Question

A hollow cubical box is 0.453 m on an edge. This box is floating in a lake with 1/4 of its height beneath the surface. The walls of the box have a negligible thickness. Water is poured into the box. What is the depth of the water in the box at the instant the box begins to sink?

Answer 1

here,

the volume of box , V = (0.453 m)^3 = 0.09296 m^3

When it submerges,

it will displace (V * density of water = 92.96 kg) of water

So the amt of water initially displaced was (1/4)* 0.09296 m^3 = 0.02324 m^3

meaning the mass of the box is = 1000 * 0.02324 = 23.24 kg

So we need to add 92.96 - 23.24 kg = 69.7 kg
of water.

for water added to the box

mass of water added , m = 69.7 kg = p * V = p * A * d

where A is the area = (0.453 m)^2  

69.7 = 1000 * 0.453^2 * d

solving for d

d = 0.34 m

the depth of the water in the box at the instant the box begins to sink is 0.34 m