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

For road safety on a banked road, the velocity of the car has following scaling relation with respect to the curvature:
\(v^2 \propto R\)
So,
\(v^2=C R---(1)\)
where \(C\) is a constant that involves the banking angle and friction.
If the new velocity is \(v^{\prime}=1.1 v\) as \(10 \%\) higher speed, then
\(v^{\prime 2}=C R^{\prime}---(2)\)
where, \(R^{\prime}\) is the new radius of curvature due to safely.
Now, take the ratio of equation (1) and (2)
\(\begin{aligned}
& \left(\frac{v^{\prime}}{v}\right)^2=\frac{R^{\prime}}{R} \\
& \Rightarrow(1.1)^2=\frac{R^{\prime}}{R} \\
& \Rightarrow \frac{(1.1)^2-1}{1}=\frac{R^{\prime}-R}{R}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{0.21}{1}=\frac{\left(R^{\prime}-R\right)}{R} \\ & \Rightarrow\left(R^{\prime}-R\right)=0.21 R=0.21 \times 20 \mathrm{~m}=4.2 \mathrm{~m}\end{aligned}\)