In a deposit of silt, the water table originally at a depth of 1 m below ground level was lowered to 3 m below ground level and got capillary saturated. The saturated unit weight of soil is $20 k N / m^{3} .$ The change in effective pressure at a depth of 2 m is $(A s s u m e γ_{w} = 10 k N / m^{3})$

10 k N / m^{2}

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20 k N / m^{2}

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30 k N / m^{2}

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10 k N / m^{2}

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Solution

The correct option is B $20 k N / m^{2}$

As soil is capillary saturated after lowering of W.T

$γ = γ_{s a t} \Rightarrow Δ σ = 0$ (No change in total stress)

$∴ Δ σ^{'} = Δ σ - Δ u (Δ σ = 0)$

Due to lowering of W.T, $@ 2 m \to Δ u = - γ_{w} \times 1 = - 10$

Due to capillary; $@ 2 m \to Δ u = - γ_{w} \times 1 = - 10$

$∴ T o t a l, Δ u = - 20 k P a$

$∴ Δ σ^{'} = - (- 20) = 20 k P a$

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In a deposit of silt, the water table originally at a depth of 1 m below ground level was lowered to 3 m below ground level and got capillary saturated. The saturated unit weight of soil is 20kN/m3. The change in effective pressure at a depth of 2 m is (Assume γw=10kN/m3)