Question
Download Solution PDFThe natural water content of a soil sample is 40%; the specific gravity is 2.7 and the void ratio is 1.2; then the degree of saturation of the soil will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
For a given soil sample,
S × e = w × G
where
S = degree of saturation, e = void ratio, w = water content
G = specific gravity of the soil solid
Calculation:
Given,
⇒ w = 40 %, G = 2.7, e = 1.2
As, S × e = w × G
\(S = \frac{{0.40\times 2.7}}{{1.2}}\)
S = 0.9
In terms of percentage S = 90 %
Last updated on Jun 4, 2025
-> OSSC JE Preliminary Fiinal Answer Key 2025 has been released. Applicants can raise objections in OSSC JE answer key till 06 June 2025.
-> OSSC JE prelim Exam 2025 was held on 18th May 2025.(Advt. No. 1233/OSSC ).
-> The OSSC JE 2024 Notification was released for 759 vacancies through OSSC CTSRE 2024.
-> The selection is based on the written test. This is an excellent opportunity for candidates who want to get a job in the Engineering sector in the state of Odisha.
-> Candidates must refer to the OSSC JE Previous Year Papers to understand the type of questions in the examination.