A motor cyclist moving with a velocity of 72 km/hour on a flat road takes a turn on the road at a point where the radius of curvature of the road is 20 meters . The acceleration due to gravity is `10m//sec^(2)`. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than

Options

A. `theta=tan^(-1)(6)`
B. `theta=tan^(-1)(2)`
C. `theta=tan^(-1)(25.92)`
D. `theta=tan^(-1)(4)`

Explanation

To solve the problem, we need to determine the maximum angle (θ) at which the motorcyclist can lean while taking a turn without skidding. We can use the relationship between the speed of the motorcyclist, the radius of curvature of the turn, and the angle of lean.

### Step-by-Step Solution:

1. **Convert the Speed from km/h to m/s:**
The speed of the motorcyclist is given as 72 km/h. We need to convert this to meters per second (m/s).
\[ \text{Speed in m/s} = \frac{72 \times 1000}{3600} = 20 \text{ m/s} \]

2. **Identify the Radius of Curvature:**
The radius of curvature (r) of the turn is given as 20 meters.

3. **Use the Formula for the Angle of Lean:**
To avoid skidding, the tangent of the angle (θ) at which the motorcyclist must lean can be calculated using the formula:
\[ \tan \theta = \frac{V^2}{g r} \]
Where:
- \( V \) = speed of the motorcyclist (20 m/s)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( r \) = radius of curvature (20 m)

4. **Substitute the Values into the Formula:**
\[ \tan \theta = \frac{(20)^2}{10 \times 20} \]
\[ \tan \theta = \frac{400}{200} = 2 \]

5. **Calculate the Angle θ:**
Now, we need to find the angle θ using the inverse tangent function:
\[ \theta = \tan^{-1}(2) \]

6. **Final Result:**
The maximum angle θ at which the motorcyclist can lean without skidding is:
\[ \theta \approx 63.43^\circ \]

### Conclusion:
The motorcyclist must not bend with respect to the vertical plane by an angle greater than approximately \( 63.43^\circ \).