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Illustrations

Example-1

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.

 

Description: 60348.png

 

Solution

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

 

ρgL3

 

= 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.
 

 

Example-2

Calculate h in the U tube shown in the figure.

 

Description: 60359.png

 

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

 

Solution

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.
 

 

Example-3

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).

Solution

Let ρ′ be the density of block.

 

∴ Volume of block V = (m′ρ′).

 

∴ Buoyant force = Weight of volume V of liquid

 

Vρg

 

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,

 

Description: 59558.png

 

∴ Description: 59565.png
 





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