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Physics · LAWS OF MOTION
हिन्दीA circular racetrack of radius \( R = 100\text{ m} \) is banked at an angle of \( \theta = 45^\circ \). The coefficient of static friction between the race-car tyres and the track is \( \mu_s = 0.5 \). Taking \( g = 10\text{ m/s}^2 \), find the ratio of the maximum permissible speed \( v_{\max} \) (to avoid slipping outwards) to the optimum speed \( v_0 \) (where no frictional force is required).
Correct answer · Option 2
\( \sqrt{3} \)
Explanation
Optimum speed \( v_0 = \sqrt{R g \tan\theta} = \sqrt{1000 \times 1} = 10\sqrt{10}\text{ m/s} \). Maximum speed \( v_{\max} = \sqrt{R g \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} = \sqrt{1000 \times \frac{0.5 + 1}{1 - 0.5}} = \sqrt{3000} = 10\sqrt{30}\text{ m/s} \). Thus, \( \frac{v_{\max}}{v_0} = \sqrt{\frac{3000}{1000}} = \sqrt{3} \).
