Question

If the super elevation of the highway provided is zero, then the design speed of highway having a curve of 200 m and coefficient of friction 0.10 is

• A. 40 kmph
• B. 50 kmph
• C. 55 kmph
• D. 60 kmph

Hint:

Remember the formula for super-elevation and its components. Pay close attention to units, converting between m/s and kmph when necessary. In design problems, always consider safety factors and practical considerations, but for objective questions, calculate based on the given formula and parameters.

Answer 1

The Correct Option is B

Step 1: Understand the formula for design speed on a horizontal curve.
The design speed ($V$) on a horizontal curve is governed by the super-elevation ($e$), the coefficient of lateral friction ($f$), and the radius of the curve ($R$). The standard formula for super-elevation is: $e + f = \frac{V^2}{gR}$ where: $V$ = design speed in m/s $g$ = acceleration due to gravity (approximately 9.81 m/s$^2$) $R$ = radius of the horizontal curve in meters
Step 2: Apply the given conditions to the formula.
We are given:

• Super elevation ($e$) = 0 (zero)
• Radius of the curve ($R$) = 200 m
• Coefficient of friction ($f$) = 0.10

Substitute these values into the formula: $0 + 0.10 = \frac{V^2}{9.81 \times 200}$ $0.10 = \frac{V^2}{1962}$
Step 3: Solve for $V$ in m/s.
$V^2 = 0.10 \times 1962$ $V^2 = 196.2$ $V = \sqrt{196.2}$ $V \approx 14.007$ m/s
Step 4: Convert the design speed from m/s to kmph.
To convert speed from m/s to kmph, multiply by $\frac{18}{5}$ (or 3.6). $V_{kmph} = V_{m/s} \times 3.6$ $V_{kmph} = 14.007 \times 3.6$ $V_{kmph} \approx 50.4252$ kmph
Step 5: Compare with the given options.
The calculated design speed is approximately 50.4 kmph. Looking at the options: (1) 40 kmph (2) 50 kmph (3) 55 kmph (4) 60 kmph The closest option to the calculated value is 50 kmph. $$\boxed{\text{50 kmph}}$$