MHTCET 22 Apr 2025 Shift 2 Paper

Concept:For a banked road with friction, the maximum safety speed depends on the radius of curvature as $v_{\max} \propto \sqrt{r}$, provided the angle of banking and friction coefficient remain constant.Explanation:The maximum safety speed is given by:$$v_{\max} = \sqrt{r g \left(\frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta}\right)}$$Since $\theta$ and $\mu_s$ do not change, the factor inside the bracket is constant.Hence, $v_{\max}$ is directly proportional to $\sqrt{r}$.So, for two different radii and speeds:$$\frac{v_2}{v_1} = \sqrt{\frac{r_2}{r_1}}$$Squaring both sides gives:$$\frac{r_2}{r_1} = \left(\frac{v_2}{v_1}\right)^2$$Given initial radius $r_1 = 20\ \text{m}$, and the speed is increased by $20\%$:$$v_2 = 1.2\, v_1$$Therefore,$$r_2 = (1.2)^2 \times 20 = 1.44 \times 20 = 28.8\ \text{m}$$Increase in radius of curvature:$$\Delta r = r_2 - r_1 = 28.8 - 20 = 8.8\ \text{m}$$Answer:The required increase in the radius of curvature is $8.8\ \text{m}$, i.e. option C.