Discussion :: Hydraulics and Fluid Mechanics - Section 7 (Q.No.1)
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The specific gravity of an oil whose specific weight is 7.85 kN/m3, is
Answer: Option A
No answer description available for this question.
|Swam said: (Nov 14, 2013)|
|7.85*10^3/9.81*10^3 = 0.8.|
|Maniakanta Ghosh said: (Apr 8, 2014)|
|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.
|Williams Tosin said: (Apr 5, 2015)|
|Wow nice workings.|
|Saravanan said: (Jun 13, 2016)|
|Thanks for explaining the solution.|
|Ravi said: (Jan 23, 2017)|
|Thank you for the explanation.|
|Saranya said: (Mar 17, 2017)|
|Thank you for all your explanation.|
|Kiran said: (Oct 26, 2017)|
|I Did not understand this.|
|Subhajit Pamja said: (Jan 18, 2018)|
|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).
|Ganesh said: (Feb 11, 2018)|
|Thanks for the explanation.|
|Ashok said: (Apr 2, 2018)|
Sp.gravity=density of oil/density of water,
|Ekene Stanley said: (Apr 15, 2018)|
|Specific weight = 7.85kn/m^3,
The density of water = 9.81,
7.85÷9.81 = 0.8anwser.
|Harsh Shukla said: (May 2, 2019)|
|Specific gravity or relative density = W1/W2.
W1 = specific weight of given liquid = 7.85 KN/m3.
W2 = specific weight of standard liquid (Water) = 9.81KN/m3.
W1/W2 = 7.85/9.81 = 0.8.
|Akashdeep Singh said: (Jul 20, 2019)|
|The formula of specific gravity:
= Specific weight of liquid/ specific weight of pure water at standard temperature.
The specific weight of water =9.81 KN/m^3.
Then according to formula:
7.85/9.81 = 0.8 Ans.
|Vimal Gautam said: (Apr 30, 2020)|
|Thanks all for the explanation.|
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