At a point through cross section in a mass of saturated and homogenous soil, seeping water follows a path inclined at $10^{\circ}$ below to the horizontal. The hydraulic gradient is 0.5, void ratio and specific gravity of grains are 0.9 and 2.75 respectively. The resultant body force per unit volume of cylinder soil element is $(γ_{w} = 10 k N / m^{3})$

10.5 k N / m^{3}

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13 k N / m^{3}

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11.213 k N / m^{3}

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9.21 k N / m^{3}

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Solution

The correct option is C $11.213 k N / m^{3}$
$∴ \sum V = W - B = W^{1}$

Given :
$G = 2.75$
$e = 0.9$

$∴ γ^{1} = (\frac{G - 1}{1 + e}) . γ_{w}$

$γ^{1} = (\frac{2.75 - 1}{1 + 0.9}) \times 10 = 9.21 k N / m^{3}$

$∴ \frac{R}{V} = \sqrt{{9.21}^{2} + 5^{2} + 2 \times 9.21 \times 5 \times cos 80^{\circ}}$

$\frac{R}{V} = 11.213 k N / m^{3}$

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The correct option is C 11.213kN/m3∴∑V=W−B=W1 Given : G=2.75 e=0.9 ∴γ1=(G−11+e).γw γ1=(2.75−11+0.9)×10=9.21kN/m3 ∴RV=√9.212+52+2×9.21×5×cos80∘ RV=11.213kN/m3