Correct Answer - Option 1 : 10
5/3
Concept:
Determination of seepage discharge when the medium is isotropic:
Seepage discharge per meter length is given by, \(q = kh\frac{{{N_f}}}{{{N_d}}}\)
Where, h = Hydraulic head or head difference between upstream and downstream level or head loss through the soil
Nf = Total number of flow channels
Nd = Total number of equipotential drops
k = Coefficient of permeability
Calculation:
Given,
Permeability of sand in horizontal direction, KHS = 3 × 10-3 cm/sec,
Permeability of sand in vertical direction, KVS = 1 × 10-3 cm/sec
Permeability of clay in horizontal direction, KHC = 9 × 10-7 cm/sec,
Permeability of clay in vertical direction, KVC = 3 × 10-9 cm/sec
Equivalent permeability of sand, (Keq)sand = \(√ {{K_{HS}} × {K_{Vs}}} = √ {3 × {{10}^{ - 3}} × 1 × {{10}^{ - 3}}} \)
(Keq)sand = √3 × 10-3 cm/sec
Equivalent permeability of clay, \({\left( {{K_{eq}}} \right)_{sand}} = √ {{K_{HC}} × {K_{VC}}} = √ {9 × {{10}^{ - 7}} × 3 × {{10}^{ - 9}}} \)
(Keq)clay = 3√3 × 10-8 cm/sec
Seepage loss is given as, \(q = kh\frac{{{N_f}}}{{{N_d}}}\)
q ∝ K
\(\frac{{{q_{sand}}}}{{{q_{clay}}}} = \frac{{{{\left( {{K_{eq}}} \right)}_{sand}}}}{{{{\left( {{K_{eq}}} \right)}_{clay}}}}\)
⇒ \(\frac{{{q_{sand}}}}{{{q_{clay}}}} = \frac{{\sqrt 3 \times {{10}^{ - 3}}}}{{3\sqrt 3 \times {{10}^{ - 8}}}} = \frac{{{{10}^5}}}{3}\)