A circular road of radius 50 m has the angel of banking equal to `30^0`. At what speed should a vehicle go on this road so that the friction is not used?

Options

A. `17 m//sec`
B. `20 m//sec`
C. `24m//sec`
D. `40 m//sec`

Explanation

To find the speed at which a vehicle should travel on a circular road with a radius of 50 m and a banking angle of \(30^\circ\) without using friction, we can use the formula derived from the conditions of banking:

### Step-by-Step Solution:

1. **Understanding the Banking Condition**:
When a vehicle goes around a banked curve without relying on friction, the centripetal force required for circular motion is provided entirely by the horizontal component of the normal force. The relationship can be expressed as:
\[ \tan(\theta) = \frac{v^2}{rg} \]
where:
- \( \theta \) is the angle of banking,
- \( v \) is the speed of the vehicle,
- \( r \) is the radius of the circular path,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

2. **Substituting the Given Values**:
We know:
- \( r = 50 \, \text{m} \)
- \( \theta = 30^\circ \)
- \( g \approx 9.81 \, \text{m/s}^2 \)

First, we calculate \( \tan(30^\circ) \):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \]

3. **Setting Up the Equation**:
Substitute the values into the equation:
\[ \frac{1}{\sqrt{3}} = \frac{v^2}{50 \times 9.81} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ v^2 = 50 \times 9.81 \times \frac{1}{\sqrt{3}} \]

5. **Calculating the Right Side**:
Calculate \( 50 \times 9.81 \):
\[ 50 \times 9.81 = 490.5 \]
Now, divide by \( \sqrt{3} \):
\[ v^2 = \frac{490.5}{\sqrt{3}} \approx \frac{490.5}{1.732} \approx 283.4 \]

6. **Finding the Speed \( v \)**:
Now take the square root to find \( v \):
\[ v = \sqrt{283.4} \approx 16.82 \, \text{m/s} \]

### Final Answer:
The speed at which the vehicle should go on this road so that friction is not used is approximately \( 16.82 \, \text{m/s} \).