Archimedes Principle, Buoyant Force (NEET LEVEL)

Illustration

What fraction of the total volume of an iceberg floating on sea-water is exposed? The density of ice is 920 kg/m³ and that of sea-water is 1030 kg/m³.

Solution:      

If V and V₁ be the total and submerged volumes of the iceberg

\frac{{{{V}_{1}}}}{V}\,\,=\,\,\frac{{\text{density of iceberg}}}{{\text{density of sea-water}}}\,\,=\,\,\frac{{920}}{{1030}}\,\,            =\,\,\frac{{92}}{{103}}

Now the fraction of volume exposed = \frac{{V\,\,-\,\,{{V}_{1}}}}{V}\,\,=\,\,1\,\,\,-\,\,\,\frac{{{{V}_{1}}}}{V} \,\,=\,\,1\,\,\,-\,\,\,\,\frac{{92}}{{103}}\,\,=\,\,\frac{{11}}{{103}} = 10.7%

Illustration

When a boulder of mass 240 kg is placed on a floating iceberg it is found that the iceberg just sinks. What is the mass of the iceberg? Take the relative density of ice as 0.9 and that of sea-water as 1.02.

Solution:      

Let M be the weight of iceberg in kg. Then its volume

V\,\,=\,\,\frac{M}{{0.9\,\,\times \,\,{{{10}}^{3}}}}\,\,{{m}^{3}}\,\,=\,\,\frac{M}{{900}}\,\,{{m}^{3}}

When this volume just sinks under sea-water, the weight of water displaced = V´ 1020 kg.

From the principle of buoyancy, this weight must be equal to the weight of iceberg and the boulder on it.

M + 240 = \frac{M}{{900}}´ 1020        = M´ \frac{{102}}{{90}}

or    M\ \left( {\frac{{102}}{{90}}\,\,-\,\,1} \right)\,\,   = 240

or    M\,\,=\,\,\frac{{240\,\,\times \,\,90}}{{12}}= 1800 kg

The mass of the iceberg = 1800 kg

Relative Density

In case of a liquid, sometimes another term relative density (R.D.) is defined. It is the ratio of density of the substance to the density of water at 4°C.

Hence, \text{R}\text{.D}\text{.}=\frac{{\text{Density of substance}}}{{\text{Density of water at 4}{}^\circ \text{C}}}

R.D. is a pure ratio. So, it has no units. It is also sometimes referred as specific gravity.

Density of water at 4°C in CGS is 1gm/cm³. Therefore, numerically the R.D. and density of substance (in CGS) are equal. In SI unit the density of water at 4°C is 1000 kg/m³.

Illustration

Relative density of an oil is 0.8. Find the absolute density of oil in CGS and SI units.

Solution:      

Density of oil (in CGS) = (R.D.) gm/cm³

= 0.8 g/cm³

 = 800 kg/m³

Density of a mixture of two or more liquids

Here, we have two cases.

Case 1:  Suppose two liquids of densities \displaystyle \rho ₁and  \displaystyle \rho ₂ having masses m₁ and m₂ are mixed together. Then the density of the mixture will be

\rho =\frac{{\text{Total mass}}}{{\text{Total volume}}}=\frac{{({{m}_{1}}+{{m}_{2}})}}{{({{V}_{1}}+{{V}_{2}})}} =\frac{{({{m}_{1}}+{{m}_{2}})}}{{\left( {\frac{{{{m}_{1}}}}{{{{\rho }_{1}}}}+\frac{{{{m}_{2}}}}{{{{\rho }_{2}}}}} \right)}}

If m₁ = m₂, then \rho =\frac{{2{{\rho }_{1}}{{\rho }_{2}}}}{{{{\rho }_{1}}+{{\rho }_{2}}}}

Case 2:  If two liquids of densities \displaystyle \rho ₁ and \displaystyle \rho ₂ having volumes V₁ and V₂ are mixed, then the density of the mixture is, \rho =\frac{{\text{Total mass}}}{{\text{Total volume}}}=\frac{{{{m}_{1}}+{{m}_{2}}}}{{{{V}_{1}}+{{V}_{2}}}}=\frac{{{{\rho }_{1}}{{V}_{1}}+{{\rho }_{2}}{{V}_{2}}}}{{{{V}_{1}}+{{V}_{2}}}}

If, V₁ = V₂, then \rho =\frac{{{{\rho }_{1}}+{{\rho }_{2}}}}{2}.

Effect of temperature on density 

As the temperature of a liquid is increased, the mass remains the same while the volume is increased and hence, the density of the liquid decreases. \left( {\text{as}\ \rho \propto \frac{1}{V}} \right).Thus,

\frac{{\rho \prime }}{\rho }=\frac{V}{{V\prime }}=\frac{V}{{V+dV}}=\frac{V}{{V+V\gamma \Delta \theta }}

or    \frac{{\rho \prime }}{\rho }=\frac{1}{{1+\gamma \Delta \theta }}

Hence, g = thermal coefficient of volume expansion

and  D \displaystyle \theta = rise in temperature

\rho \prime =\frac{\rho }{{1+\gamma \Delta \theta }}.

Effect of pressure on Density 

As pressure is increased, volume decreases and hence density will increase. Thus, \rho \propto \frac{1}{V}

\frac{{\rho \prime }}{\rho }=\frac{V}{{V\prime }}=\frac{V}{{V+dV}}=\frac{V}{{V-\left( {\frac{{dp}}{B}} \right)V}}          or         \frac{{\rho \prime }}{\rho }=\frac{1}{{1-\frac{{dp}}{B}}}

Here, dp = change in pressure and B = Bulk modulus of elasticity of the liquid

Therefore, \rho \prime =\frac{\rho }{{1-\frac{{dp}}{B}}}.

Illustration

The volume of a body is V and its density is d’. Density of water is d and d’>d. The body is submerged inside water and is lifted through a height h in water. Which of the following is correct?

(a) The potential energy of the body increases by hVdg

(b) P.E. increases by hV(d’–d)g

(c) P.E. increases by hV(d’ + d)g

(d) P.E. decreases

Solution:         

The resultant downward force on the sinking particle = Vd‘g – Vdg = V(d’ – d)g. When it is lifted up work is done against this force, W = hV(d’ – d)g = P.E. gained

∴  (b)