# Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion

1.

The specific gravity of an oil whose specific weight is 7.85 kN/m3, is

 [A]. 0.8 [B]. 1 [C]. 1.2 [D]. 1.6

Answer: Option A

Explanation:

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) 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.

 Ekene Stanley said: (Apr 15, 2018) Specific weight = 7.85kn/m^3, The density of water = 9.81, weight/volume, 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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