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

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(C) The maximum safe speed $v$ on a banked road is given by the formula $v = \sqrt{rg(	an 	heta + \mu) / (1 - \mu 	an 	heta)}$.Assuming the fricti

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$8.8$

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(C) The maximum safe speed $v$ on a banked road is given by the formula $v = \sqrt{rg(	an 	heta + \mu) / (1 - \mu 	an 	heta)}$.Assuming the friction coefficient $\mu$ and banking angle $	heta$ remain constant,the speed $v$ is directly proportional to the square root of the radius $r$,i.e.,$v \propto \sqrt{r}$.Let the initial radius be $r_1 = 20 \ m$ and initial speed be $v_1$. The new speed is $v_2 = v_1 + 0.20v_1 = 1.2v_1$.Since $v \propto \sqrt{r}$,we have $v_2 / v_1 = \sqrt{r_2 / r_1}$.Substituting the values: $1.2 = \sqrt{r_2 / 20}$.Squaring both sides: $1.44 = r_2 / 20$.$r_2 = 1.44 	imes 20 = 28.8 \ m$.The increase in the radius of curvature is $\Delta r = r_2 - r_1 = 28.8 \ m - 20 \ m = 8.8 \ m$.

$A$ 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 $20 \%$,without changing the angle of banking,the increase in the radius of curvature will be (Assume friction is same on the road). (in $m$)

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