Civil Engineering - UPSC Civil Service Exam Questions - Discussion

23. 

If a soil sample of weight 0.18 kg having a volume of 10-4 m3 and dry unit wt. of 1600 kg/m3 is mixed with 0.02 kg of water then the water content in the sample will be

[A]. 30%
[B]. 25%
[C]. 20%
[D]. 15%

Answer: Option B

Explanation:

No answer description available for this question.

Santu Naskar said: (May 31, 2017)  
The mass of soil solid =1600*10^-4= 0.16 then water content =0.18/0.16 -1=0.125 then mass of water in the sample =0.125*0.16=0.02 kg when again 0.02 kg water is mixed then total wt of water in the sample become =0.02 + 0.02 = 0.04.
w.c = 0.04 * 100/0.16 = 25%

Sharma said: (May 30, 2018)  
As Yd = Y/(1+w),
Yd = 1600,
Y = (0.18+0.02)/10^-4 = 2000.

So w = (Y/Yd)-1,
(2000/1600) -1 = 0.25 i.e 25%.

Navya_J said: (Apr 19, 2021)  
In qstn there is W_samlpe and V_sample so that we can find Yb.
So,
Yb = 0.18/10^-4.
= 1800 kg/m^3 ..and there adding water that is 0.02 kg so (0.02/10^-4)=200.
Yb =1800+200=2000 kg/m^3,
From the formula of;
Yd =Yb/(1+w),
1600=2000/(1+w),
w= 0.25,
= 25%.

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