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

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

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

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

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

answer is B.

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Detailed Solution

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

$= \sqrt{μ r g} = \sqrt{0.1 \times (3 m) \times (10 m s^{- 2})} = \sqrt{3} m s^{- 1}$

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