The slope of the smooth banked horizontal road is p . If the radius of the curve be r , the maximum velocity with which a car can negotiate the curve is given by

The correct option is $\sqrt{prg}$ (which matches Option 2: √(prg)).

Step-by-Step Derivation and Explanation:

1. Understanding Banked Curve Physics:
When a car negotiates a curved road of radius $r$ at a velocity $v$, it experiences a centripetal force directed toward the center of the curve. On a banked, smooth (frictionless) road, the horizontal component of the normal force provides the necessary centripetal force, while the vertical component balances the weight of the car.

2. Defining the Forces:
Let $\theta$ be the angle of banking. The forces acting on the car of mass $m$ are:
• The gravitational force ($mg$) acting vertically downwards.
• The normal force ($N$) acting perpendicular to the banked surface.

3. Resolving the Normal Force:
Resolving the normal force $N$ into vertical and horizontal components:
• Vertical component: $Ncos\left(\theta \right)=mg$ (Balances the weight)
• Horizontal component: $Nsin\left(\theta \right)=\frac{m{v}^2}{r}$ (Provides the centripetal force)

4. Relating Angle to Velocity:
Dividing the horizontal component equation by the vertical component equation gives:
$tan\left(\theta \right)=\frac{v^2}{rg}$

5. Substituting the Slope:
The slope of the road is given as $p$. Geometrically, the slope of the banking is represented by the tangent of the banking angle:
$p=tan\left(\theta \right)$
Substituting $p$ into the relation:
$p=\frac{v^2}{rg}$

6. Solving for Maximum Velocity:
Rearranging the equation to solve for the maximum velocity $v$:
$v^2=prg$
Taking the square root of both sides:
$v=\sqrt{prg}$