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A road is banked at an angle of 30^(...

A road is banked at an angle of `30^(@)` to the horizontal for negotiating a curve of radius `10 sqrt(3)` m. At what velocity will a car experience no friction while negotiating the curve? Take `g=10 ms^(-2)`

A

54 kmph

B

75 kmph

C

36 kmph

D

18 kmph

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the velocity at which a car experiences no friction while negotiating a banked curve, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Data:** - Angle of banking, \( \theta = 30^\circ \) - Radius of the curve, \( r = 10\sqrt{3} \, \text{m} \) - Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) 2. **Understand the Forces Acting on the Car:** - The normal force \( N \) acts perpendicular to the surface of the road. - The gravitational force \( mg \) acts vertically downwards. - The car requires a centripetal force to negotiate the curve, which is provided by the horizontal component of the normal force. 3. **Resolve the Normal Force:** - The normal force can be resolved into two components: - \( N \cos \theta \) (vertical component) - \( N \sin \theta \) (horizontal component) 4. **Set Up the Equations:** - The vertical forces must balance: \[ N \cos \theta = mg \] - The horizontal forces provide the centripetal force: \[ N \sin \theta = \frac{mv^2}{r} \] 5. **Divide the Two Equations:** - Dividing the equation for the horizontal forces by the equation for the vertical forces, we get: \[ \frac{N \sin \theta}{N \cos \theta} = \frac{\frac{mv^2}{r}}{mg} \] - This simplifies to: \[ \tan \theta = \frac{v^2}{rg} \] 6. **Substitute Known Values:** - For \( \theta = 30^\circ \), \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). - Substitute \( r = 10\sqrt{3} \) and \( g = 10 \): \[ \frac{1}{\sqrt{3}} = \frac{v^2}{(10\sqrt{3})(10)} \] 7. **Solve for \( v^2 \):** - Rearranging gives: \[ v^2 = \frac{10\sqrt{3} \cdot 10}{\sqrt{3}} = 100 \] 8. **Calculate \( v \):** - Taking the square root gives: \[ v = \sqrt{100} = 10 \, \text{m/s} \] 9. **Convert to km/h:** - To convert \( v \) to kilometers per hour: \[ v = 10 \, \text{m/s} \times \frac{18}{5} = 36 \, \text{km/h} \] ### Final Answer: The velocity at which the car experiences no friction while negotiating the curve is **36 km/h**.
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