Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 7 (Q.No. 1)
1.
The specific gravity of an oil whose specific weight is 7.85 kN/m3, is
Discussion:
14 comments Page 1 of 2.
Swam said:
1 decade ago
7.85*10^3/9.81*10^3 = 0.8.
Maniakanta ghosh said:
1 decade ago
Specific weight of oil/specific weight of water.
I) SPECIFIC WEIGHT OF WATER = 9.81 KN/m3.
7.85 /9.81 = 0.8 unit less.
I) SPECIFIC WEIGHT OF WATER = 9.81 KN/m3.
7.85 /9.81 = 0.8 unit less.
Williams Tosin said:
1 decade ago
Wow nice workings.
Saravanan said:
9 years ago
Thanks for explaining the solution.
Ravi said:
9 years ago
Thank you for the explanation.
Saranya said:
9 years ago
Thank you for all your explanation.
Kiran said:
8 years ago
I Did not understand this.
(1)
Subhajit Pamja said:
8 years ago
Specific Wt=wt /unit volume
= w/v= mg/v.
We can say m/v *g and that is = rho(p)* g
Given, oil specific wt (pg)=7.85 KN/m^3.
And we have specific gravity= sp.wt of the liquid/sp.wt of pure water at 4°c.
So, specific Wt. Of water =9.81kN/m^3.
And the and is 7.8/9.81=0.79=>0.8(unitless).
= w/v= mg/v.
We can say m/v *g and that is = rho(p)* g
Given, oil specific wt (pg)=7.85 KN/m^3.
And we have specific gravity= sp.wt of the liquid/sp.wt of pure water at 4°c.
So, specific Wt. Of water =9.81kN/m^3.
And the and is 7.8/9.81=0.79=>0.8(unitless).
Ganesh said:
8 years ago
Thanks for the explanation.
Ashok said:
8 years ago
Specific weight=ρ*g=7.85*100.
=>ρ=(7.85*1000)/9.81=800.
Sp.gravity=density of oil/density of water,
=800/1000=0.8.
=>ρ=(7.85*1000)/9.81=800.
Sp.gravity=density of oil/density of water,
=800/1000=0.8.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers