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Byju's Answer

Standard XII

Physics

Applications of Radius of Curvature

The maximum v...

Question

The maximum velocity with which a driver must drive his car on a flat curved road of radius of curvature $150 m$ and coefficient of friction $0.6$ , to avoid the skidding off his car is (take $g = 10 m / s^{2}$ )

60 m / s

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50 m / s

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40 m / s

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30 m / s

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Solution

The correct option is D $30 m / s$
The maximum safe speed for a vehicle taking a turn on a curved banked road
is given by $v = \sqrt{(μ * r g)} = \sqrt{(0.6 * 150 * 10)} = 30 m / s$

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Similar questions

Q. The minimum velocity (in

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) with which a car driver must traverse a flat curve of radius

150 m

and coefficient of friction

0.6

to avoid skidding is

Q. The minimum velocity (in m/s) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is:

The maximum velocity (in $m s^{- 1}$ ) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

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Q. A car is moving with a speed of

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The maximum velocity with which a driver must drive his car on a flat curved road of radius of curvature 150m and coefficient of friction 0.6, to avoid the skidding off his car is (take g=10m/s2)

The correct option is D 30m/sThe maximum safe speed for a vehicle taking a turn on a curved banked roadis given by v=√(μ∗rg)=√(0.6∗150∗10)=30m/s

The maximum velocity with which a driver must drive his car on a flat curved road of radius of curvature $150 m$ and coefficient of friction $0.6$ , to avoid the skidding off his car is (take $g = 10 m / s^{2}$ )

The correct option is D $30 m / s$
The maximum safe speed for a vehicle taking a turn on a curved banked road
is given by $v = \sqrt{(μ * r g)} = \sqrt{(0.6 * 150 * 10)} = 30 m / s$