A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume of the submerged part of her body?

1. By Archimedes' Principle the weight of water displaced must also be 610 N hence mass of water displaced = 610/9.8 = 62.44 kg. The volume of the water displaced = 62.24/1000 = 0.06224 m^3

2 Let the mass of cork be M kg. As it is floating the mass of water displaced must also be equal to M kg. Let the volume of cork be V. The volume of water displaced is V/10 m^3. so we have

M/[V/10] = 1000 kg m^-3 or 10 (M/V) = 1000 kg m^-3 or density of cork d is given by

d = M/V = 100 kg m^-3

3. The pressure P' at the bottom would be P + 2.5x0.81x980 + 6.5x1x980 dyne / cm^2 with P as atmospheric pressure which is 1012928 dyne / cm2 or

P' = 1012928 + 8354.5 = 1021273.5 dyne / cm^2