Q126 / 180 · NEET 2026 (Re-NEET) · Physics
A car travels on a circular racetrack of radius $50$ m, which is banked at an angle $\theta$. If the car travels at a speed $10$ ms$^{-1}$, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be $10$ ms$^{-2}$, the value of $\theta$ is :
- A$\tan^{-1}\left(2\sqrt{3}\right)$
- Correct answerB$\tan^{-1}\left(\dfrac{1}{5}\right)$
- C$\tan^{-1}\left(\dfrac{2}{5}\right)$
- D$\tan^{-1}\left(\sqrt{3}/2\right)$
Solution
For minimum wear and tear on the tyres, the frictional force should be zero. The required centripetal force is provided entirely by the horizontal component of the normal reaction.
The optimum speed $v$ on a banked road is given by:
$v = \sqrt{R g \tan \theta}$
Squaring both sides:
$v^2 = R g \tan \theta$
Substituting the given values $v = 10$ ms$^{-1}$, $R = 50$ m, and $g = 10$ ms$^{-2}$:
$(10)^2 = 50 \times 10 \times \tan \theta$
$100 = 500 \tan \theta$
$\tan \theta = \dfrac{100}{500} = \dfrac{1}{5}$
$\theta = \tan^{-1}\left(\dfrac{1}{5}\right)$
Answer: $\tan^{-1}\left(\dfrac{1}{5}\right)$
The optimum speed $v$ on a banked road is given by:
$v = \sqrt{R g \tan \theta}$
Squaring both sides:
$v^2 = R g \tan \theta$
Substituting the given values $v = 10$ ms$^{-1}$, $R = 50$ m, and $g = 10$ ms$^{-2}$:
$(10)^2 = 50 \times 10 \times \tan \theta$
$100 = 500 \tan \theta$
$\tan \theta = \dfrac{100}{500} = \dfrac{1}{5}$
$\theta = \tan^{-1}\left(\dfrac{1}{5}\right)$
Answer: $\tan^{-1}\left(\dfrac{1}{5}\right)$
Hence, the correct answer is option: 'B'.