Statement I : A cyclist is moving on an unbanked road with a speed of 7 kmh⁻¹ and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s²)
Statement II : If the road is banked at an angle of 45°, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh⁻¹ without slipping.
In the light of the above statements, choose the correct answer from the options given below.

Solution

Physics sir · 3 steps

Find the maximum safe speed in each case, convert it to km/h, and compare it with the given speed.

1. Unbanked road. On a flat road only static friction supplies the centripetal force, so mv²/r ≤ μ mg. That gives vₘₐₓ = √(μ r g) = √(0.2 × 2 × 9.8) = √3.92 = 1.98 m s⁻¹, which is about 7.13 km h⁻¹.
2. Banked road. On a banked road friction acts down the slope at the highest speed, and the normal force also helps provide the centripetal force. With tan 45° = 1, the maximum speed comes from this formula:

   vₘₐₓ² = rg (tan θ + μ)/(1 - μ tan θ) = 2 × 9.8 × 1.2/0.8 = 29.4 m² s⁻²

   Statement	Maximum safe speed	Given speed	Verdict
   I (flat, μ = 0.2)	1.98 m s⁻¹ ≈ 7.13 km h⁻¹	7 km h⁻¹	below the limit, no slip
   II (banked 45°)	5.42 m s⁻¹ ≈ 19.5 km h⁻¹	18.5 km h⁻¹	below the limit, no slip

3. So. Both speeds are within the safe range. For Statement II, the minimum speed is vₘᵢₙ² = 19.6 × 0.8/1.2 = 13.1, so vₘᵢₙ ≈ 13 km h⁻¹, and 18.5 lies safely above it. Both statements are true, which is option (b).

The idea to keep. On a banked road with friction, the safe speeds form a range, from √(rg(tan θ-μ)/(1+μ tan θ)) up to √(rg(tan θ+μ)/(1-μ tan θ)). Always convert km/h to m/s before you compare.

Answer from the NTA final answer key. Solution by PrepWiser.