Question

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

• A. $25^\circ$
• B. $60^\circ$
• C. $45^\circ$
• D. $30^\circ$

Hint:

For banking problems:
- The equation $\tan \theta = \frac{v^2}{g R}$ determines the angle.
- No friction is needed if speed matches the ideal banking angle.

Answer 1

The Correct Option is D

Step 1: Use the banking formula
The angle of banking is given by:
$$\tan \theta = \frac{v^2}{g R}$$
Step 2: Substitute given values
$$\tan \theta = \frac{(20)^2}{(10) (40\sqrt{3})}$$
$$\tan \theta = \frac{400}{400\sqrt{3}}$$
Step 3: Solve for $\theta$
$$\tan \theta = \frac{1}{\sqrt{3}}$$
- Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we get:
$$\theta = 30^\circ$$