Question

calculate the buoyant force of a floating body 95% submerged in water. density of water is 1000 Kg/m^3

Answer 1

Answer:

density = 1000 kgm⁻³

g⃗ = 9.8 ms²

95 % submerged in water

volume of the body = 100 - 95

= 5 m³

buoyant force = density × gravitational accelaration × volume

= 1000×9.8×5

= 49,000 N

Answer 2

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Calculate the buoyant force of a floating body 95% submerged in water. density of water is 1000 Kg/m³

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Given,

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Density of water, $\tt\red{ p = 1000 kg.m^{-3} }$

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From Archimedes principle formula,

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${\large{\sf{\pink{Fb = ρ × g × V}}}}$

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Or

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${\large{\sf{\pink{Vb × ρb × g = ρ × g × V}}}}$

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Where,

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ρ,g, and V are the density, acceleration due to gravity, and volume of the water

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Vb, ρb, and g are the volume, density, and acceleration due to gravity of body immersed

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Rearranging the equation,

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${\large{\tt{\blue{ρb=VρVb}}}}$

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Since 95% of the body is immersed,

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$\implies\tt 0.95 × Vb = V$

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$\tt\large\red {∴ ρb = 950 kg.m^{-3}}$

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