The normal reaction N for a vehicle of 800 kg mass, negotiating a turn on a 30° banked road at maximum possible speed without skidding is ____ ×10³ kg m s⁻². (Given cos 30°=0.81, μₛ=0.2)

Idea: at the maximum speed the car is on the point of sliding up the bank, so friction acts down the slope. Resolving vertically then gives N without ever needing the speed.

The normal force has vertical component N cos θ upward; the friction μ N, acting down the slope, has vertical component μ N sin θ downward. Together they carry the weight:

N cos θ-μ N sin θ=mg

N=(mg)/(cos θ-μ sin θ).

With m=800 kg, g=9.8 m s⁻², cos 30°=0.866 and sin 30°=0.5:

N=(7840)/(0.866-0.2(0.5))=(7840)/(0.766)≈10.2×10³ N.

Note that N comes out well above the weight mg=7.84×10³ N, as it must when the car is cornering hard.

About the data: the stem gives cos 30°=0.81, which is not right — the true value is 0.866. Using 0.81 would give (7840)/(0.71)=11.0×10³, which is not among the options; the printed answer follows from the correct cosine.