An-80-kg-man-floats-with-4-of-his-volume-above-the-surface-in-fresh-water-what-is-his-volume-what-volume-would-be-above-the-surface-in-Sea-water-How-great-is-the-upthrust-on-him-in-air-due-to-the-ai

Question Number 222373 by Tawa11 last updated on 24/Jun/25

An 80 kg man floats with 4% of his volume
above the surface in fresh water.what is his
volume?
what volume would be above the surface in
Sea water?How great is the upthrust on him
in air due to the air he displaces?
density of sea water =1030kgm^-3,
density of fresh water=1000kgm^-3?

Answered by mr W last updated on 24/Jun/25

V=((80)/(0.96×1000))≈0.083m^(−3) 1−((0.96×1000)/(1030))≈0.068, i.e. 6.8% of his volume is above the sea water surface. upthrust in air: 0.083×1.225≈0.10 kg

$${V}=\frac{\mathrm{80}}{\mathrm{0}.\mathrm{96}×\mathrm{1000}}\approx\mathrm{0}.\mathrm{083}{m}^{−\mathrm{3}} \\ $$$$\mathrm{1}−\frac{\mathrm{0}.\mathrm{96}×\mathrm{1000}}{\mathrm{1030}}\approx\mathrm{0}.\mathrm{068},\:{i}.{e}.\:\mathrm{6}.\mathrm{8\%}\:{of} \\ $$$${his}\:{volume}\:{is}\:{above}\:{the}\:{sea}\:{water} \\ $$$${surface}. \\ $$$${upthrust}\:{in}\:{air}: \\ $$$$\mathrm{0}.\mathrm{083}×\mathrm{1}.\mathrm{225}\approx\mathrm{0}.\mathrm{10}\:{kg} \\ $$

Commented by Tawa11 last updated on 24/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by fantastic last updated on 25/Jun/25

I am answering you with some explanation (1) The weight of the man w_(man) =80 Kg-wt Let the volume of the man be V_(man) m^3 4% of his volume is outside the water so 96% is inside the water So the volume of water displaaced by the man is =((96)/(100))Vm^3 . So the weight of the displaced water =((96V)/(100))1000 Kg-wt=960V Kg-wt As the man is floating so 960V=80 So V=0.083^. /0.083333333333... m^3 (2) Let x% be inside of sea water So ((Vx)/(100))m^3 is the volume of displaced sea water So the weight=((Vx)/(100))×1030 Kg-wt So ((Vx)/(100))×1030=80 or (((0.083^. ×x)/(100)))×1030=80 x≈93.2038834951456% So the out side % will be ≈6.7961165048544% So ≈6.8% will be outside

$${I}\:{am}\:{answering}\:{you}\:{with}\:{some}\:{explanation} \\ $$$$\left(\mathrm{1}\right) \\ $$$${The}\:{weight}\:{of}\:{the}\:{man}\:{w}_{{man}} =\mathrm{80}\:{Kg}-{wt} \\ $$$${Let}\:{the}\:{volume}\:{of}\:{the}\:{man}\:{be}\:{V}_{{man}} {m}^{\mathrm{3}} \\ $$$$\mathrm{4\%}\:{of}\:{his}\:{volume}\:{is}\:{outside}\:{the}\:{water} \\ $$$${so}\:\mathrm{96\%}\:{is}\:{inside}\:{the}\:{water} \\ $$$${So}\:{the}\:{volume}\:{of}\:{water}\:{displaaced}\:{by}\:{the}\:{man}\:{is} \\ $$$$=\frac{\mathrm{96}}{\mathrm{100}}{Vm}^{\mathrm{3}} \:. \\ $$$${So}\:{the}\:{weight}\:{of}\:{the}\:{displaced}\:{water} \\ $$$$=\frac{\mathrm{96}{V}}{\mathrm{100}}\mathrm{1000}\:{Kg}-{wt}=\mathrm{960}{V}\:{Kg}-{wt} \\ $$$${As}\:{the}\:{man}\:{is}\:{floating}\:{so} \\ $$$$\mathrm{960}{V}=\mathrm{80} \\ $$$${So}\:{V}=\mathrm{0}.\mathrm{08}\overset{.} {\mathrm{3}}/\mathrm{0}.\mathrm{083333333333}…\:\:{m}^{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right) \\ $$$${Let}\:{x\%}\:{be}\:{inside}\:{of}\:{sea}\:{water} \\ $$$${So}\:\frac{{Vx}}{\mathrm{100}}{m}^{\mathrm{3}} \:{is}\:{the}\:{volume}\:{of}\:{displaced}\:{sea}\:{water} \\ $$$${So}\:{the}\:{weight}=\frac{{Vx}}{\mathrm{100}}×\mathrm{1030}\:{Kg}-{wt} \\ $$$${So}\: \\ $$$$\frac{{Vx}}{\mathrm{100}}×\mathrm{1030}=\mathrm{80} \\ $$$${or}\:\left(\frac{\mathrm{0}.\mathrm{08}\overset{.} {\mathrm{3}}×{x}}{\mathrm{100}}\right)×\mathrm{1030}=\mathrm{80} \\ $$$${x}\approx\mathrm{93}.\mathrm{2038834951456\%} \\ $$$${So}\:{the}\:{out}\:{side}\:\%\:{will}\:{be} \\ $$$$\approx\mathrm{6}.\mathrm{7961165048544\%} \\ $$$${So}\:\approx\mathrm{6}.\mathrm{8\%}\:{will}\:{be}\:{outside} \\ $$

Commented by Tawa11 last updated on 24/Jun/25

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

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An-80-kg-man-floats-with-4-of-his-volume-above-the-surface-in-fresh-water-what-is-his-volume-what-volume-would-be-above-the-surface-in-Sea-water-How-great-is-the-upthrust-on-him-in-air-due-to-the-ai

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