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 (a) μ = (v₀² + rgtan θ)/(rg + v₀²tan θ) (b) μ = (v₀² - rgtan θ)/(rg - v₀²tan θ) (c) μ = (v₀² - rgtan θ)/(rg + v₀²tan θ) (d) μ = (v₀² + rgtan θ)/(rg - v₀²tan θ)

Answer: (c). At maximum speed friction acts down the slope: v₀² = rg(μ + tan θ)/(1 - μtan θ). v₀² - μ v₀²tan θ = μ rg + rgtan θ μ(rg + v₀²tan θ) = v₀² - rgtan θ So the answer is (c) μ = (v₀² - rgtan θ)/(rg + v₀²tan θ).

A car of mass ' m ' moves on a banked road having radius ' r ' and… | JEE Main 2025 (24 Jan, Shift 1)

Q7JEE Main 2025 · 24 JAN · SHIFT 1Difficult

Banking of Roads & Bridges

A car of mass ' $m$ ' moves on a banked road having radius ' $r$ ' and banking angle $\theta$ . To avoid slipping from banked road, the maximum permissible speed of the car is $v_0$ . The coefficient of friction $\mu$ between the wheels of the car and the banked road is

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Answer: (c)

At maximum speed friction acts down the slope: $v_0^2 = rg\frac{\mu + \tan\theta}{1 - \mu\tan\theta}$ .
$v_0^2 - \mu v_0^2\tan\theta = \mu rg + rg\tan\theta$
$\mu(rg + v_0^2\tan\theta) = v_0^2 - rg\tan\theta$
So the answer is (c) $\mu = \frac{v_0^2 - rg\tan\theta}{rg + v_0^2\tan\theta}$ .

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