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

Step 1: Read what minimum wear and tear means.
Tyres wear out because of friction. If friction is not needed at all to keep the car on the track, the wear is minimum. This is the ideal banking condition.
Step 2: Identify the force providing centripetal force.
With no friction, only the horizontal part of the normal reaction pulls the car toward the centre. Balancing forces gives the ideal banking relation.
Step 3: Write the banking formula.
\[ \tan\theta = \frac{v^2}{Rg} \]
Step 4: List the given data.
Here $v = 10$ m/s, $R = 50$ m, and $g = 10$ m/s$^2$.
Step 5: Substitute the values.
\[ \tan\theta = \frac{10^2}{50\times 10} = \frac{100}{500} = \frac15 \]
Step 6: Solve for the angle.
Taking the inverse tangent gives $\theta = \tan^{-1}\left(\dfrac15\right)$, which is option B.
\[ \boxed{ \theta = \tan^{-1}\left(\frac15\right) } \]