Question

An empty container ship has a fourth of its body under water. Safety regulations state that atleast a third of its body must be above the water. How much goods can it carry?
The weight of the ship alone is $W$.

Answer 1

Assuming the symbols:
$W$ is the weight of the ship without any goods in it,
$w$ is the maximum weight that can be added to the ship,
${\rho }_w$ is the density of the seawater,
$V$ is the volume the ship can take up
$V_1$ is the volume displaced by the ship without any goods in it,
$V_2$ is the volume displaced by the ship with the goods in it and,
$g$ is the acceleration due to gravity.

From the question, we know that $V_1=\frac{V}{4}$ and $V_2=\frac{2V}{3}$

According to Archimedes' principle, the weight of the ship alone is:
$W={\rho }_{w}g{V}_1$ ------------- (1)
$W+w={\rho }_{w}g{V}_2$ ------- (2)

Dividing the equation (2) by (1),

$1+\frac{w}{W}=\frac{V_2}{V_1}=\frac{2V/3}{V/4}$

$⟹1+\frac{w}{W}=\frac{8}{3}$

$⟹w=\frac{5W}{3}$

Hence. the ship can carry five-thirds of its own weight.