# 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%

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%.