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(B) The necessary centripetal force for a car on a curved unbanked road is provided by the static friction force between the tyres and the road.For sa

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$25 \,m$

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(B) The necessary centripetal force for a car on a curved unbanked road is provided by the static friction force between the tyres and the road.For safe turning,the centripetal force must be less than or equal to the maximum static friction force:$\frac{mV^2}{r} \leq \mu mg$To find the maximum radius $r$ for a given speed $V$,we use the limiting case:$\frac{mV^2}{r} = \mu mg$$r = \frac{V^2}{\mu g}$Given values are $V = 10 \,ms^{-1}$,$\mu = 0.4$,and $g = 10 \,ms^{-2}$.Substituting these values into the equation:$r = \frac{10^2}{0.4 	imes 10}$$r = \frac{100}{4}$$r = 25 \,m$Thus,the maximum radius of curvature is $25 \,m$.

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 a curved unbanked road. Then the maximum radius of curvature of the road is (acceleration due to gravity $= 10 \,ms^{-2}$)

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