Question

A car travels on a circular racetrack of radius \(50\,\text{m}\), which is banked at an angle \(\theta\). If the car travels at a speed \(10\,\text{m s}^{-1}\), then the wear and tear on its tyres is minimum. Taking \(g=10\,\text{m s}^{-2}\), the value of \(\theta\) is:

• A. \(\tan^{-1}(2\sqrt3)\)
• B. \(\tan^{-1}\left(\frac15\right)\)
• C. \(\tan^{-1}\left(\frac25\right)\)
• D. \(\tan^{-1}\left(\frac{\sqrt3}{2}\right)\)

Hint:

For a perfectly banked road, \[ \tan\theta=\frac{v^2}{Rg} \] No friction is needed and tyre wear becomes minimum.

Answer 1

The Correct Option is B

Concept:

• Minimum wear and tear implies that friction is not required.

• Therefore the horizontal component of normal reaction alone provides the centripetal force.

• For ideal banking, \[ \tan\theta=\frac{v^2}{Rg} \]

Step 1: Write the banking condition
\[ \tan\theta = \frac{v^2}{Rg} \] Given \[ v=10\,\text{m s}^{-1} \] \[ R=50\,\text{m} \] \[ g=10\,\text{m s}^{-2} \]

Step 2: Substitute numerical values
\[ \tan\theta = \frac{10^2}{50\times10} \] \[ = \frac{100}{500} \] \[ = \frac15 \]

Step 3: Find the angle
\[ \theta = \tan^{-1} \left( \frac15 \right) \] Hence, \[ \boxed{ \theta= \tan^{-1} \left( \frac15 \right) } \] \[ \boxed{\text{Option (B)}} \]