A circular road of radius R is banked for a speed v = 40 km/hr. A car of mass m attempts to go on the circular road, the friction coefficient between the tyre and road is negligible:

Moderate

A

the car cannot make a turn without skidding
B

if the car runs at a speed less than 40 km/hr, it will slip up the slope
C

if the car runs at the correct speed of 40 km/hr, the force by the road on the car is equal to $\frac{{mv}^{2}}{r}$
D

if the car runs at the correct speed of 40 km,/hr, the force by the road on the car is greater than mg as well as greater than $\frac{{mv}^{2}}{r}$

Solution

Car can make a turn without skidding at 40 km/hr. The car will slip down the slope if it runs at a speed less than 40 km/hr.
$N \sin θ = \frac{{mv}^{2}}{r} \Rightarrow N = \frac{1}{\sin θ} (\frac{{mv}^{2}}{r})$
and $N \cos θ = mg \Rightarrow N = \frac{mg}{\cos θ}$
$∵ \sin θ and \cos θ < 1$
So $N > mg and N > \frac{{mv}^{2}}{r}$

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