Question
Download Solution PDFCalculate the approximate mass density of oil with 4.5 m3 of volume and 40 kN of weight. (Consider g = 10 m/s2)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Specific gravity or Relative density:
The ratio of density or specific weight of a subject matter to that of a standard matter is known as specific gravity relative density.
\({\rm{S}} = \frac{{{{\rm{\rho }}_{{\rm{subject}}}}}}{{{{\rm{\rho }}_{{\rm{standard}}}}}}{\rm{or}}\frac{{{{\rm{\gamma }}_{{\rm{subject}}}}}}{{{{\rm{\gamma }}_{{\rm{standard}}}}}}\)
Specific weight or weight density:
Specific weight or weight density of a fluid is the ratio between the weight of a fluid to its volume.
\({\rm{\gamma }} = \frac{{{\rm{Weight\;of\;a\;fluid}}}}{{{\rm{Volume\;of\;a\;fluid}}}} = \frac{{\rm{W}}}{{\rm{V}}}\)
where weight = mass × gravitational acceleration = m × g
\({\rm{\gamma }} = \frac{{{\rm{m}} \times {\rm{g}}}}{{\rm{V}}} = \frac{{\rm{m}}}{{\rm{V}}} \times {\rm{g}} = {\rm{\rho }} \times {\rm{g}}\)
Calculation:
Given,
W = 40 kN, V = 4.5 m3
Mass = W/g = 40000/10 = 4000 kg
Mass density = Mass/Volume = 4000/4.5 = 889 kg/m3
Last updated on May 28, 2025
-> SSC JE notification 2025 for Civil Engineering will be released on June 30.
-> Candidates can fill the SSC JE CE application from June 30 to July 21.
-> 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.