MGW = ρb V gWhen kept in a fluid, the APPARENT weight is given byWa = W – Fb = ρbV g – ρf V g = (ρb – ρv)V g Calculation:GivenWeight in air = 400 N⇒ (ρb – ρa)V g = 400 - (1)Weight in water = 225 N⇒ (ρb – ρW)V g = 225 - (2)From equation 1 and 2,\(\FRAC{{{\rho _b} - {\rho _a}}}{{{\rho _b} - {\rho _w}}} = \frac{{400}}{{225}}\) We have ρa = 1.18 kg/m3, ρw = 1000 kg/m3,\(\Rightarrow \frac{{{\rho _b} - 1.18}}{{{\rho _b} - 1000}} = \frac{{400}}{{225}} \Rightarrow {\rho _b} = 2284.2\;kg/{m^3}\) Relative density of body = 2284.2/1000 = 2.28 kg/m3 ≈ 2.3 kg/m3

"> MGW = ρb V gWhen kept in a fluid, the APPARENT weight is given byWa = W – Fb = ρbV g – ρf V g = (ρb – ρv)V g Calculation:GivenWeight in air = 400 N⇒ (ρb – ρa)V g = 400 - (1)Weight in water = 225 N⇒ (ρb – ρW)V g = 225 - (2)From equation 1 and 2,\(\FRAC{{{\rho _b} - {\rho _a}}}{{{\rho _b} - {\rho _w}}} = \frac{{400}}{{225}}\) We have ρa = 1.18 kg/m3, ρw = 1000 kg/m3,\(\Rightarrow \frac{{{\rho _b} - 1.18}}{{{\rho _b} - 1000}} = \frac{{400}}{{225}} \Rightarrow {\rho _b} = 2284.2\;kg/{m^3}\) Relative density of body = 2284.2/1000 = 2.28 kg/m3 ≈ 2.3 kg/m3

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A stone weighs 400 N in air and when immersed in water it weighs 225 N. If the specific weight of water is 9810 N/m3, the relative density of the stone will be nearly

Fluid Mechanics Buoyancy And Floatation in Fluid Mechanics . 7 months ago

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Concept:The original weight of the body is given by MGW = ρb V gWhen kept in a fluid, the APPARENT weight is given byWa = W – Fb = ρbV g – ρf V g = (ρb – ρv)V g Calculation:GivenWeight in air = 400 N⇒ (ρb – ρa)V g = 400 - (1)Weight in water = 225 N⇒ (ρb – ρW)V g = 225 - (2)From equation 1 and 2,\(\FRAC{{{\rho _b} - {\rho _a}}}{{{\rho _b} - {\rho _w}}} = \frac{{400}}{{225}}\) We have ρa = 1.18 kg/m3, ρw = 1000 kg/m3,\(\Rightarrow \frac{{{\rho _b} - 1.18}}{{{\rho _b} - 1000}} = \frac{{400}}{{225}} \Rightarrow {\rho _b} = 2284.2\;kg/{m^3}\) Relative density of body = 2284.2/1000 = 2.28 kg/m3 ≈ 2.3 kg/m3

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