Question
Download Solution PDFThe maximum dry density and optimum moisture content of a soil is given by 1.65 gm/cc and 20.5% respectively. What is the percentage of air content of soil at OMC, if the specific gravity of particles is given by 2.65?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFUsing the relation,
\({\gamma _d}_{max} = \frac{{G{\gamma _w}}}{{1 + e}}\)
\(1.65 = \frac{{2.65 \times 1}}{{1 + e}}\)
e = 0.606
Using the relation
se = wG
s = \(\frac{{0.205 \times 2.65}}{{0.606}}\)
s = 0.896
ac + s = 1
where, ac = air content
ac = 0.104 = 10.4 %
Last updated on Jul 1, 2025
-> SSC JE notification 2025 for Civil Engineering has been released on June 30.
-> Candidates can fill the SSC JE CE application from June 30 to July 21.
-> SSC JE Civil Engineering written exam (CBT-1) will be conducted on 21st to 31st October.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.
-> The Staff Selection Commission conducts the SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.