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.

Options

A. The car can make a turn without skidding.
B. If the car turns at a speed less than 40 km/hr, it will slip down
C. 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)+((mv^(2))/r)^(2))`
D. 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 graeter than `(mv^(2))/r`

Explanation

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
`N sin theta =(mv^(2))/r. N=sqrt((mg)^(2)+((mv^(2))/r))`