The coefficient of friction between the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding is then
(Take g= 10 $m s^{- 2}$ )

\sqrt{15} m s^{- 1}

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\sqrt{3} m s^{- 1}

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\sqrt{30} m s^{- 1}

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\sqrt{10} m s^{- 1}

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Solution

The correct option is B $\sqrt{3} m s^{- 1}$
Here, $r = 3 m, μ = 0.1, g = 10 m s^{- 2}$
The maximum speed with which a cyclist can take a turn without skidding is
$= \sqrt{μ r g}$
$= \sqrt{.01 \times (3 m) (10 m s^{- 2})}$

$= \sqrt{3} m s^{- 1}$

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Similar questions

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 $(a s s u m e g = 10 m s^{- 2})$

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The coefficient of friction between the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding is then(Take g= 10 ms−2)

The correct option is B √3ms−1Here,r=3m, μ=0.1,g=10 ms−2The maximum speed with which a cyclist can take a turn without skidding is=√μrg=√.01×(3m)(10ms−2)=√3ms−1

The coefficient of friction between the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding is then
(Take g= 10 $m s^{- 2}$ )

The correct option is B $\sqrt{3} m s^{- 1}$
Here, $r = 3 m, μ = 0.1, g = 10 m s^{- 2}$
The maximum speed with which a cyclist can take a turn without skidding is
$= \sqrt{μ r g}$
$= \sqrt{.01 \times (3 m) (10 m s^{- 2})}$

$= \sqrt{3} m s^{- 1}$