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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 ms−120 \text{ ms}^{-1}

B

30 ms−130 \text{ ms}^{-1}

C

5 ms−15 \text{ ms}^{-1}

D

10 ms−110 \text{ ms}^{-1}

Step-by-Step Solution

  1. Concept: On a frictionless banked road, the horizontal component of the normal force provides the necessary centripetal force required for circular motion. The optimum speed v0v_0 (where no friction is required) depends only on the radius RR, gravity gg, and banking angle θ\theta [NCERT Physics Class 11, Laws of Motion, Section 4.10, Eq 4.22].
  2. Formula: v=Rgtan⁡θv = \sqrt{Rg \tan\theta}
  3. Given Data:
  • Radius, R=90 mR = 90 \text{ m}
  • Banking angle, θ=45∘\theta = 45^\circ
  • Acceleration due to gravity, g≈10 m/s2g \approx 10 \text{ m/s}^2 (standard approximation for integer answers).
  • Mass m=1000 kgm = 1000 \text{ kg} (Note: Mass cancels out in the derivation and is not needed for the calculation).
  1. Calculation: Substituting the values into the formula: v=90×10×tan⁡(45∘)v = \sqrt{90 \times 10 \times \tan(45^\circ)} Since tan⁡(45∘)=1\tan(45^\circ) = 1: v=900×1v = \sqrt{900 \times 1} v=30 m/sv = 30 \text{ m/s}
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