## What is Archimede’s Principle ?

When a body is partially or fully immersed in a fluid at rest, then the body experiences an upward force called buoyancy. Buoyancy is equal to the weight of the fluid displaced by the body.

Suppose that a uniform cylindrical block of mass M and of uniform cross-sectional area A is kept submerged in a liquid of density ρ. The material of the block has density ρ_{o} .

Suppose that the total volume of the block is V and the submerged volume of the block is V_{S} (say).

The weight of the block is W_{b} = Mg = Vρ_{o}g

The volume of the fluid displaced is equal to the submerged volume of the block = V_{S}.

Therefore, the weight of the displaced fluid is

F_{B} = V_{s}ρg , acting vertically upward.

The maximum possible buoyant force F_{B/max} that can act on the block is equal to the weight of maximum volume of the fluid displaced, which occurs when the body is completely immersed within the fluid (by force if necessary).

F_{B/max} = Vρg

(a) If F_{B/max} ≥ W_{b} , then the block will float in the fluid.

(i) More specifically when F_{B/max} = W_{b}, it just floats just submerged,

(ii) When F_{B/max} > W_{b} , then it floats partially submerged.

(b) If F_{B/max} < W_{b}

then, the block will sink in the fluid. If there is no fluid friction (viscosity) then the equation of motion of the block is given by

W_{b} – F_{B/max} = Ma

Solved Example : A body of mass 2 kg and volume .001/m^{3} is in a fluid of density (4/3) gm/cm^{3}. The fluid is at rest. Determine whether the body will float or sink in the fluid. If it floats, then, what fraction is immersed of volume in the fluid? If sinks, then find out the acceleration.

Solution : The force acting on the body

F_{b} = Vρg = (0.001) (4/3) × 1000 (9.8)

$\large F_b = \frac{4\times 9.8}{3}$ = 13.06 N

⇒ Weight of the body (W = mg = 2 × 9.8 = 19.6 N ) is greater than F_{b}

⇒ The body sinks completely.

Its acceleration $\large = \frac{W-F_b}{m} $

$\large = \frac{19.6- \frac{4\times 9.8}{3}}{2} $

= 3.3 m/s^{2}

Solved Example : A rubber ball of mass 10 gm and volume 15 cm^{3} is dipped in water to a depth of 10 m. Assuming density of water uniform throughout the depth, find

(a) the acceleration of the ball, and

(b the time taken by it to reach the surface if it is released from rest (take g = 980 cm/s^{2}).

Solution: The maximum buoyant force on the ball is

F_{B} = V ρ_{w}g = 15 × 1 × 980 dyne = 14700 dyne.

The weight of the ball is mg = 10 × g = 10 × 980 dyne.

The net upward force, F = (15 × 980 – 10 × 980) dyne

= 5 × 980 dyne = 4900 dyne.

(a) Therefore, acceleration of the ball upward

$\large a = \frac{F}{m} = \frac{5 \times 980}{10}$

a = 490 cm/s^{2} = 4.9 m/s^{2}.

(b) Time taken by it reach the surface is

$\large t = \sqrt{\frac{2 h}{a}} $

$\large t = \sqrt{\frac{2 \times 10}{4.9}} $

= 2.02 sec

## What is the Centre of Buoyancy & Equilibrium ?

The resultant force of buoyancy acts at a fixed point within the body and is known as the centre of buoyancy.

If the centre of gravity and centre of buoyancy lie on the same vertical line then the body floats in fluid in equilibrium.

If the centre of buoyancy is above the centre of gravity the equilibrium is stable, otherwise it is unstable.

Solved Example : A vessel contains oil of density 0.8 gm/cm^{3} over mercury of density 13.6 gm/cm^{3}. A homogeneous sphere floats with half of its volume immersed in mercury and other half in oil. What is the density of the material of the sphere ?

Solution: Let the volume of the sphere be 2 V and density ρ .

According to Archimedes’ principle, 2 V. ρ g = V ρ_{mercury} g + V ρ_{oil} g.

Therefore, $\large \rho = \frac{\rho_{mercury} + \rho_{oil}}{2} $

$\large \rho = \frac{13.6 + 0.8}{2}$

= 7.2 g/cm^{3}

Solved Example: A uniform wooden bar of length L and mass M is hinged on a vertical wall of a container containing water, at one end [in figure]. 3/5th part of the bar is submerged in water. Find the ratio of densities of the liquid and the bar.

Solution: Let ρ = density of liquid & σ = density of bar .

Taking the moment of W = mg and F_{b} = V ρ g about O, we obtain

$\large V \rho g (\frac{7}{10} l sin\theta) = ( \frac{5}{3}V ) \sigma g (\frac{l}{2} sin\theta)$

$\large \frac{7 \rho}{10} = \frac{5 \sigma}{6} $

$\large \frac{\rho}{\sigma} = \frac{25}{21}$

Solved Example: A cube of 1 kg is floating at the interface of two liquids of density ρ_{1} = 3/4 gm/cm^{3} and ρ_{2} = 2.5 gm/cm^{3}. The cube has 3/5 part in ρ_{2} and 2/5 part in ρ_{1}. Find the density of the cube.

Solution: The weigh of the cube = (force acting by the liquid of density ρ1) + (force acting by the liquid of density of ρ2)

Let mass of the block = m

$\large m g = V_1 \rho_1 g + V_2 \rho_2 g $

$\large m = V_1 \rho_1 + V_2 \rho_2 $

$\large A L \rho = A(\frac{2}{5}L)\rho_1 + A(\frac{3}{5}L)\rho_2$

$\large \rho = \frac{2\rho_1 + 3 \rho_2}{5}$

$\large \rho = \frac{2(3/4) + 3 \times 2.5 }{5}$

$\large \rho = \frac{1.5 + 7.5}{5} = \frac{9}{5}$

= 1.8 g/cm^{3}

### Also Read :

How to calculate Pressure & Variation of Pressure with depth ? What is Pascal’s law & Its Application ? How to Measure Pressure with Barometer & Manometer ? |