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 the road is negligible. Choose the correct alternatives :

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The car can make a turn without skidding.

If the car turns at a speed less than 40 km/hr, it will slip down

If the car turns at the constant speed of 40 km/hr, the force by the road on the car is equal to $\sqrt{(mg)^{2} + {(\frac{{mv}^{2}}{r})}^{2}}$

If the car turns 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 $\frac{{MV}^{2}}{r}$

answer is B, C, D.

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Detailed Solution

When speed of car is 40 km/hr, car can make a turn without skidding. If speed is less than 40 km/hr than tendency of slipping is downward so it will slip down. If speed is greater than 40 km/hr than tendency of slipping upward so it will slip up. If the car’s turn at correct speed 40 km/hr
$Nsin θ = \frac{{mv}^{2}}{r}, N = \sqrt{(mg)^{2} + (\frac{{mv}^{2}}{r})}$