A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45°, the speed of the car is: (g = 10 ms–2)A 5ms–1 B 10ms–1 C 20ms–1 D 30ms–1

The problem involves a car moving on a banked curve. The forces acting on the car are gravity, the normal force from the road, and the centripetal force that keeps the car moving in a circle. Since the road is frictionless, the frictional force is zero.

The force due to gravity acts vertically downwards and is equal to mg, where m is the mass of the car and g is the acceleration due to gravity.

The normal force from the road acts perpendicular to the road surface.

The centripetal force required to keep the car moving in a circle of radius r at speed v is given by mv^2/r.

Since there is no friction, the component of the normal force along the slope of the road must provide the centripetal force.

So, we can write: Nsinθ = mv^2/r, where θ is the banking angle.

The component of the normal force perpendicular to the slope of the road must balance the weight of the car, so we can write: Ncosθ = mg.

Dividing these two equations gives: tanθ = v^2/rg.

Substituting the given values: tan45° = v^2/(90m * 10m/s^2).

Since tan45° = 1, we get: v^2 = 900 m^2/s^2.

Taking the square root of both sides gives: v = 30 m/s.

So, the speed of the car is 30 m/s. The correct answer is D.