An automobile enters a turn of radius `R`. If the road is banked at an angle of `45^(@)` and the coefficient of friction is 1, the minimum and maximum speed with which the automobile can negotiate the turn without skidding is `:`

Options

A. `sqrt((rg)/(2))and sqrt(rg)`
B. `sqrt(rg)/(2)and sqrt(rg)`
C. `sqrt(rg)/(2)and 2sqrt(rg)`
D. 0 and inifinite

Explanation

To solve the problem of determining the minimum and maximum speed of an automobile negotiating a banked turn, we can use the following formulas derived from the principles of circular motion and friction.

### Given:
- Radius of the turn, \( R \)
- Angle of banking, \( \theta = 45^\circ \)
- Coefficient of friction, \( \mu = 1 \)

### Step 1: Calculate Minimum Speed (\( V_{min} \))

The formula for the minimum speed on a banked curve is given by:

\[ V_{min}^2 = Rg \left( \tan \theta - \mu \right) \div \left( 1 + \mu \tan \theta \right) \]

#### Substituting the values:
- Since \( \theta = 45^\circ \), we have \( \tan 45^\circ = 1 \).
- The coefficient of friction \( \mu = 1 \).

Now substituting these values into the formula:

\[ V_{min}^2 = Rg \left( 1 - 1 \right) \div \left( 1 + 1 \cdot 1 \right) \]

This simplifies to:

\[ V_{min}^2 = Rg \cdot 0 \div 2 = 0 \]

Thus,

\[ V_{min} = 0 \, \text{m/s} \]

### Step 2: Calculate Maximum Speed (\( V_{max} \))

The formula for the maximum speed on a banked curve is given by:

\[ V_{max}^2 = Rg \left( \tan \theta + \mu \right) \div \left( 1 - \mu \tan \theta \right) \]

#### Substituting the values:
Using \( \tan 45^\circ = 1 \) and \( \mu = 1 \):

\[ V_{max}^2 = Rg \left( 1 + 1 \right) \div \left( 1 - 1 \cdot 1 \right) \]

This simplifies to:

\[ V_{max}^2 = Rg \cdot 2 \div 0 \]

Since division by zero is undefined, we conclude that:

\[ V_{max} = \infty \, \text{m/s} \]

### Final Results:
- Minimum speed \( V_{min} = 0 \, \text{m/s} \)
- Maximum speed \( V_{max} = \infty \, \text{m/s} \)