A road is 10 text{ m} wide. Its radius of cur | vedclass

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(D) For a banked road, the optimum velocity $v$ is given by the formula $	an \alpha = \frac{v^2}{Rg}$.Here, $\alpha$ is the banking angle, $R$ is the

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$8.5$

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(D) For a banked road, the optimum velocity $v$ is given by the formula $	an \alpha = \frac{v^2}{Rg}$.Here, $\alpha$ is the banking angle, $R$ is the radius of curvature, and $g$ is the acceleration due to gravity.From the geometry of the road, $	an \alpha = \frac{h}{w}$, where $h$ is the height difference between the outer and inner edges and $w$ is the width of the road.Equating the two expressions for $	an \alpha$: $\frac{v^2}{Rg} = \frac{h}{w}$.Rearranging for $v$: $v = \sqrt{\frac{Rgh}{w}}$.Given: $R = 50 	ext{ m}$, $h = 1.5 	ext{ m}$, $w = 10 	ext{ m}$, and $g = 9.8 	ext{ m/s}^2$.Substituting the values: $v = \sqrt{\frac{50 	imes 9.8 	imes 1.5}{10}} = \sqrt{5 	imes 9.8 	imes 1.5} = \sqrt{73.5} \approx 8.57 	ext{ m/s}$.Comparing this with the given options, the most suitable velocity is $8.5 	ext{ m/s}$.

$A$ road is $10 \text{ m}$ wide. Its radius of curvature is $50 \text{ m}$. The outer edge is above the inner edge by a distance of $1.5 \text{ m}$. This road is most suited for the velocity $[g = 9.8 \text{ m/s}^2]$: (in $\text{ m/s}$)

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