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An iron cube of side L metre is immersed in water (ρ = 103 kg m–3). Find the resultant thrust on the cube due to water and prove Archimedes principle.


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Thrust on the sides cancel out.


Thrust on the top surface = Pressure  × Area


F1 = ρgh1 × L2


Thrust on the bottom surface = Pressure × Area


F2 = ρgh2 × L2


Since h2 > h1F2 > F1


∴ Net upward thrust = F2 − F1


F = ρgL2 (h2 − h1) = ρgL2 × L




= 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.


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(ρoil = 900 kg m–3ρliq = 1600 kg m–3ρHg = 13,600 kg m–3)



For equilibrium, pressure at C = Pressure at D


PAB + PBC = PED (P = ρgh)


hρoilg + (0.2 − h) = 0.2 ρliqg


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




Net upward force F = Buoyant force − Weight of block


Vρg − mg


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


W = Force  Displacement = (Vρg − mg)h


This work done appears as KE in the block. Due to this, if the block rises to a height h1, then KE


mgh1. Therefore,


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∴ Description: 59565.png

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