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

(1) Translatory equilibrium : When a body of density \[\rho \] and volume V is immersed in a liquid of density \[\sigma \], the forces acting on the body are

Weight of body \[W=mg=V\rho g,\] acting vertically downwards through centre of gravity of the body.

Upthrust force = \[V\sigma g\] acting vertically upwards through the centre of gravity of the displaced liquid i.e., centre of buoyancy.  

(i) A body will float in liquid only and only if \[\rho \le \sigma \]

(ii) In case of floating as weight of body = upthrust

So \[{{W}_{App}}\] = Actual weight  ? upthrust = 0

(iii) In case of floating \[V\rho g={{V}_{in}}\sigma \,g\]

So the equilibrium of floating bodies is unaffected by variations in g though both thrust and weight depend on g.

(2) Rotatory Equilibrium : When a floating body is slightly tilted from equilibrium position, the centre of buoyancy B shifts. The vertical line passing through the new centre of buoyancy B¢ and initial vertical line meet at a point M called meta-centre. If the meta-centre M is above the centre of gravity the couple due to forces at G (weight of body W) and at \[{B}'\] (upthrust) tends to bring the body back to its original position. So for rotational equilibrium of floating body the meta-centre must always be higher than the centre of gravity of the body.

However, if meta-centre goes below CG, the couple due to forces at G and \[{B}'\] tends to topple the floating body.

That is why a wooden log cannot be made to float vertical in water or a boat is likely to capsize if the sitting passengers stand on it. In these situations CG becomes higher than MG and so the body will topple if slightly tilted.

(3) Application of floatation

(i) When a body floats then the weight of body = Upthrust

\[V\rho g={{V}_{in}}\sigma g\]\[\Rightarrow \] \[{{V}_{in}}=\left( \frac{\rho }{\sigma } \right)\,V\]    

\[\therefore \] \[{{V}_{\text{out}}}=V-{{V}_{in}}=\left( 1-\frac{\rho }{\sigma } \right)\,V\]

i.e., Fraction of volume outside the liquid  \[{{f}_{\text{out}}}=\frac{{{V}_{\text{out}}}}{\text{V}}=\left[ 1-\frac{\rho }{\sigma } \right]\]

(ii) For floatation \[V\rho ={{V}_{in}}\sigma \] \[\Rightarrow \] \[\rho =\frac{{{V}_{in}}}{V}\,\sigma \]\[={{f}_{in}}\sigma \]

If two different bodies A and B are floating in the same liquid then \[\frac{{{\rho }_{A}}}{{{\rho }_{B}}}=\frac{{{({{f}_{in}})}_{A}}}{{{({{f}_{in}})}_{B}}}\]

(iii) If the same body is made to float in different liquids of densities \[{{\sigma }_{A}}\] and \[{{\sigma }_{B}}\] respectively.

\[V\rho ={{({{V}_{in}})}_{A}}{{\sigma }_{A}}={{({{V}_{in}})}_{B}}{{\sigma }_{B}}\] \[\therefore \]  \[\,\frac{{{\sigma }_{A}}}{{{\sigma }_{B}}}=\frac{{{({{V}_{in}})}_{B}}}{{{({{V}_{in}})}_{A}}}\]

(iv) If a platform of mass M and cross-section A is floating in a liquid of density \[\sigma \] with its height h inside the liquid

\[Mg=hA\sigma g\]                                                                ...(i)

Now if a body of mass m is placed on it and the platform sinks by y then \[(M+m)g=(y+h)\,A\sigma g\]                                               ...(ii)

Subtracting equation (i) from (ii),

\[mg=A\sigma \,y\,g\], i.e., \[W\propto y\]                                         ...(iii)

So we can determine the weight of a body by placing it on a floating platform and noting the depression of the platform in the liquid by it.