The coefficient of static friction between the road and tyres of a car is 0.4. The maximum permissible speed of the car is 10 ms^-1 on curved unbanked road. Then the maximum radius of curvature of the road is (acceleration due to gravity = 10ms^-1)

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$10 \sqrt{5} m$

25 m

$20 \sqrt{2} m$

30 m

answer is B.

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Detailed Solution

The necessary centripetal force is provided by friction force

So, $\frac{{mV}^{2}}{r} = μ mg$

$\Rightarrow V = \sqrt{μ rg}$

$\Rightarrow 10 = \sqrt{0. 4 × r × 10}$

$\Rightarrow 100 = 4r$

$\Rightarrow r = 25 m$

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