A car is negotiating a curved road of radius R. The road is banked at an angle theta . The coefficient of friction between the

Q: A car is negotiating a curved road of radius $R$. The road is banked at an angle $\theta .$ The coefficient of friction between the tyres of the car and the road is $\mu _s.$ The maximum safe velocity on this road is

a $\begin{array}{l} For\,vertical\,equilibrium\,on\,the\,road,\\ N\cos \theta = mg + f\sin \theta \\ mg = N\cos \theta - f\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\ Centripetal\,force\,for\,safe\,turning\,,\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N\sin \theta + f\cos \theta = \frac{{m{v^2}}}{R}\,\,\,\,\,\,...\left( {ii} \right)\\ From\,eqns.\,\left( i \right)\,and\,\left( {ii} \right),\,we\,get\\ \,\,\,\,\,\,\,\,\,\,\,\frac{{{v^2}}}{{Rg}} = \frac{{N\sin \theta + f\cos \theta }}{{N\cos \theta - f\sin \theta }} \end{array}$ $\begin{array}{l} \Rightarrow \,\frac{{{v^2}_{\max }}}{{Rg}} = \frac{{N\sin \theta + {\mu _s}N\cos \theta }}{{N\cos \theta - {\mu _s}N\sin \theta }}\\ \,\,\,\,\,\,{v_{\max }} = \sqrt {Rg\left( {\frac{{{\mu _s} + \tan \theta }}{{1 - {\mu _s}\tan \theta }}} \right)} \end{array}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A car is negotiating a curved road of radius $R$. The road is banked at an angle $\theta .$ The coefficient of friction between the tyres of the car and the road is $\mu _s.$ The maximum safe velocity on this road is

A$\sqrt {gR\frac{{{\mu _s} + tan\theta }}{{1 - {\mu _s}tan\theta }}} \;\;\;\;$
B$\;\sqrt {\frac{g}{R}\frac{{{\mu _s} + tan\theta }}{{1 - {\mu _s}tan\theta }}} $
C$\;\frac{g}{{{R^2}}}\frac{{{\mu _s} + tan\theta }}{{1 - {\mu _s}tan\theta }}$
D$\;\sqrt {g{R^2}\frac{{{\mu _s} + tan\theta }}{{1 - {\mu _s}tan\theta }}} $

NEET 2016, Diffcult

STD 11 Science
NEET
STD 11 - 4.2. friction
Physics

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