A car of 800 kg is taking turn on a banked road of radius 300 m and angle of banking 30°. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: (g=10 m / s², √3=1.73)

Solution

Physics sir · 4 steps

At top speed the car wants to slide up the bank, so friction points down the slope and helps gravity.

1. Direction of friction. At the maximum safe speed the car tends to slip outward, up the incline. Static friction μ N therefore acts down the slope, adding to the inward pull.
2. Balance the forces. Vertically, N cos θ = mg + μ N sin θ. Horizontally, N sin θ + μ N cos θ = mv²/r. Dividing one by the other removes N and m.

   vₘₐₓ² = rg (tan θ + μ)/(1 - μ tan θ)

3. Put in numbers. tan 30° = 1/1.73 = 0.578, so the fraction is 0.778/(1 - 0.1156) = 0.778/0.884 = 0.880. With rg = 300 × 10 = 3000 m² s⁻², vₘₐₓ² = 2640 m² s⁻².
4. So. vₘₐₓ = √2640 ≈ 51.4 m s⁻¹, which is option (a). The mass of 800 kg is not needed.

The idea to keep. For maximum speed on a banked road, friction acts down the slope: use +μ on top and -μ tan θ below. For minimum speed the signs flip. Mass always cancels.

Answer from the NTA final answer key. Solution by PrepWiser.