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Standard XII

Physics

Net Acceleration for Non Uniform Circular Motion

The coefficie...

Question

The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m without skidding is $(a s s u m e g = 10 m s^{- 2})$

$40 m s^{- 1}$

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$20 m s^{- 1}$

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$15 m s^{- 1}$

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$10 m s^{- 1}$

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Solution

The correct option is D
$10 m s^{- 1}$

$v = \sqrt{μ r g} = \sqrt{0.25 \times 40 \times 10} = 10 m / s$

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The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m without skidding is (assume g=10ms−2)

The correct option is D 10 ms−1 v=√μrg=√0.25×40×10=10 m/s

The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m without skidding is $(a s s u m e g = 10 m s^{- 2})$

The correct option is D
$10 m s^{- 1}$

$v = \sqrt{μ r g} = \sqrt{0.25 \times 40 \times 10} = 10 m / s$