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Physics · LAWS OF MOTION
हिन्दीA banked circular track is designed such that \( \tan\theta = \frac{1}{3} \). The coefficient of static friction between tyres and the track is \( \mu_s = \frac{1}{3} \). What is the percentage increase in the maximum permissible speed \( v_{\max} \) over the optimum speed \( v_0 \) (where friction is zero)?
Correct answer · Option 3
50%
Explanation
\( v_0 = \sqrt{R g \tan\theta} = \sqrt{\frac{R g}{3}} \). \( v_{\max} = \sqrt{R g \frac{\tan\theta + \mu_s}{1 - \mu_s \tan\theta}} = \sqrt{R g \frac{1/3 + 1/3}{1 - 1/9}} = \sqrt{R g \frac{2/3}{8/9}} = \sqrt{\frac{3 R g}{4}} \). Ratio \( \frac{v_{\max}}{v_0} = \sqrt{\frac{3/4}{1/3}} = \frac{3}{2} = 1.5 \). Percentage increase = \( (1.5 - 1) \times 100\% = 50\% \).
