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)
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Question

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)

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G1 Physics · Accepted Answer

For maximum speed on a banked road with friction: v(max)² = rg[(tanθ + μₛ)/(1 - μₛ tanθ)] Given: r = 300 m θ = 30° μₛ = 0.2 g = 10 m/s² tan30° ≈ 1/√3 ≈ 0.577 Therefore: v(max)² = 300 × 10 × [(0.577 + 0.2)/(1 - 0.2 × 0.577)] This gives: v(max) ≈ 51.4 m/s