Practice question

Given: A 1250 kg car turns on a banked road ( θ = 15°, μ_s = 0.3 ) with radius 45 m . What is the maximum speed without slipping? (Take g = 10 m/s², tan 15° approx 0.268 ) These values define the system as per NCERT data. Formula: Maximum speed: v_{max = sqrtrg μ_s + tanθ/1 - μ_s tanθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Numerator: μ_s + tanθ = 0.3 + 0.268 = 0.568 . Denominator: 1 - 0.3 × 0.268 = 1 - 0.0804 = 0.9196 . v_{max² = 45 × 10 × 0.568/0.9196 approx 450 × 0.6175 approx 277.875 . v_{max = sqrt277.875 approx 16.67 m/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.