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.
Akashdeep Singh said:
6 years ago
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.
= 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.
(4)
Harsh Shukla said:
6 years ago
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.
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.
(2)
Kiran said:
8 years ago
I Did not understand this.
(1)
Ekene Stanley said:
8 years ago
Specific weight = 7.85kn/m^3,
The density of water = 9.81,
weight/volume,
7.85÷9.81 = 0.8anwser.
The density of water = 9.81,
weight/volume,
7.85÷9.81 = 0.8anwser.
(1)
Vimal gautam said:
6 years ago
Thanks all for the explanation.
(1)
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.
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