Question

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

• A. The car cannot 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 correct speed of 40 km/hr, the force by the road on the car is equal to ${mv}^2/r$
• 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 greater than ${mv}^2/r$

Answer 1

The correct option is 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 greater than ${mv}^2/r$

The car can turn without skidding at 40 km/h, thus option (a) is wrong.

Since $\mu =0$, so only force on the car by the road is the Normal force and no frictional force. So for speed less than 40 km/h it will slip down. Option (b) is correct.

As stated above only force by the road on the car is normal force $\left(N\right)$horizontal component of which $\left(Nsin\theta \right)$ produces a centripetal acceleration$=\frac{v^2}{r}$ . So $Nsin\theta =\frac{m{v}^2}{r}⟹N=\frac{m{v}^2}{rsin\theta }$ Where θ is angle of banking. Thus option (c) is wrong.

As above,$N=\frac{m{v}^2}{rsin\theta }$ , so N is always greater than $\frac{m{v}^2}{r}$ because $sin\theta$ is always $<1$. Similarly $Ncos\theta =mg⟹N=\frac{mg}{cos\theta }$ So $N$ is always $>mg$ because $cos\theta$ is always $<1.$ Thus option (d) is correct.

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