Question

A swimmer is peacefully floating motionless in a lake. Determine the fraction of the

total volume of her body which remains visible above the water, given that the density

of her body is 920 kg/m3, and the density of the lake water is 1025 kg/m3.

Answer 1

Archimedes principle says buoyant force acting on an object immersed in water is equal to weight of displaced liquid.

Assuming the fraction of volume of the body that is dipped in the liquid is f.So, if v is the volume of swimmer, then volume inside water(i.e the volume of water displaced) is given by fv.

Now, forces acting on the swimmer include:

1. Force due to gravity of earth=mass*gravitational acceleration=mg, where m = mass of the swimmer, g= gravitational acceleration.

Also, mass=density*volume=>m=d_s*v, where d_s is density of swimmer, v is his volume.

So, gravitational force=mg=d_s*vg (downwards)

2.Buoyant force= weight of displaced liquid=mass of displaced liquid*g; g= gravitational acceleration

mass of displaced liquid=volume of displaced liquid*density of liquid=fv*d_w where d_w is the density of water.

So, buoyant force= fv*d_w*g (downwards)

Since,there is no vertical acceleration for the swimmer, these forces must balance each other to satisfy newton's second law.

d_s*vg = fv*d_w*g=>f=d_s/ d_w

So, fraction of the total volume of swimmer's body that is above water=1-f=1-d_s/ d_w

Here, d_s=920 kg/m³ and d_w=1025 kg/m³

So,fraction of the total volume of swimmer's body that is above water=

1-f=1-920/1025=0.1024 which is equal to 0.10 (rounded off)