Q 11-04-212JEE MainMedium
A circular track of radius $20$ m is banked at an angle of $37^\circ$. The coefficient of friction between the tyres and the road is $0.5$. The maximum speed with which a car can go round the track without skidding is ______ m/s. (Take $g = 10\ \text{m/s}^2$ and $\tan37^\circ = \dfrac34$)
Numerical value type. Enter your answer.
Answer: 20
Concept. At the maximum speed the car tends to slide up the slope, so friction acts down the slope. Resolving gives
$$v_{\max}^2 = rg\,\frac{\tan\theta + \mu}{1 - \mu\tan\theta}$$
Step 1. Bracket.
$$\frac{0.75 + 0.5}{1 - 0.5 \times 0.75} = \frac{1.25}{0.625} = 2$$
Step 2. Speed.
$$v_{\max}^2 = 20 \times 10 \times 2 = 400 \;\Rightarrow\; v_{\max} = 20\ \text{m/s}$$
Answer. $20$.
Solution by Sreeraj P, M.Sc Physics