The intensity of passive earth pressure at a depth of 5 m in a cohesion less soil with an angle of internal friction of 30- when water rises to ground level is kN-m2- Take saturated unit weight of sand as 21 kN-m3and -w - 9-81 kN-m3-

Coefficient of passive earth pressure, k_P = 1 + sinϕ1 sinϕ = 1 + sin 30^∘1 sin 30^∘ = 3 Intensity of passive earth pressure, P_p = k_p γ h + γ_w h = k_p (γ ...

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Standard XII

Physics

Ideal Gas Equation

The intensity...

Question

The intensity of passive earth pressure at a depth of 5 m in a cohesion less soil with an angle of internal friction of $30^{\circ}$ when water rises to ground level is $k N / m^{2}$ . Take saturated unit weight of sand as $21 k N / m^{3}$ and $γ_{w} = 9.81 k N / m^{3}$ .

216.9

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Solution

The correct option is A 216.9
Coefficient of passive earth pressure,
$k_{P} = \frac{1 + sin ϕ}{1 - sin ϕ} = \frac{1 + sin 30^{\circ}}{1 - sin 30^{\circ}} = 3$
Intensity of passive earth pressure,
$P_{p} = k_{p} γ h + γ_{w} h$
$= k_{p} (γ_{s a t} - γ_{w}) h + γ_{w} h$
$= 3 \times (21 - 9.81) \times 5 + 9.81 \times 5$
$= 216.9 k N / m^{2}$

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The intensity of passive earth pressure at a depth of 5 m in a cohesion less soil with an angle of internal friction of 30∘ when water rises to ground level is kN/m2. Take saturated unit weight of sand as 21kN/m3and γw=9.81kN/m3. 216.9

The correct option is A 216.9Coefficient of passive earth pressure, kP=1+sinϕ1−sinϕ=1+sin30∘1−sin30∘=3 Intensity of passive earth pressure, Pp=kpγh+γwh =kp(γsat−γw)h+γwh =3×(21−9.81)×5+9.81×5 =216.9kN/m2

The intensity of passive earth pressure at a depth of 5 m in a cohesion less soil with an angle of internal friction of $30^{\circ}$ when water rises to ground level is $k N / m^{2}$ . Take saturated unit weight of sand as $21 k N / m^{3}$ and $γ_{w} = 9.81 k N / m^{3}$ .

216.9