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Q.

A motorcyclist moving with a velocity of 72 km/h on a flat road takes a turn on the road at a point, where the radius of curvature of the road is 20 m. The acceleration due to gravity is 10 m/s². In order to avoid sliding, he must bend with respect to the vertical plane by an angle

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a

$θ = \tan^{- 1} (4)$

b

$θ = \tan^{- 1} (6)$

c

$θ = \tan^{- 1} (2)$

d

$θ = 45 °$

answer is C.

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Detailed Solution

v = 72 km/h = 72 $\times \frac{5}{18}$ m/s = 20 m/s

tan $θ$ = $\frac{v^{2}}{R g}$

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