Discussion :: GATE Exam Questions - Section 6 (Q.No.31)
|Mohan said: (Sep 4, 2015)|
|How to solve this question? Kindly give explanation.|
|Navneet said: (Nov 15, 2015)|
|(16*100)/17 = 94.12.|
|Fazil said: (May 9, 2018)|
|Thank you so much @Navneet.|
|Manish said: (Aug 6, 2018)|
Could you please give the full solution clearly?
|Vaish said: (Feb 23, 2019)|
|It should be 106.25 m3.|
|Sweetie said: (Oct 11, 2020)|
|V1/V2 = yd2/yd1.
V1/100 = 16/17.
V1 = 94.11m^3.
|Heramb said: (Oct 24, 2020)|
Please listen, Here, I solved in detail.
A] Dry density is given;
I) Borrow pit dry density =17Kn/m3
II) Embankment dry density =16Kn/m3
B] To find out weight of soil
I) Borrow pit
W=Dry density X volume
W= 17 X 1
W= 17KN/m3 ..[Takevolume=1.standardcase]
W=Dry density X volume.
W= 16 X 1.
W=16KN/m3..[Take Volume=1.standard case]
C]. To find out the volume for 1 m3.
VOLUME OF borrow pit= Weight of soil of Embankment /Dry density of borrow pit
VOLUME OF borrow pit= 16/17
1 m3 of Embankment =0.94012 m3 of Borrow Pit.
100 m3 of Embankment =[ x] m3 of Borrow Pit.
x *1 = 100 * 0.94012,
Therefore, volume is 94.012% of Borrow Pit.
Post your comments here:
Email : (optional)
» Your comments will be displayed only after manual approval.