Earthwork in railway embankment — prismoidal volume with varying height: A railway embankment 150 m long has formation width 11 m and side slopes 2 horizontal : 1 vertical. Ground falls 1 in 50 along the alignment. Formation falls 1 in 150 (same direction). If height at zero chainage is 0.5 m, compute the earthwork by the prismoidal formula.

Introduction / Context:
For linear earthworks with linearly varying cross-sections, the prismoidal formula gives an accurate volume using areas at the two ends and the mid-section. In rail embankments, cross-sectional area depends on height, formation width, and side slope.

Given Data / Assumptions:

• Length L = 150 m.
• Formation width b = 11 m; side slope s = 2 (H:V = 2:1).
• Ground fall = 1/50; formation fall = 1/150 (same direction).
• Height at start (x = 0) = h0 = 0.5 m.

Concept / Approach:

Height relative to ground increases with distance at the rate (ground fall − formation fall). Cross-sectional area for an embankment is A = bh + sh^2. Apply the prismoidal formula V = L/6 * (A0 + 4Am + A1).

Step-by-Step Solution:

Verification / Alternative check:

End-area method gives V ≈ L * (A0 + A1)/2 = 150(6 + 40)/2 = 3450 m³; prismoidal correction subtracts 200 m³ to account for curvature of area-height relation, yielding 3250 m³ as above.

Why Other Options Are Wrong:

• 3225 and 3275 m³ are near-misses not matching exact prismoidal computation.
• 3300 m³ is higher; likely based on average area without correction.

Common Pitfalls:

• Using bh + 2sh^2 instead of bh + s*h^2 (the standard derivation already accounts for both sides).
• Forgetting the difference of ground and formation gradients when computing height variation.

Final Answer:

3250 m3.