JEE Main & Advanced Physics Fluid Mechanics, Surface Tension & Viscosity / द्रव यांत्रिकी, भूतल तनाव और चिपचिपापन Density

In a fluid, at a point, density \[\rho \] is defined as: \[\rho =\underset{\Delta V\to 0}{\mathop{\lim }}\,\frac{\Delta m}{\Delta V}=\frac{dm}{dV}\]

(1) In case of homogenous isotropic substance, it has no directional properties, so is a scalar.

(2) It has dimensions \[[M{{L}^{-3}}]\] and S.I. unit kg/\[{{m}^{3}}\] while C.G.S. unit g/cc with \[1g/cc={{10}^{3}}kg/{{m}^{3}}\]

(3) Density of substance means the ratio of mass of substance to the volume occupied by the substance while density of a body means the ratio of mass of a body to the volume of the body. So for a solid body.

Density of body = Density of substance

While for a hollow body, density of body is lesser than that of substance \[[\text{As}\,\,{{V}_{\text{body}}}>{{V}_{\text{sub}\text{.}}}]\]

(4) When immiscible liquids of different densities are poured in a container, the liquid of highest density will be at the bottom while that of lowest density at the top and interfaces will be plane.

(5) Sometimes instead  of density we use the term relative density or specific gravity which is defined as :

\[RD=\frac{\text{Density of body}}{\text{Density of water }}\]

(6) If \[{{m}_{1}}\] mass of liquid of density \[{{\rho }_{1}}\] and \[{{m}_{2}}\] mass of density \[{{\rho }_{2}}\] are mixed, then as

\[m={{m}_{1}}+{{m}_{2}}\] and \[V=({{m}_{1}}/{{\rho }_{1}})+({{m}_{2}}/{{\rho }_{2}})\]                              

[As \[V=m/\rho \]]

\[\rho =\frac{m}{V}=\frac{{{m}_{1}}+{{m}_{2}}}{({{m}_{1}}/{{\rho }_{1}})+({{m}_{2}}/{{\rho }_{2}})}=\frac{\sum {{m}_{i}}}{\sum ({{m}_{i}}/{{\rho }_{i}})}\]

If \[{{m}_{1}}={{m}_{2}}\]      \[\rho =\frac{2{{\rho }_{1}}{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}}=\]Harmonic mean

(7) If \[{{V}_{1}}\] volume of liquid of density \[{{\rho }_{1}}\] and \[{{V}_{2}}\] volume of liquid of density \[{{\rho }_{2}}\] are mixed, then as:

\[m={{\rho }_{1}}{{V}_{1}}+{{\rho }_{2}}{{V}_{2}}\] and \[V={{V}_{1}}+{{V}_{2}}\]             [As \[\rho =m/V\]]

If \[{{V}_{1}}={{V}_{2}}=V\]   \[\rho =({{\rho }_{1}}+{{\rho }_{2}})/2\] = Arithmetic Mean

(8) With rise in temperature due to thermal expansion of a given body, volume will increase while mass will remain unchanged, so density will decrease, i.e.,

\[\frac{\rho }{{{\rho }_{0}}}=\frac{(m/V)}{(m/{{V}_{0}})}=\frac{{{V}_{0}}}{V}=\frac{{{V}_{0}}}{{{V}_{0}}(1+\gamma \Delta \theta )}\]        

[As \[V={{V}_{0}}(1+\gamma \Delta \theta )\]]

or  \[\rho =\frac{{{\rho }_{0}}}{(1+\gamma \Delta \theta )}\tilde{}{{\rho }_{0}}(1-\gamma \Delta \theta )\]

(9) With increase in pressure due to decrease in volume, density will increase, i.e.,

\[\frac{\rho }{{{\rho }_{0}}}=\frac{(m/V)}{(m/{{V}_{0}})}=\frac{{{V}_{0}}}{V}\]                 [As\[\rho =\frac{m}{V}\]]

But as by definition of bulk-modulus

\[B=-{{V}_{0}}\frac{\Delta p}{\Delta V}\] i.e., \[V={{V}_{0}}\left[ 1-\frac{\Delta p}{B} \right]\]

So  \[\rho ={{\rho }_{0}}{{\left( 1-\frac{\Delta p}{B} \right)}^{-1}}\tilde{-}{{\rho }_{0}}\left( 1+\frac{\Delta p}{B} \right)\]