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If the banking angle of curved road is given by $\tan^{- 1} (\frac{3}{5})$ and the radius of curvature of the road is 6 m, then the safe driving speed should not exceed ( g = 10 m/s^-2)

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86.4 km/h

43.2 km/h

21.6 km/h

30.4 km/h

answer is C.

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Detailed Solution

$tanθ = \frac{v^{2}}{rg} \frac{3}{5} = \frac{v^{2}}{6 \times 10} v = 6 m / s = \frac{6 \times 18}{5} = 21.6 kmph$

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