Question
Download Solution PDFA soil sample has a total unit weight of \(16.97 \text{KN}/\text{m}^3\) and a void ratio of 0.84 . Determine the moisture content.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
The following relationship will be used for solving the above problem
\(\gamma=\frac{G\gamma_w(1+w)}{1+e}\)
where e = void ratio, w = moisture content, G = Specific Gravity: \(\gamma\) is specific unit weight of soil; \(\gamma_w\) is the specific unit weight of water
Calculation:
Given: e = 0.84: \(\gamma=16.97 {kN}/{m}^3\); \(\gamma_w=9.81 {kN}/{m}^3\); G = 2.7
So, \(16.97=\frac{2.7\times9.81(1+w)}{1+0.84}\)
\(1.18=1+w\)
w = 0.18 = 18%
Last updated on Jul 1, 2025
-> JKSSB Junior Engineer recruitment exam date 2025 for Civil and Electrical Engineering has been rescheduled on its official website.
-> JKSSB JE exam will be conducted on 31st August (Civil), and on 24th August 2025 (Electrical).
-> JKSSB JE application form correction facility has been started. Candidates can make corrections in the JKSSB recruitment 2025 form from June 23 to 27.
-> JKSSB JE recruitment 2025 notification has been released for Civil Engineering.
-> A total of 508 vacancies has been announced for JKSSB JE Civil Engineering recruitment 2025.
-> JKSSB JE Online Application form will be activated from 18th May 2025 to 16th June 2025
-> Candidates who are preparing for the exam can access the JKSSB JE syllabus PDF from official website of JKSSB.
-> The candidates can check the JKSSB JE Previous Year Papers to understand the difficulty level of the exam.
-> Candidates also attempt the JKSSB JE Mock Test which gives you an experience of the actual exam.