Question

A turn of radius $20m$ is banked for the vehicles going at a speed of $10{ms}^{-1}$. If the coefficient of static friction between the road and the tyre is $0.4$, what is the possible speed of a vehicle so that it does not slips down?
(Take $g=10{ms}^{-2}$)

• A. $50k/h$
• B. $48k/h$
• C. $54k/h$
• D. $10k/h$

Answer 1

The correct option is C $54k/h$

$\begin{array}{c}Given,\\ V=36km/h=10m/s\\ r=20m\\ \mu =0.4\\ Theroadisbankedwithanangle,\\ \theta ={\tan }^{-1}\left(\frac{v^2}{g}\right)={\tan }^{-1}\left(\frac{100}{20\times 10}\right)={\tan }^{-1}\left(\frac{1}{2}\right)or\tan \theta =0.5\\ So,R_1-mg\cos \theta -\frac{m{v}_1^2}{r}\sin \theta =\mathrm{0........}\left(1\right)\\ and\mu R_1+mg\sin \theta -\frac{m{v}_1^2}{r}\cos \theta =\mathrm{0.........}\left(2\right)\\ solvingtheequationweget,\\ V_1=\sqrt{rg\frac{\tan \theta -\mu }{1+\mu \tan \theta }}\\ =\sqrt{20\times 10\times \frac{0.1}{1.2}}\\ =4.082m/s=14.7km/h\\ so,thepossibelspeedarebetween14.7km/hand54km/h\end{array}$