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Soil Mechanics and Foundation Engineering - Engineering

Q1:

The weight of a pycnometer containing 400 g sand and water full to the top is 2150 g. The weight of pycnometer full of clean water is 1950 g. If specific gravity of the soil is 2.5, the water content is

A 5%

B 10%

C 15%

D 20%

ANS:D - 20%

[{Wt of sample/(difference of big weight)} *(G-1)/G] - 1.

1. wt of sample + pycnometer + water.
2. Wt of water + pycnometer.

Water content = [{wt of sample/(difference of both weights)*((G-1)/G)} - 1].

WC= [{400/(2150-1950)*((2.5-1)/2.5) } - 1].
= {(400/200)*1.5/2.5} - 1.
6/5 - 1 = 0.2.
Since WC in percentage; So, 0.2*100= 20%. Let us assume that the weight of the empty pycnometer is 100 gm, i.e M1=100 gm.
According to the question, the weight of sand is 400 gm then, M2=100 + 400 = 500 gm.
Also, M3= 2150 gm and M4=1950 gm.
wc ={(M2-M1)/(M3-M4)*(Gs-1)/Gs}-1.
= 1.2 - 1,
= 0.2 ,
wc = 20%. W1 =w of pycnometer empty.
W2 = w of pycnometer +soil of the sample.
W3 = w of soil+water+pyconometer.
W4 = w of water+pyconometer.(full of water).

Assume the weight of pycnometer is 100 because , its CONSTANT in w1 w2 w3 and w4
So, w1=100 and w2=100+400 = 500,
w4 and w3 is given;
wc ={(w2-w1)/(w3-w4)*(Gs-1)/Gs}-1.
wc = 20%.