Estimate the earthwork quantity by traphezoidal formula method, for the construction of an approach road, whose length is 500m, base width of embankment = 10m, height of embankment = 70 cm and side slope = 1H : 2V.

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  1. 4500 cu. m
  2. 5000 cu. m
  3. 3377.5 cu. m
  4. 2500 cu. m

Answer (Detailed Solution Below)

Option 3 : 3377.5 cu. m
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Explanation:

F1 SSC Priyas 3 9 24 D1

Given Data:

Length of road (L) = 500 m

Base width of embankment (B) = 10 m

Height of embankment (H) = 0.70 m

h = Distance between the cross-sections = 0.7/2 = 0.35

Side slope = 1H:2V

Top width= 10- 2x0.35=9.3 m

\(\text{Area at Top } (A_1): \\ \text{Area } A_1 = \text{Top width} \times \text{Length} \\ A_1 = 9.3 \times 500 = 4650 \, \text{m}^2 \)

\(\text{Area at Bottom } (A_2): \\ \text{Area } A_2 = \text{Base width} \times \text{Length} \\ A_2 = 10 \times 500 = 5000 \, \text{m}^2 \)

\(\text{Mean Area } (A_M): \\ \text{Area } A_M = \frac{A_1 + A_2}{2} \times \text{length of road} \\ A_M = \frac{9.3 + 10}{2} \times 500 = 4825 \, \text{m}^2 \)

\(\text{Total Volume Calculation } (V): \\ \text{Volume } V = \frac{h}{2} \times \left[ A_1 + 2 \times A_M + A_2 \right] \\ V = \frac{0.35}{2} \times \left[ 4650 + 2 \times 4825 + 5000 \right] = 3377.5 \, \text{m}^3 \text{ or cu. m} \)

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