A car is driven on a banked road of radius of | vedclass

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(A) For a car on a banked road,the maximum safe speed $v$ is given by the relation $v = \sqrt{Rg \frac{\mu + 	an 	heta}{1 - \mu 	an 	heta}}$.Since

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$4.2$

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(A) For a car on a banked road,the maximum safe speed $v$ is given by the relation $v = \sqrt{Rg \frac{\mu + 	an 	heta}{1 - \mu 	an 	heta}}$.Since the angle of banking $	heta$ and the coefficient of friction $\mu$ remain unchanged,we have $v^2 \propto R$,or $v^2 = C R$,where $C$ is a constant.Let the initial speed be $v$ and the initial radius be $R = 20 \ m$.The new speed is $v' = v + 0.10v = 1.1v$.Using the relation $v^2 = CR$,we have $v'^2 = CR'$,where $R'$ is the new radius.Dividing the two equations: $\frac{v'^2}{v^2} = \frac{R'}{R}$.Substituting the values: $(1.1)^2 = \frac{R'}{R} \Rightarrow 1.21 = \frac{R'}{R}$.Therefore,$R' = 1.21 R = 1.21 	imes 20 \ m = 24.2 \ m$.The increase in the radius of curvature is $\Delta R = R' - R = 24.2 \ m - 20 \ m = 4.2 \ m$.

$A$ car is driven on a banked road of radius of curvature $20 \ m$ with maximum safe speed. In order to increase its safe speed by $10 \%$,the increase in the radius of curvature will be (Angle of banking and friction is unchanged in both the cases.) (in $m$)

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