Question

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

Answer 1

The process of raising the outer edge of the curved road above the inner edge is known as banking of roads. The reason why the roads are banked is so that the vehicle can go round the curved track at a reasonable speed without skidding.

Complete step by step solution:
When a vehicle goes round a curved road, it requires some centripetal force. While rounding the curve, the wheels of the vehicle have a tendency to leave the curved path and regain its original path i.e., straight line. Now the force of friction between wheels and the roads opposes this tendency. This force of friction therefore, acts towards the centre of the circular track and provides the necessary centripetal force.
In the case of vertical equilibrium,
$N\cos \theta = mg + {f_1}\sin \theta$
$mg = N\cos \theta - {f_1}\sin \theta ........(1)$
In the case of horizontal equilibrium,
$N\sin \theta + {f_1}\cos \theta = \dfrac{{m{v^2}}}{R}........(2)$
On dividing equation (1) and (2), we get,
$\dfrac{{{v^2}}}{{Rg}} = \dfrac{{\sin \theta + {\mu _s}\cos \theta }}{{\cos \theta - {\mu _s}\sin \theta }}.........({f_1} \propto {\mu _s})$
By keeping $v$ on the LHS and taking all the other terms on the other side, we get,
$v = \sqrt {Rg\left( {\dfrac{{\sin \theta + {\mu _s}\cos \theta }}{{\cos \theta - {\mu _s}\sin \theta }}} \right)}$
On dividing both the numerator and denominator by $\cos \theta$ ,
$v = \sqrt {Rg\left( {\dfrac{{\tan \theta + {\mu _s}}}{{1 - {\mu _s}\tan \theta }}} \right)}$
Now, this can be written as,
$v = \sqrt {gR\dfrac{{{\mu _s} + \tan \theta }}{{1 - {\mu _s}\tan \theta }}}$
So, the correct answer is $(B)\sqrt {gR\dfrac{{{\mu _s} + \tan \theta }}{{1 - {\mu _s}\tan \theta }}}$

Note:
If we make any attempt to run a vehicle at a greater speed, the vehicle is likely to skid and go out of track. In order that the vehicle can go round the curved track at a reasonable speed without skidding, the sufficient centripetal force is managed for it by raising the outer edge of the track a little above the inner edge. This process is called the banking of the circular track or banking of roads.