The radius of curvature of a road at a certain turn is 50m . The width of the road is 10m and its outer edge is 1.5m higher than the inner edge. The safe speed for such an inclination will be

The correct option is 8.6 m/s.

Understanding the Concept of Banking of Roads:
When a vehicle takes a turn on a circular path, it requires a centripetal force directed toward the center of the curve. On a flat road, this force is provided solely by the friction between the tires and the road surface, which may not always be reliable. To ensure safe turning at higher speeds, the outer edge of the road is raised above the inner edge. This inclination is called the banking of the road, and the angle it makes with the horizontal is the banking angle ($\theta$).

For a road banked at an angle $\theta$ with a radius of curvature $r$, the optimum safe speed $v$ (where no frictional force is needed to sustain the circular motion) is given by the formula:

$v=\sqrt{r·g·tan\left(\theta \right)}$

where:
• $r$ is the radius of curvature of the road,
• $g$ is the acceleration due to gravity (approximately $9.8\text{m/s}{2}^{}$),
• $\theta$ is the angle of banking.

Step-by-Step Derivation & Calculation:

Step 1: Identify the given values from the problem.
• Radius of curvature, $r=50\text{m}$
• Width of the road (hypotenuse of the incline triangle), $w=10\text{m}$
• Height of the outer edge above the inner edge, $h=1.5\text{m}$

Step 2: Find the trigonometric components for the banking angle $\theta$.
From the geometry of the inclined road, we can represent the cross-section as a right-angled triangle where the height is $h$ and the hypotenuse is the width $w$:

$sin\left(\theta \right)=\frac{h}{w}=\frac{1.5}{10}=0.15$

For small angles of inclination, the approximation $tan\left(\theta \right)\approx sin\left(\theta \right)$ is highly accurate. Let us use:

$tan\left(\theta \right)\approx 0.15$

Step 3: Calculate the safe speed $v$.
Substitute the values into the formula:

$v=\sqrt{50·9.8·0.15}$

Simplify the expression inside the square root:

$50·9.8=490$

$490·0.15=73.5$

Now, find the square root of the result:

$v=\sqrt{73.5}\approx 8.57\text{m/s}$

Rounding to one decimal place gives:

$v\approx 8.6\text{m/s}$

Thus, the safe speed for the inclination is 8.6 m/s.