A

Given Data:
Radius of the turn, r = 20 m
Speed, v = 36 km/h = 36 \times \frac{1000}{3600} = 10 \text{ m/s}
Coefficient of static friction, \mu = 0.4
Convert speed to m/s:
v = 36 km/h = 10 m/s

Step 1: Calculate the angle of banking (\theta)
The banking angle can be calculated using the formula for the banking of roads:
\[ \tan(\theta) = \frac{v^2}{rg} \]
where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
\[ \tan(\theta) = \frac{(10)^2}{20 \times 9.81} = \frac{100}{196.2} \approx 0.509\]

Now, \( \theta \approx \tan^{-1}(0.509) \approx 27.0^\circ \)

Step 2: Calculate the maximum and minimum speeds
The forces acting on the vehicle are:
1. Gravitational force (mg)
2. Normal force (N)
3. Frictional force (f)

The maximum speed for the vehicle not to skid is given by:
\[ v_{max} = \sqrt{r(g(\mu + \tan(\theta)))} \]
Substituting the values:
\[ v_{max} = \sqrt{20 \times 9.81 \times (0.4 + 0.509)} = \sqrt{20 \times 9.81 \times 0.909} \approx \sqrt{178.54} \approx 13.35 \text{ m/s} \approx 48.06 \text{ km/h} \]

The minimum speed is:
\[ v_{min} = \sqrt{r(g(\tan(\theta) - \mu))} \]
Substituting the values:
\[ v_{min} = \sqrt{20 \times 9.81 \times (0.509 - 0.4)} = \sqrt{20 \times 9.81 \times 0.109} \approx \sqrt{21.42} \approx 4.63 \text{ m/s} \approx 16.66 \text{ km/h} \]

Step 3: Conclusion
The possible speeds of a vehicle for it to neither slip down nor skid up on a banked turn of radius 20 m is between:
\( 16.66 \text{ km/h} \) and \( 48.06 \text{ km/h} \).