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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.]
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[Angle of banking and friction is unchanged in both the cases.]
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Verified Answer
The correct answer is:
4.2 m
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)
$\begin{aligned}
& \left(\frac{v^{\prime}}{v}\right)^2=\frac{R^{\prime}}{R} \\
& \Rightarrow(1.1)^2=\frac{R^{\prime}}{R} \\
& \Rightarrow \frac{(1.1)^2-1}{1}=\frac{R^{\prime}-R}{R}
\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{0.21}{1}=\frac{\left(R^{\prime}-R\right)}{R} \\ & \Rightarrow\left(R^{\prime}-R\right)=0.21 R=0.21 \times 20 \mathrm{~m}=4.2 \mathrm{~m}\end{aligned}$
$v^2 \propto R$
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)
$\begin{aligned}
& \left(\frac{v^{\prime}}{v}\right)^2=\frac{R^{\prime}}{R} \\
& \Rightarrow(1.1)^2=\frac{R^{\prime}}{R} \\
& \Rightarrow \frac{(1.1)^2-1}{1}=\frac{R^{\prime}-R}{R}
\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{0.21}{1}=\frac{\left(R^{\prime}-R\right)}{R} \\ & \Rightarrow\left(R^{\prime}-R\right)=0.21 R=0.21 \times 20 \mathrm{~m}=4.2 \mathrm{~m}\end{aligned}$
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