Q: The optimum (safe) speed on a road of radius r banked at angle θ, neglecting friction, is:

v_0 = \sqrt{rg\tan\theta}. For a frictionless banked road, v₀ = √(rg tanθ).

Q: The angle of banking θ required for a vehicle moving at speed v on a curve of radius r is:

\theta = \tan^{-1}\!\left(\dfrac{v^2}{rg}\right). From tanθ = v²/rg, the banking angle is θ = tan⁻¹(v²/rg).

Q: The maximum safe speed on a road of radius r banked at angle θ with coefficient of friction μ is:

v_{max} = \sqrt{\dfrac{rg(\tan\theta + \mu)}{1 - \mu\tan\theta}}. When friction acts down the incline (maximum speed): v_max = √[rg(tanθ + μ)/(1 − μ tanθ)].

Q: The minimum safe speed on a banked road (radius r, angle θ, friction μ) is:

v_{min} = \sqrt{\dfrac{rg(\tan\theta - \mu)}{1 + \mu\tan\theta}}. When friction acts up the incline (minimum speed): v_min = √[rg(tanθ − μ)/(1 + μ tanθ)].

Question 1

Roads are banked at curves mainly to:

Accepted Answer

Provide the necessary centripetal force without depending on friction. Banking makes a component of the normal reaction provide the centripetal force, so the vehicle can turn safely even without friction.

Question 2

The optimum (safe) speed on a road of radius r banked at angle θ, neglecting friction, is:

Accepted Answer

v_0 = \sqrt{rg	an	heta}. For a frictionless banked road, v₀ = √(rg tanθ).

Question 3

The angle of banking θ required for a vehicle moving at speed v on a curve of radius r is:

Accepted Answer

heta = 	an^{-1}\!\left(\dfrac{v^2}{rg}ight). From tanθ = v²/rg, the banking angle is θ = tan⁻¹(v²/rg).

Question 4

On a frictionless banked road, the centripetal force is provided by:

Accepted Answer

The horizontal component of the normal reaction (N sinθ). N cosθ balances the weight, while the horizontal component N sinθ supplies the centripetal force.

Question 5

When a vehicle moves at exactly the optimum speed on a banked road:

Accepted Answer

No frictional force is needed, reducing wear on the tyres. At the optimum speed the horizontal component of N alone provides the centripetal force, so friction is not required.

Question 6

On a banked road with friction, a vehicle can safely negotiate the curve for speeds:

Accepted Answer

Between a minimum value v_min and a maximum value v_max. Friction allows a range of safe speeds from v_min (friction acts up the slope) to v_max (friction acts down the slope).

Question 7

The maximum safe speed on a road of radius r banked at angle θ with coefficient of friction μ is:

Accepted Answer

v_{max} = \sqrt{\dfrac{rg(	an	heta + \mu)}{1 - \mu	an	heta}}. When friction acts down the incline (maximum speed): v_max = √[rg(tanθ + μ)/(1 − μ tanθ)].

Question 8

The minimum safe speed on a banked road (radius r, angle θ, friction μ) is:

Accepted Answer

v_{min} = \sqrt{\dfrac{rg(	an	heta - \mu)}{1 + \mu	an	heta}}. When friction acts up the incline (minimum speed): v_min = √[rg(tanθ − μ)/(1 + μ tanθ)].

Question 9

In the "well of death" (death well), the centripetal force needed to keep the rider on the vertical wall is provided by:

Accepted Answer

The normal reaction of the wall (directed horizontally towards the centre). The wall pushes the rider horizontally inward; this normal reaction provides the centripetal force (N = mv²/r).

Question 10

In the well of death, the weight of the rider is balanced by:

Accepted Answer

The force of friction between the tyres and the wall. The vertical weight mg is balanced by the upward friction (μN); this gives the minimum speed v_min = √(rg/μ_s).

Question 11

The angle of banking of a road for a given speed and radius is:

Accepted Answer

Independent of the mass of the vehicle. Since tanθ = v²/rg, the banking angle does not depend on the mass of the vehicle.

Question 12

For a road of width w banked so that its outer edge is raised by height h, the angle of banking θ (for small θ) satisfies:

Accepted Answer

an	heta \approx \dfrac{h}{w}. The outer edge is raised by h over the width w, so tanθ ≈ h/w (for small banking angles).

Question 13

A curve of radius 90 m is banked at 45°. The optimum speed (g = 10 m/s²) is:

Accepted Answer

30 m/s. v₀ = √(rg tanθ) = √(90 × 10 × tan45°) = √(900 × 1) = 30 m/s.

Question 14

The angle of banking for a curve of radius 50 m to be safe at 10 m/s (g = 10 m/s²) is about:

Accepted Answer

tan⁻¹(0.2) ≈ 11.3°. tanθ = v²/rg = 10²/(50 × 10) = 100/500 = 0.2, so θ = tan⁻¹(0.2) ≈ 11.3°.

Question 15

A road of radius 20 m is banked at an angle whose tangent is 0.5. The optimum speed (g = 10 m/s²) is:

Accepted Answer

10 m/s. v₀ = √(rg tanθ) = √(20 × 10 × 0.5) = √100 = 10 m/s.

Question 16

A vehicle moves at 20 m/s on a curve of radius 80 m. The required angle of banking (g = 10 m/s²) is:

Accepted Answer

tan⁻¹(0.5) ≈ 26.6°. tanθ = v²/rg = 20²/(80 × 10) = 400/800 = 0.5, θ = tan⁻¹(0.5) ≈ 26.6°.

Question 17

A curve of radius 60 m is banked at 30° (tan30° ≈ 0.577). The optimum speed (g = 10 m/s²) is about:

Accepted Answer

≈ 18.6 m/s. v₀ = √(rg tanθ) = √(60 × 10 × 0.577) = √346 ≈ 18.6 m/s.

Question 18

In a well of death of radius 4 m, the coefficient of friction between the tyres and the wall is 0.4. The minimum speed to avoid slipping down (g = 10 m/s²) is:

Accepted Answer

10 m/s. v_min = √(rg/μ) = √(4 × 10/0.4) = √100 = 10 m/s.

Question 19

A rider in a well of death of radius 2.5 m needs a minimum speed when μ = 0.25 (g = 10 m/s²). This minimum speed is:

Accepted Answer

10 m/s. v_min = √(rg/μ) = √(2.5 × 10/0.25) = √100 = 10 m/s.

Question 20

In a well of death of radius 4.5 m, the minimum safe speed is found to be 15 m/s (g = 10 m/s²). The coefficient of friction is:

Accepted Answer

0.2. v_min = √(rg/μ) ⇒ μ = rg/v² = (4.5 × 10)/15² = 45/225 = 0.2.

Banking of Roads & Death Well — MH-CET Physics MCQs with Solutions

Questions with solutions