Discussion :: Hydraulics and Fluid Mechanics - Section 7 (Q.No.1)
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1. | The specific gravity of an oil whose specific weight is 7.85 kN/m3, is |
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Answer: Option A Explanation: No answer description available for this question.
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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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