# Illustrations

An iron cube of side *L* metre is immersed in water (ρ = 10^{3} kg m^{–3}). Find the resultant thrust on the cube due to water and prove Archimedes principle.

Thrust on the sides cancel out.

Thrust on the top surface = Pressure × Area

*F*_{1} = ρ*gh*_{1} × *L*^{2}

Thrust on the bottom surface = Pressure × Area

*F*_{2} = ρ*gh*_{2 }× *L*^{2}

Since *h*_{2} > *h*_{1}, *F*_{2} > *F*_{1}

∴ Net upward thrust = *F*2 − *F*1

*F* = ρ*gL*^{2} (*h*_{2} − *h*_{1}) = ρ*gL*^{2 }× *L*

= ρ*gL*^{3}

= Weight of liquid which was in place of iron.

∴ Loss of weight of iron = Net upward thrust = Weight of liquid displaced.

This is Archimedes principle.

Calculate *h* in the U tube shown in the figure.

(ρ_{oil} = 900 kg m^{–3}, ρ_{liq} = 1600 kg m^{–3}, ρ_{Hg} = 13,600 kg m^{–3})

For equilibrium, pressure at C = Pressure at D

*P*_{AB} + *P*_{BC} = *P*_{ED} (∴*P* = ρ*gh*)

*h*ρ_{oil}*g* + (0.2 − *h*) = 0.2 ρ_{liq}*g*

*h *× 900 + (0.2 − *h*)13,600 = 0.2 × 1600

Simplifying, *h* = 0.189 m or *h* = 18.9 cm.

A wooden block of mass *m* is submerged in a liquid of density *ρ**, to a depth **h* and released. To what height will it jump up above the surface of water? (Neglect the resistance offered by water and air).

Let ρ′ be the density of block.

∴ Volume of block *V* = (*m*′ρ′).

∴ Buoyant force = Weight of volume *V* of liquid

= *V*ρ*g*

Net upward force *F *= Buoyant force − Weight of block

= *V*ρ*g *− *mg*

Work done by this force as the block comes to the surface is given by

*W* = Force Displacement = (*V*ρ*g *− *mg*)*h*

This work done appears as KE in the block. Due to this, if the block rises to a height *h*_{1}, then KE

= *mgh*_{1}. Therefore,

∴