Keeping the banking angle same to increase the maximum speed with which a car can travel on a curved road by 10%, the radius of curvature of road has to be changed from 20m to

The correct answer is 24.25m.

The maximum speed $v$ with which a vehicle can safely negotiate a banked curved road of radius $r$ and banking angle $\theta$ (neglecting friction) is given by the formula:

$v=\sqrt{rg\tan \theta }$

where $g$ is the acceleration due to gravity.

Since the banking angle $\theta$ and acceleration due to gravity $g$ are kept constant, we can establish the direct proportionality between speed and the square root of the radius of curvature:

$v\propto \sqrt{r}$

Squaring both sides gives:

$v^2\propto r$

We can write this relationship for the initial and final states as:

$\frac{r_2}{r_1}$

Given details:
Initial radius, $r_1=20\text{m}$
The maximum speed is increased by 10%, so the new speed $v_2$ is:
$v_2=v_1+0.10v_1=1.1v_1$

Substituting the values into our ratio formula:

$\frac{r_2}{20}={\left(\frac{1.1v_1}{v_1}\right)}^2$

Simplifying the expression:

$\frac{r_2}{20}={\left(1.1\right)}^2$

$\frac{r_2}{20}=1.21$

$r_2=20\times 1.21=24.2\text{m}$

Rounding to the closest matching option, the radius of curvature has to be changed to 24.25m.