Question

A sample of soil was prepared by mixing a quantity of dry soil with 25% of water by mass. Find mass of the wet mixture required to produce cylindrical specimen of 15 cm depth and 5% air content. If specify gravity of solids is 2.68 and dia cylinder is 20 cm.

• A. 7200 gm
• B. 9260 gm
• C. 9060 gm
• D. 9300 gm

Answer 1

The correct option is B 9260 gm

Given data

$w=0.25$

$V=\frac{\pi }{4}\times {20}^2\times 15=4712.38c{m}^3$

$G=2.68$

$a_c=0.05$

as we know

$S\times e=wG$

$\therefore S+a_c=1$
$S=0.95$

$0.95\times e=0.25\times 2.68$

$e=0.7052$

$e=\frac{V_V}{V_S}=0.7052$

$V_S=\frac{V_T}{1+e}=2763.43c{m}^3$

$V_V=1948.77c{m}^3$

$V_a=97.43c{m}^3$

$V_w=1851.33c{m}^3$

$W_{Total}=W_W+W_S$

$W_{Total}={\rho }_W\times V_W+G\times {\rho }_W\times V_S$

$=1\times 1851.33+2.68\times 1\times 2763.43=9257.32gm$