Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 27 (Q.No. 23)
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
Discussion:
3 comments Page 1 of 1.
Navya_j said:
4 years ago
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%.
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%.
(9)
Sharma said:
7 years ago
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%.
Yd = 1600,
Y = (0.18+0.02)/10^-4 = 2000.
So w = (Y/Yd)-1,
(2000/1600) -1 = 0.25 i.e 25%.
(5)
Santu naskar said:
8 years ago
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%
w.c = 0.04 * 100/0.16 = 25%
(1)
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