- Find the moisture content, using dry weight per trial based on the table. - Solve the mean, sd, and rsd of the measurements. DescriptionTrial 1Trial 2Trial 3Trial 4Weight of empty watch glass (g)38.115138.147438.947338.7400Weight of watch glass and soil sample (g)90.144085.755486.698085.1235Weight of watch glass and dry soil67.017064.191565.111063.8859

For Trial 1: Weight of empty watch glass, say W1 (g) = 38.1151 Weight of watch glass and soil…

Answered: - Find the moisture content, using dry…

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter7: Permeability

Section: Chapter Questions

Problem 7.23P

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PLS ANSWER ASAPA WET OF SOIL SAMPLE HAS A VOLUME OF 4.85 x 10^-4 m3 AND WEIGHS 8.5N. AFTER OVEN DRYING, THE WEIGHT REDUCES TO 7.5N. IF sg = 2.7DETERMINE THE UNIT WEIGHT. DETERMINE THE MOISTURE CONTENT. DETERMINE THE DEGREE OF SATURATIONAnswer in 4 decimal places with unit
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 25 and the in situ moisture content is 50%. Determine the liquidity index. a.0.91 b.1.25 c.0.56 d.0.59 With FBD if necessary
Calculate the coefficient of permeability of a soil sample of 6 cm is height and 50cm2 in cross sectional area, if the quantity of water equal to 450ml passed down in 10minutes under an effective constant head of 40cm. On oven drying, the test specimen weigh 495g. Take Gs as 2.65, calculate the seepage velocity of water during test.
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 20 and the in situ moisture content is 35%. Determine the liquidity index. a.0.56 b.0.91 c.1.25 d.0.59
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 25 and the in situ moisture content is 45%. Determine the liquidity index. a. 0.56 b. 0.91 c. 1.25 d. 0.59
A WET OF SOIL SAMPLE HAS A VOLUME OF 4.85 x 10^-4 m3 AND WEIGHS 8.5N. AFTER OVEN DRYING, THE WEIGHT REDUCES TO 7.5N. IF sg = 2.7DETERMINE THE UNIT WEIGHT. DETERMINE THE MOISTURE CONTENT. DETERMINE THE DEGREE OF SATURATIONAnswer in 4 decimal places with unit
41 The data from a falling head test is shown. Diameter of standpipe: 4 cm. Initial Head: 92.9 cm. Duration of test: 12 min. Length of soil sample: 33.27 cm. Diameter of permeameter: 43 cm. If the soil permeability is 0.0028 cm/min, determine the final head (in cm.) of the permeability test. Round off to two decimal places.
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL=25 and the in situ moisture content is 45%. Determine the liquid limit. O a. 41 O b. 47 O c. 43 O d. 50
Subj- civil Q. An undisturbed soil sample that is cylindrical in shape and measures 3 in (76.2 mm) in diameter by 6 in (152.4 mm) in length after oven drying weighs 2.57 lb (1.166 kg bulk) as opposed to 3.04 lb (1.38 kg mass). What are the densities and the weights in (a) wet and dry units? Additionally, find out the moisture content.
For a falling-head permeability test, the following are given:length of specimen = 380 mmarea of specimen = 6.5 cm2k = 0.175 cm/minWhat should be the area of the standpipe for the head to dropfrom 650 cm to 300 cm in 8 min?
2) Using the following results: weight of dried pan and sand Is 1.0376 g. weight of pan and liquid sample is 4.6274 g. and weight of the pan and dried sample is 1.7321 g. What was the moisture content of the sample and what is the percent total solids?
pls ans soon as posssible During a constant head permeability test on a sand sample, 260ml of water was collected in 2 min. If the sample had a length of 100 mm, a diameter of 40 mm and a maintained head of 200 mm, what is the coefficient of permeability?

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Description	Trial 1	Trial 2	Trial 3	Trial 4
Weight of empty watch glass (g)	38.1151	38.1474	38.9473	38.7400
Weight of watch glass and soil sample (g)	90.1440	85.7554	86.6980	85.1235
Weight of watch glass and dry soil	67.0170	64.1915	65.1110	63.8859