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Uniform circular motion on level and banked roads: NEET MCQs with answers

50 questions on this topic of Laws of Motion (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A car is moving with a constant speed v on a rough circular banked road of banking angle θ. If the car is moving at the optimum speed v₀ = √(R · g · tan θ), the frictional force acting on the tyres of the car is:

  1. Directed up the inclined banking plane
  2. Zero
  3. Directed down the inclined banking plane
  4. Directly proportional to the mass of the car and directed horizontally inward
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Answer: (B) Zero

At the optimum speed v₀ = √(R · g · tan θ), the horizontal component of the normal reaction alone provides the required centripetal force (N sin θ = m v₀² / R) while the vertical component balances the weight (N cos θ = mg). Thus, no tendency to slip exists, and the frictional force is zero.

medium

2. A road has a radius of curvature of 80 m. What should be the banking angle θ so that a car travelling at 72 km/h can negotiate the bend safely without relying on friction? (Take g = 10 m/s²)

  1. tan⁻¹(0.4)
  2. tan⁻¹(0.25)
  3. tan⁻¹(0.5)
  4. tan⁻¹(0.8)
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Answer: (C) tan⁻¹(0.5)

The design speed is v = 72 km/h = 72 × (5/18) = 20 m/s. The optimum banking angle without friction is given by tan θ = v² / (R · g). Substituting the values: tan θ = (20)² / (80 × 10) = 400 / 800 = 0.5. Hence, θ = tan⁻¹(0.5).

hard

3. A racing track of radius 100 m is banked at an angle of 45°. If the coefficient of friction between the wheels and the track is 0.2, the ratio of maximum safe speed to minimum safe speed is (take g = 10 m/s²):

  1. √1.5
  2. √2
  3. 1.5
  4. 2.0
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Answer: (C) 1.5

The maximum speed is v_max = √[r·g(tan θ + μ)/(1 - μ tan θ)] and minimum speed is v_min = √[r·g(tan θ - μ)/(1 + μ tan θ)]. For θ = 45°, tan 45° = 1. Thus v_max / v_min = √[((1 + 0.2)/(1 - 0.2)) / ((1 - 0.2)/(1 + 0.2))] = √[(1.2 / 0.8) / (0.8 / 1.2)] = √[(1.5) / (1/1.5)] = √(1.5²) = 1.5.

medium

4. A road curve of radius 40 m is banked at 45°. Taking g = 10 m/s², the speed at which a car can take the curve without relying on friction is:

  1. 6.32 m/s
  2. 28.3 m/s
  3. 20 m/s
  4. 40 m/s
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Answer: (C) 20 m/s

For a banked road without friction, tan θ = v²/(rg). With θ = 45°, v = √(rg) = √(40 × 10) = 20 m/s.

easy

5. A car takes a turn of radius 20 m on a level, unbanked road where the coefficient of static friction between the tyres and the road is 0.8. Taking g = 10 m/s², what is the maximum speed at which the car can negotiate the curve without skidding?

  1. 16 m/s
  2. 4 m/s
  3. 10 m/s
  4. 12.65 m/s
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Answer: (D) 12.65 m/s

For a level road, the maximum safe speed without skidding is given by v_max = √(μ_s · R · g). Substituting the given values: v_max = √(0.8 × 20 × 10) = √160 ≈ 12.65 m/s.

hard

6. A curved road of radius 50 m is banked at an angle θ = 37° (tan 37° = 0.75). The coefficient of static friction between the tyres and road is μ_s = 0.5. Taking g = 10 m/s², the maximum permissible speed of a vehicle on this road without skidding outward is:

  1. 38.2 m/s
  2. 24.5 m/s
  3. 20.0 m/s
  4. 31.6 m/s
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Answer: (D) 31.6 m/s

The maximum speed is given by v_max = √[ R · g · (tan θ + μ_s) / (1 - μ_s · tan θ) ]. Here, v_max = √[ 50 × 10 × (0.75 + 0.5) / (1 - 0.5 × 0.75) ] = √[ 500 × 1.25 / 0.625 ] = √[ 500 × 2 ] = √1000 ≈ 31.6 m/s.

medium

7. A car is moving around a curved horizontal track of radius R = 100 m. The coefficient of static friction is μ_s = 0.6. If g = 9.8 m/s², what is the ratio of the centripetal acceleration at maximum safe speed to the acceleration due to gravity?

  1. 0.36
  2. 1.67
  3. 0.78
  4. 0.60
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Answer: (D) 0.60

On a level road, maximum centripetal acceleration is a_c = v_max² / R = μ_s · g. Therefore, the ratio a_c / g = (μ_s · g) / g = μ_s = 0.60.

hard

8. A highway curve of radius 125 m is banked at an angle of 45°. If the coefficient of static friction between the tyres and the road is 0.2, what is the minimum safe speed at which a car can negotiate the curve without sliding down? (Take g = 10 m/s²)

  1. 35.4 m/s
  2. 28.9 m/s
  3. 15.0 m/s
  4. 25.0 m/s
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Answer: (B) 28.9 m/s

The minimum safe speed on a banked road with friction is given by v_min = √[R · g · (tan θ - μ) / (1 + μ · tan θ)]. Here, tan 45° = 1, μ = 0.2, R = 125 m, and g = 10 m/s². So, v_min = √[125 × 10 × (1 - 0.2) / (1 + 0.2)] = √[1250 × 0.8 / 1.2] = √[1000 / 1.2] = √833.3 ≈ 28.9 m/s.

medium

9. A curve of radius R on a frictionless road is banked at an angle θ. A car of mass m passes through this curve at the exact design speed v₀. The magnitude of the normal force exerted by the road on the car is:

  1. m · g / sin θ
  2. m · g · cos θ
  3. m · g · sin θ
  4. m · g / cos θ
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Answer: (D) m · g / cos θ

Since there is no vertical acceleration, the vertical forces balance: N cos θ = mg. Therefore, the normal force exerted by the road is N = mg / cos θ. (Its horizontal component N sin θ provides the centripetal force mv₀²/R).

medium

10. A road curve of radius 10 m is banked at 45°. Taking g = 10 m/s², the speed at which a car can take the curve without relying on friction is:

  1. 3.16 m/s
  2. 10 m/s
  3. 20 m/s
  4. 14.1 m/s
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Answer: (B) 10 m/s

For a banked road without friction, tan θ = v²/(rg). With θ = 45°, v = √(rg) = √(10 × 10) = 10 m/s.

medium

11. A road curve of radius 90 m is banked at 45°. Taking g = 10 m/s², the speed at which a car can take the curve without relying on friction is:

  1. 42.4 m/s
  2. 30 m/s
  3. 90 m/s
  4. 9.49 m/s
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Answer: (B) 30 m/s

For a banked road without friction, tan θ = v²/(rg). With θ = 45°, v = √(rg) = √(90 × 10) = 30 m/s.

medium

12. A vehicle of mass m moves along a banked road of radius r with banking angle θ at the optimum speed v₀ = √(r · g · tan θ). The magnitude of the normal reaction exerted by the road on the vehicle is:

  1. m·g / cos θ
  2. m·g · cos θ
  3. m·g · sin θ
  4. m·g / sin θ
Show answer and explanation

Answer: (A) m·g / cos θ

At the optimum speed v₀, there is no tendency to slip, so static friction is zero. Resolving forces vertically gives N cos θ = mg, which directly yields N = mg / cos θ.

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