JEE Main 2025 · 24 January · morning · Single correct · +4 / −1

A car of mass ' m ' moves on a banked road having radius ' r ' and banking angle θ. To avoid slipping from banked road, the maximum permissible speed of the car is v₀. The coefficient of friction μ between the wheels of the car and the banked road is

  1. (a) μ=(v₀²+r g tan θ)/(r g+v₀² tan θ)
  2. (b) μ=(v₀²-r g tan θ)/(rg-vₒ² tan θ)
  3. (c) μ=(v₀²-r g tan θ)/(r g+v₀² tan θ)
  4. (d) μ=(vₒ²+r g tan θ)/(r g-vₒ² tan θ)
Answer and solution

Answer (c) μ=(v₀²-r g tan θ)/(r g+v₀² tan θ)

Solution

Physics sir · 5 steps

At the top speed the car is about to slide up, so friction points down the slope. Write both force equations.

  1. Direction of friction. At the maximum speed the car tends to slide outward, that is, up the slope. So friction f = μ N acts down the slope, along with the normal force N.
  2. Resolve the forces. Vertical: N cos θ - μ N sin θ = mg. Horizontal, towards the centre: N sin θ + μ N cos θ = mv₀²/r.
  3. Divide the equations. Dividing the horizontal equation by the vertical one cancels N and m. Then divide the top and bottom by cos θ.

    v₀²/rg = (tan θ + μ)/(1 - μ tan θ)

  4. Collect the μ terms. Cross-multiply: v₀² - μ v₀²tan θ = rg tan θ + μ rg. Move the μ terms to one side: v₀² - rg tan θ = μ(rg + v₀²tan θ).
  5. So. μ = (v₀² - rg tan θ)/(rg + v₀²tan θ), which is option (c).

The idea to keep. On a banked road, maximum speed means friction acts down the slope and minimum speed means it acts up. Fix that direction first, then divide the horizontal equation by the vertical one.

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

Study Circular Motion

Asked about 70 times in JEE Main since 2020. This question is from 2025.