Soil Mechanics and Foundation Engineering

Q1: A soil has bulk density 2.30 g/cm3 and water content 15 per cent, the dry density of the sample, is

A 1.0 g/cm2

B 1.5 g/cm3

C 2.0 g/cm3

D 2.5 g/cm3

ANS:C - 2.0 g/cm3

Yd = Y/1+w.
= 2.4/1+.2.
= 2. The water content, W = 15%,
Bulk density, Ym = 2.30 gm/cc,
Dry density, Yd =?

We know that, Yd = 100/(100+W)*Ym.
= 100/(100+15)*2.30,
= 2.0 gm/cc We know that Bulk unit weigh[Yb] =Total weight/Total volume.
So, Yb = W/V =Ww + Ws/V.

NOW, Take common Ws, Hence Ws[Ww/ Ws + 1]/V.
NOW Ws / V = Yd [dry unit weight] .

Hence Yb = Yd [w = 1] because Ww/Ws = w[water content].
Now Yd = Yb /[1 + w] = 2.30/[1 + 0.15] =2 g/centi meter cube. Dry density = Grw/1+e.
Bulk density = Grw (1+w)/1+e.
rt = rd*(1+w).
rd = 2.3/1+.15 = 2gm/cubic centimeter.



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