QJEEM24Ph7ee2: JEE Main 2024 Physics - Class 11 JEEM24Ph7ee2 Solution: A car of 800   kg is taking turn on a banked road of…

When a car takes a turn on a banked road, the forces involved are the gravitational force, the normal force from the surface, and frictional force (if any). The net force provides the necessary centripetal force for the circular motion. The angle of banking and static friction contribute to the maximum speed the car can achieve without slipping.

The forces acting on the car are as follows:

1. The normal force ($$N$$) acts perpendicular to the surface of the road.

2. Gravitational force ($$mg$$) acts downward.

3. Frictional force ($$f$$), which can provide additional centripetal force if needed. It acts parallel to the surface of the road, towards the center of the circle.

The normal force and the gravitational force components can be resolved into two directions: perpendicular and parallel to the road surface. The maximum speed is achieved when all available forces (normal, frictional) are utilized to provide the necessary centripetal force ($$F_c$$) without slipping.

The centripetal force required for circular motion is given by:

$F_c = \frac{mv^2}{r}$

Where:

• $$m$$ = mass of the car = $$800kg$$
• $$v$$ = speed of the car
• $$r$$ = radius of the turn = $$300m$$

On a banked curve, the maximum velocity can be calculated using the formula:

$v = \sqrt{rg(\tan\theta + \mu)\Big/\big(1-\mu\tan\theta)}$

Where:

• $$\theta$$ = angle of banking = $${30}^{\circ }$$
• $$\mu$$ = coefficient of static friction = $$0.2$$
• $$g$$ = acceleration due to gravity = $$10m/s^2$$

First, calculate the tangent of the angle:

$\tan30^{\circ} = \frac{1}{\sqrt{3}} = \frac{1}{1.73}$

Substitute all the values into the formula:

$v = \sqrt{300 \times 10 \times \left(\frac{1}{1.73} + 0.2\right)\Big/\big(1 - 0.2 \times \frac{1}{1.73}\big)}$

$v = \sqrt{3000 \times \left(\frac{1}{1.73} + 0.2\right)\Big/\big(1 - \frac{0.2}{1.73}\big)}$

$v = \sqrt{3000 \times \frac{1 + 1.73 \times 0.2}{1.73 - 0.2}}$

$v = \sqrt{3000 \times \frac{1 + 0.346}{1.73 - 0.2}}$

$v = \sqrt{3000 \times \frac{1.346}{1.53}}$

$v = \sqrt{3000 \times 0.8797}$

$v = \sqrt{2639.1} \approx 51.37 \, m/s$

Hence, the closest option to the calculated maximum speed without slipping is:

Option A: 51.4 m/s