Question

A car moves at a speed of 20 ms\(^{-1}\) on a banked track and describes an arc of a circle of radius \(40\sqrt{3}\) m. The angle of banking is (g = 10 ms\(^{-2}\))

• A. 25°
• B. 60°
• C. 45°
• D. 30°
• E. 40°

Hint:

For banked roads, directly use \(\tan\theta = \frac{v^2}{rg}\).

Answer 1

The Correct Option is D

Concept: For a frictionless banked road: \[ \tan \theta = \frac{v^2}{rg} \]

Step 1: Substitute given values. \[ v = 20, \quad r = 40\sqrt{3}, \quad g = 10 \]

Step 2: Apply formula. \[ \tan \theta = \frac{20^2}{40\sqrt{3} \times 10} = \frac{400}{400\sqrt{3}} = \frac{1}{\sqrt{3}} \]

Step 3: Find angle. \[ \theta = 30^\circ \]

Step 4: Conclusion. \[ \boxed{30^\circ} \]