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Archimedes’ principle. Up thrust is an upward force due to the fluid resisting being compressed. When any object is immersed or submerged into fluid its weight appears to have been reduced because it experiences an up thrust from the fluid.

Archimedes principle states that when a body is either wholly or partly submerged in a fluid the up thrust is equal to the weight of fluid displaced. i.e

**Experiment: To verify Archimedes principle**

**Weight Wa of object in air**

An object is weighed in air using a spring balance to obtain Wa.

**Weight Ww of object in water**

The object is weighed when completely immersed in water using a spring balance to obtain Ww and the displacement water collected in beaker as shown below.

**Weight of displaced water**

By using a spring balance the beaker is weighed with the displaced water when it’s empty

**Up thrust “U” = Wa – Ww**

It is found that weight of displaced water is equal to up thrust. Thus Archimedes’ principle.

**Calculations involving Archimedes principle**

For any calculation involving Archimedes’ principle the following should be noted:

i) The body should be completely immersed or submerged.

ii) The weight of the body when completely immersed or submerged is called its apparent weight.

The apparent weight is less than the weight of the body because when the body is immersed it experiences an up thrust.

Where 𝑉𝑓 𝑎𝑛𝑑 𝜌𝑓 are the volume and density of the displaced fluid.

For a body completely immersed or submerged fully, according to the displacement method,

Where “V” is volume of displaced fluid “𝜌” is density of fluid.

**Example: 1**

A glass blocks weight 25N. When wholly immersed in water, the block appears to weigh 15N. Calculate the Up thrust.

**Solution**

Wa=25N; Wf=15N;

Upthrust=Wa−Wf

=25−15

Upthrust=10N

**Example 2:**

A metal weighs 20 N in air and 15N when fully immersed in water Calculate the:

i) Up thrust;

ii) Weight of displaced water

iii) Volume of Displaced Water

**Solution**

**Example 3:**

An iron cube of volume 800cm3 is totally immersed in

(a)Water (b) oil of density 0.8gcm-3.Calculate the up thrust in each case. Density of water =1000 kgm-3

𝑉𝑏=800𝑐𝑚−3 = 800/(100x100x100) kgm-3;

**Solution**

(a) Upthrust in water

ρf =1gcm-3=1000 kgm-3

Vf=Vb=800𝑐𝑚−3 = 800/(100x100x100) kgm-3;

Up thrust = weight of displaced water

𝑈𝑝𝑡ℎ𝑟𝑢𝑠𝑡 =𝑉𝑓𝜌𝑓𝑔

Note: the greater the density, the greater the up thrust. The apparent weight of a body is less in fluids of greater density.

**Example 3:**

An iron cube, mass 480g and density 8g/cm3 is suspended by a string so that it is half immersed in oil of density 0.9g/cm3. Find the tension in string.

**Application of Archimedes principle**

**(a) Relative density of a solid**

By Archimedes principle, the apparent weight is equal to the weight of water displaced by the solid. The volume of this water displaced is the same as the volume of the solid.

But apparent loss in weight of solid in water = Wa – Ww

RelativeDensity(R.D)

**Example**

A glass block weighs 25N. When wholly immersed in water the block appears to weigh 15N. Calculate the relative density.

**Solution**

**(b) Relative density of liquid**

This is determined by using a solid. This solid sinks in water and in the liquid for which the relative density is to be determined.

A solid of weight Wa is weighed when completely immersed in the liquid to obtain Wl. The solid is then weighed when completely immersed in water to obtain Ww.

So ℎ𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑞𝑢𝑖𝑑 (𝑅.𝐷) is given by;

**Example**

A metal weighs 25N in air. When completely immersed in liquid it weighs 15N and it weighs 20N when completely immersed in water. Calculate the relative density of the liquid.

**Solution**