Determine the equation for the volume of body’s partially part immersed in a fluid for the floating body.
.When any body float on the surface of liquid then weight of body $=$ weight of displaced liquid
by the body.
$\mathrm{V} \rho g=\mathrm{V}^{\prime} \rho_{l} g$ (where $\mathrm{V}=$ volume of body)
$\mathrm{V}^{\prime}=$ volume of partially part of immerged in liquid
$=$ volume of displaced liquid
$\rho=$ Density of body
$\rho_{l}=$ Density of liquid
$\therefore \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\rho}{\rho^{\prime}}=\frac{\text { Volume of immerged part of body }}{\text { Total volume of body }}$
$=\frac{\text { Density of body }}{\text { Density of liquid }}$
A solid cube and a solid sphere both made of same material are completely submerged in water but to different depths. The sphere and the cube have same surface area. The buoyant force is
A body floats in a liquid contained in a beaker. The whole system as shown falls freely under gravity. The upthrust on the body due to the liquid is
A pan balance has a container of water with an overflow spout on the right-hand pan as shown. It is full of water right up to the overflow spout. A container on the left-hand pan is positioned to catch any water that overflows. The entire apparatus is adjusted so that it’s balanced. A brass weight on the end of a string is then lowered into the water, but not allowed to rest on the bottom of the container. What happens next ?
A solid sphere of radius $r$ is floating at the interface of two immiscible liquids of densities $\rho_1$ and $\rho_2\,\, (\rho_2 > \rho_1),$ half of its volume lying in each. The height of the upper liquid column from the interface of the two liquids is $h.$ The force exerted on the sphere by the upper liquid is $($ atmospheric pressure $= p_0\,\,\&$ acceleration due to gravity is $g) $
Write and prove Archimedes principle.