For a district road with a bituminous pavement, having a horizontal curve of radius 1000 m and a design speed of 75 km/h, what is the appropriate super-elevation to provide?

Introduction / Context:Super-elevation offsets part of the lateral acceleration on a curve. IRC practice commonly uses e = V^2 / (225 * R) when V is in km/h and R in metres, with e expressed as a fraction. This makes it straightforward to compute the required cross slope for a given design speed and radius.

Given Data / Assumptions:

• V = 75 km/h.
• R = 1000 m.
• Use e = V^2 / (225 * R), and compare with practical limits (e.g., maximum e for terrain/class).

Concept / Approach:Plug the values into the standard expression to obtain e as a fraction, then convert to a 1 in N form for the options. Also ensure it does not exceed the adopted maximum for the road category and terrain.

Step-by-Step Solution:Compute e = V^2 / (225 * R).V^2 = 75^2 = 5625.Denominator = 225 * 1000 = 225000.e = 5625 / 225000 = 0.025 = 2.5%.Convert to 1 in N: N = 1 / 0.025 = 40 ⇒ 1 in 40.

Verification / Alternative check:e = 2.5% is well within typical maximum limits for bituminous roads in this speed and radius range, confirming feasibility.

Why Other Options Are Wrong:

• 1 in 50, 1 in 60, 1 in 70: lower e than required; would not sufficiently counteract lateral acceleration at 75 km/h on R = 1000 m.
• None of these: incorrect because 1 in 40 matches the computation.

Common Pitfalls:

• Mistaking units (using V in m/s in the 225-formula set in km/h).
• Ignoring terrain-based caps (not an issue here since e is modest).

Final Answer:1 in 40 (2.5%)