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NEET PHYSICSEasy

A motor cyclist moving with a velocity of 72 km/hour72 \text{ km/hour} on a flat road takes a turn on the road at a point where the radius of curvature of the road is 20 meters20 \text{ meters}. The acceleration due to gravity is 10 m/sec210 \text{ m/sec}^2. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than:

A

θ=tan⁡−1(6)\theta = \tan^{-1}(6)

B

θ=tan⁡−1(2)\theta = \tan^{-1}(2)

C

θ=tan⁡−1(25.92)\theta = \tan^{-1}(25.92)

D

θ=tan⁡−1(4)\theta = \tan^{-1}(4)

Step-by-Step Solution

  1. Convert Units: First, convert the velocity from km/hr to m/s. v=72×518=20 m/sv = 72 \times \frac{5}{18} = 20 \text{ m/s}
  2. Identify the Physics Principle: When a cyclist takes a turn, they lean inward to generate the necessary centripetal force from the horizontal component of the normal reaction. The angle θ\theta with the vertical is given by the relationship: tan⁡θ=v2rg\tan \theta = \frac{v^2}{rg} [Source 41]
  3. Substitute Values:
  • Velocity v=20 m/sv = 20 \text{ m/s}
  • Radius r=20 mr = 20 \text{ m}
  • Gravity g=10 m/s2g = 10 \text{ m/s}^2 tan⁡θ=(20)220×10\tan \theta = \frac{(20)^2}{20 \times 10} tan⁡θ=400200=2\tan \theta = \frac{400}{200} = 2
  1. Solve for Angle: θ=tan⁡−1(2)\theta = \tan^{-1}(2) (Note: Option 3 corresponds to the error of failing to convert km/hr to m/s: 722/200=25.9272^2/200 = 25.92).
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