The intensity of active earth pressure at a depth of 10 metres in dry cohesionless sand with an angle of internal friction of 30° and with a weight of 1.8 t/m³, is

ANS:C - 6 t/m²

The Pressure intensity at any depth "h" is;

P = Ka*Y*h --------> (1)
Ka = (1-sinφ)/(1+sinφ), φ = 30° & sin(30°) = 1/2
Ka = (1-1/2)/(1+1/2) = (1/2)/(3/2) = 1/3.
Ka = 1/3.
Y = 1.8 t/m3 & h = 10 m.

Put all these into equation (1).
So, (1) =>

P = (1/3)*1.8*10,
P = 0.6 * 10,
P = 6 t/m2.

So, [C] is Correct Option. The intensity of active earth pressure at H= 10 m is given by the formula,

Pa = KaYH.

The angle of friction is used to calculate coeff.of active earth pressure,

Ka = (1-sinφ)/(1+sinφ) = (1-sin30)/(1+sin30) = 0.3333333.
Y = 1.8 t/m3.
H = 10m.

Pa = 0.333333*1.8*10 = 6 t/m2. $ = 30deg.
k = (1-sin$) / (1+sin$).
k = 0.3333333.
y = 1.8.
h = 10.

I/p = YH.
I/p = 0.3333333*1.8*10 = 5.999999~6.