A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μ s. The maximum safe velocity on this road is (a) √(g/R(μs+tan θ)/(1-μstan θ)) (b) √(g/R²(μs+tan θ)/(1-μstan θ)) (c) √(gR²(μs+tan θ)/(1-μstan θ)) (d) √(gR(μs+tan θ)/(1-μstan θ))

Answer: (d). At maximum speed, friction acts down the slope. Vertical: Ncos θ = mg + μs Nsin θ Horizontal: Nsin θ + μs Ncos θ = mv²/R Dividing: v²/Rg = (sin θ + μscos θ)/(cos θ - μssin θ) = (μs + tan θ)/(1 - μstan θ) vmax = √(gR (μs + tan θ)/(1 - μstan θ)), so the answer is (d).

A car is negotiating a curved road of radius R. The road is banked at… | NEET 2016 (Phase 1)

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Circular Motion (Related PYQs)

A car is negotiating a curved road of radius R. The road is banked at an angle $\theta$ . The coefficient of friction between the tyres of the car and the road is $\mu$ s. The maximum safe velocity on this road is

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Answer: (d)

At maximum speed, friction acts down the slope. Vertical: $N\cos\theta = mg + \mu_s N\sin\theta$
Horizontal: $N\sin\theta + \mu_s N\cos\theta = \frac{mv^2}{R}$
Dividing: $\frac{v^2}{Rg} = \frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta} = \frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}$
$v_{max} = \sqrt{gR\,\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}}$ , so the answer is (d).

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