A smooth circular road of radius r is banked for a speed `v=40 km//hr`. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible. The correct statements are:

Options

A. The car cannot make a turn without skidding.
B. If the car turns at a speed less than `40 km//hr`, it slips down.
C. f the car turns at the correct speed of `40 km//hr`, the force by the road on the car is equal to `mv^(2)//r`.
D. If the car turns at the correct speed of `40 km//hr`, the force by the road on the car is greater than mg as well as greater than `mv^(2)//r`.

Answer

To solve the problem of a car navigating a banked circular road without friction, we can analyze the forces acting on the car and derive the necessary conditions for it to maintain a circular motion. Here’s a step-by-step breakdown:

### Step 1: Understand the scenario
The road is banked at an angle θ for a specific speed (v = 40 km/h). The car of mass m is attempting to navigate this circular path. Since the friction coefficient is negligible, we will rely on the banking angle and the normal force to provide the necessary centripetal force.

### Step 2: Convert speed to SI units
Convert the speed from km/h to m/s:
\[ v = 40 \text{ km/h} = \frac{40 \times 1000}{3600} \text{ m/s} = \frac{40000}{3600} \approx 11.11 \text{ m/s} \]

### Step 3: Identify forces acting on the car
1. **Weight (mg)**: Acts downwards.
2. **Normal force (N)**: Acts perpendicular to the surface of the banked road.

### Step 4: Resolve the normal force into components
The normal force can be resolved into two components:
- **Vertical component**: \( N \cos \theta \) (balances the weight of the car)
- **Horizontal component**: \( N \sin \theta \) (provides the centripetal force required for circular motion)

### Step 5: Set up equations based on forces
1. **Vertical force balance**:
\[ N \cos \theta = mg \]
2. **Centripetal force requirement**:
\[ N \sin \theta = \frac{mv^2}{r} \]

### Step 6: Analyze the conditions for different speeds
- If the car travels at a speed less than 40 km/h, the required centripetal force cannot be provided by the normal force alone, leading to a tendency to slip downwards. Thus, the statement that "if the car turns at a speed less than 40 km/h it slips down" is **correct**.

- If the car travels at exactly 40 km/h, we can substitute \( N \) from the vertical force balance into the centripetal force equation. From the first equation, we have:
\[ N = \frac{mg}{\cos \theta} \]
Substituting into the second equation:
\[ \frac{mg}{\cos \theta} \sin \theta = \frac{mv^2}{r} \]
This leads to:
\[ mg \tan \theta = \frac{mv^2}{r} \]
This indicates that the banking angle is indeed designed for this speed.

### Step 7: Evaluate the statements
1. **The car cannot make a turn without skidding**: **Incorrect** (the road is banked for the speed).
2. **If the car turns at a speed less than 40 km/h, it slips down**: **Correct**.
3. **If the car turns at 40 km/h, the force by the road on the car is equal to \( \frac{mv^2}{r} \)**: **Incorrect** (the normal force is greater).
4. **If the car turns at 40 km/h, the force by the road on the car is greater than both mg and \( \frac{mv^2}{r} \)**: **Correct** (since \( N \) is greater than both due to the banking angle).

### Conclusion
The correct statements are:
- The car slips down if it turns at a speed less than 40 km/h.
- The force by the road on the car is greater than both mg and \( \frac{mv^2}{r} \) when turning at 40 km/h.