Question

A hollow cubical box is 0.30 m on an edge. This box is floating in a lake with one-third 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

Answer:

Explanation:

Given

length of cubical box $h=0.3 m$

If density of object $\rho _o$ and density of lake liquid $\rho _l$

when it is in equilibrium one-third of its height

Buoyancy force will be equal to weight of cubical box

$\rho \times h^3\times g=\rho _l\times h^2\times \frac{h}{3}\times g$

therefore $\frac{\rho _o}{\rho _w}=\frac{1}{3}$

When water start Pouring in it then height of liquid at which box started to sink

Let H be that height

$\rho \times h^3\times g+\rho \times h^2\times H\times g=\rho _l\times h^2\times h\times g$

cancel out the common terms and divide by density of lake

$\frac{\rho _o}{\rho _l}\times h+H=h$

$H=h-\frac{h}{3}=\frac{2}{3}h$

$H=\frac{2}{3}\times 0.3=0.2\ m$