Question

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?

Answer 1

Answer:

The volume of the submerged part of her body is $0.0622m^{3}$

Explanation:

Let's define the buoyant force acting on a submerged object.

In a submerged object acts a buoyant force which can be calculated as :

$B=$ρ.V.g

Where ''B'' is the buoyant force

Where ''ρ'' is the density of the fluid

Where ''V'' is the submerged volume of the object

Where ''g'' is the acceleration due to gravity

Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.

We can write :

$W_{girl}=B$ (I)

Where ''W'' is weight

⇒ If we consider ρ = $1000\frac{kg}{m^{3}}$ (water density) and $g=9.81\frac{m}{s^{2}}$ and replacing this values in the equation (I) ⇒

$B=W_{girl}$

$B=610N$

ρ.V.g = 610N

$1000\frac{kg}{m^{3}}.V.(9.81\frac{m}{s^{2}})=610N$ (II)

The force unit ''N'' (Newton) is defined as

$N=kg.\frac{m}{s^{2}}$

Using this in the equation (II) :

$(9810\frac{N}{m^{3}}).V =610N$

$V=\frac{610N}{9810\frac{N}{m^{3}}}$

$V=0.0622m^{3}$

We find that the volume of the submerged part of her body is $0.0622m^{3}$