A car is taking turn on a circular path of radius R. If the coefficient of friction between the tyres and road is `mu`, the maximum velocity for no slipping is

Options

A. `muRg`
B. `2muRg`
C. `(muRg)^(1//2)`
D. `(2muRg)^(1//2)`

Explanation

To solve the problem of finding the maximum velocity for a car taking a turn on a circular path without slipping, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Car**:
- When a car is turning on a circular path, the primary forces acting on it are:
- The gravitational force (weight) acting downwards: \( mg \)
- The frictional force acting towards the center of the circular path, which provides the necessary centripetal force for circular motion.

2. **Centripetal Force Requirement**:
- For an object moving in a circle of radius \( R \) with velocity \( v \), the required centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{R} \]
- Here, \( m \) is the mass of the car, \( v \) is the velocity, and \( R \) is the radius of the circular path.

3. **Frictional Force**:
- The maximum frictional force \( F_f \) that can act between the tires and the road is given by:
\[ F_f = \mu mg \]
- Here, \( \mu \) is the coefficient of friction, and \( g \) is the acceleration due to gravity.

4. **Setting Up the Inequality**:
- For the car to turn safely without slipping, the centripetal force must not exceed the maximum frictional force. Therefore, we can set up the inequality:
\[ \frac{mv^2}{R} \leq \mu mg \]

5. **Canceling Mass**:
- Since \( m \) appears on both sides of the inequality, we can cancel it out (assuming \( m \neq 0 \)):
\[ \frac{v^2}{R} \leq \mu g \]

6. **Solving for Maximum Velocity**:
- Rearranging the inequality gives us:
\[ v^2 \leq \mu g R \]
- Taking the square root of both sides, we find the maximum velocity \( v_{\text{max}} \):
\[ v_{\text{max}} = \sqrt{\mu g R} \]

### Final Answer:
The maximum velocity for the car to take a turn on a circular path of radius \( R \) without slipping is:
\[ v_{\text{max}} = \sqrt{\mu g R} \]