Question

In an immersion measurement of a boy’s density, he is found to have an apparent weight of 6.86-N when completely submerged in water with lungs empty. The mass of the boy is 68.0-kg in air. The density of water is 1000-kg/m^3.

a)Determine the density of the boy;
b)Will the boy float in sea water (density of sea water is 1029-kg/m^3)? If yes, what percentage of his body volume will be above sea water?

Answer 1

6.86 = m g - 1000 g V
g = 9.81 m/s^2
6.86 = 9.81 (68) - 9810 V
9810 V = 660
V = 66/981 m^3
m = 68
so m/V = (68/66)981 = 1011 kg/m^3
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In seawater
volume of boy = 66/981
so mass of water displacedd submerged =
1029*66/981 = 69.2 kg
BUT
boy only has mass of 68 kg
so he floats
fraction above = (69.2 - 68) /68