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NEET PHYSICSEasy

A car of mass 1000 kg1000 \text{ kg} negotiates a banked curve of radius 90 m90 \text{ m} on a frictionless road. If the banking angle is 45∘45^{\circ}, the speed of the car is:

A

20 m s−120 \text{ m s}^{-1}

B

30 m s−130 \text{ m s}^{-1}

C

5 m s−15 \text{ m s}^{-1}

D

10 m s−110 \text{ m s}^{-1}

Step-by-Step Solution

  1. Identify the Condition: The car travels on a banked road where friction is negligible (frictionless). The necessary centripetal force is provided solely by the horizontal component of the normal reaction force.
  2. Formula: For a vehicle moving on a frictionless banked curve, the optimum speed vv is given by the relation: v=Rgtan⁡θv = \sqrt{Rg \tan \theta} where RR is the radius of curvature, gg is the acceleration due to gravity, and θ\theta is the angle of banking [NCERT Physics Class 11, Laws of Motion, Section 4.10, Eq. 4.22].
  3. Substitution: Given R=90 mR = 90 \text{ m}, θ=45∘\theta = 45^{\circ}, and taking g=10 m s−2g = 10 \text{ m s}^{-2} (standard approximation for this calculation): v=90×10×tan⁡45∘v = \sqrt{90 \times 10 \times \tan 45^{\circ}}
  4. Calculation: Since tan⁡45∘=1\tan 45^{\circ} = 1: v=900×1=30 m s−1v = \sqrt{900 \times 1} = 30 \text{ m s}^{-1} (Note: The mass of the car is not required for this calculation).
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