Question

A car is moving with a speed of 5 m/s on a circular road having radius of curvature 10 m and coefficient of friction 0.5. For the safe journey of the car, the banking angle of the road should be:

• A. \( \tan^{-1} \left( \frac{1}{2} \right) \)
• B. \( \tan^{-1}
• C. \)
• D. \( \tan^{-1} \left( \frac{1}{4} \right) \)

Hint:

The banking angle of a road is determined by the formula \( \tan \theta = \frac{v^2}{r g} + \mu \), where \( v \) is the speed, \( r \) is the radius, \( g \) is gravity, and \( \mu \) is the coefficient of friction.

Answer 1

The Correct Option is A

We use the formula for the banking angle: \[ \tan \theta = \frac{v^2}{r g} + \mu \] Substituting the values \( v = 5 \, \text{m/s}, r = 10 \, \text{m}, g = 10 \, \text{m/s}^2, \mu = 0.5 \): \[ \tan \theta = \frac{25}{100} + 0.5 = 0.75 \] Taking the inverse tangent: \[ \theta = \tan^{-1}(0.75) \approx \tan^{-1}\left( \frac{1}{2} \right) \] Hence, the correct answer is \( \tan^{-1} \left( \frac{1}{2} \right) \).