A car is moving along a circular path having coefficient of friction 0.5 and radius of curvature 16.2 m. Then, the maximum velocity of the car that can travel without falling outward is (Acceleration due to gravity $= 10 m s^{- 2}$ )

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$18 m s^{- 1}$

$32.4 k m h^{- 1}$

$18 k m h^{- 1}$

$32.4 m s^{- 1}$

answer is B.

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Detailed Solution

Safe turning velocity for a car on level road is
$v_{\max} = \sqrt{u r g} = \sqrt{0.5 \times 1.62 \times 10}$
$= \sqrt{81} = 9 m s^{- 1} = 9 \times \frac{18}{5} k m / h = 32.4 k m / h$

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