin \theta = \frac{mv^2}{r} \quad \text{(2)} \]

### Step 6: Divide the equations to eliminate \( T \)
Dividing equation (2) by equation (1):
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{\frac{mv^2}{r}}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{v^2}{rg} \]

### Step 7: Substitute the known values
Substituting \( v = 10 \text{ m/s} \), \( r = 10\sqrt{3} \text{ m} \), and \( g = 10 \text{ m/s}^2 \):
\[ \tan \theta = \frac{10^2}{10\sqrt{3} \cdot 10} = \frac{100}{100\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 8: Find the angle \( \theta \)
From \( \tan \theta = \frac{1}{\sqrt{3}} \), we know:
\[ \theta = 30^\circ \]

### Step 9: Find the angle made by the rod with the track
The angle made by the rod with the vertical is \( \theta \), but we need the angle made with the track. Since the rod makes an angle \( \theta \) with the vertical, the angle with the horizontal (the track) is:
\[ 90^\circ - \theta = 90^\circ - 30^\circ = 60^\circ \]

### Final Answer
The angle made by the rod with the track is \( 60^\circ \).

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