Answer :
Answer:
Explanation:
Given
length of cubical box [tex]h=0.3 m[/tex]
If density of object [tex]\rho _o[/tex] and density of lake liquid [tex]\rho _l[/tex]
when it is in equilibrium one-third of its height
Buoyancy force will be equal to weight of cubical box
[tex]\rho \times h^3\times g=\rho _l\times h^2\times \frac{h}{3}\times g[/tex]
therefore [tex]\frac{\rho _o}{\rho _w}=\frac{1}{3}[/tex]
When water start Pouring in it then height of liquid at which box started to sink
Let H be that height
[tex]\rho \times h^3\times g+\rho \times h^2\times H\times g=\rho _l\times h^2\times h\times g[/tex]
cancel out the common terms and divide by density of lake
[tex]\frac{\rho _o}{\rho _l}\times h+H=h[/tex]
[tex]H=h-\frac{h}{3}=\frac{2}{3}h[/tex]
[tex]H=\frac{2}{3}\times 0.3=0.2\ m[/tex]