1. √(gR²μ_s+tanθ/1-μ_stanθ) 2. √(gRμ_s+tanθ/1-μ_stanθ) 3. √(g/Rμ_s+tanθ/1-μ_stanθ) 4. √(g/R²μ_s+tanθ/1-μ_stanθ)

Answer: 2. √(gRμ_s+tanθ/1-μ_stanθ) Key idea: at maximum speed friction acts down the slope, along with the normal reaction. Vertical: Ncosθ=mg+μ_sNsinθ . Horizontal: Nsinθ+μ_sNcosθ=mv²/R . Dividing: v²/Rg=tanθ+μ_s/1-μ_stanθ , so v=√(gRμ_s+tanθ/1-μ_stanθ) .

Topic: Dynamics of uniform circular motion: level and banked road· Laws of Motion

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 :

1. \(\sqrt{gR^2\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}}\)

2. \(\sqrt{gR\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}}\)

3. \(\sqrt{\frac{g}{R}\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}}\)

4. \(\sqrt{\frac{g}{R^2}\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}}\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

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Answer: 2. \(\sqrt{gR\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}}\)

Key idea: at maximum speed friction acts down the slope, along with the normal reaction.

Vertical: \(N\cos\theta=mg+\mu_sN\sin\theta\).
Horizontal: \(N\sin\theta+\mu_sN\cos\theta=\dfrac{mv^2}{R}\).
Dividing: \(\dfrac{v^2}{Rg}=\dfrac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}\), so \(v=\sqrt{gR\dfrac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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 :

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