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Question: With what terminal velocity will an air bubble of diameter \(0.8mm\) rise in a liquid of density \(9...

With what terminal velocity will an air bubble of diameter 0.8mm0.8mm rise in a liquid of density 900kg m3900kg{\text{ }}{m^{ - 3}} and viscosity 0.1N sm20.1N{\text{ s}}{{\text{m}}^{ - 2}} ? What will be the terminal velocity of the same bubble in water? Ignore density of air.

Explanation

Solution

For this question, you have to know concepts related to terminal velocity which is defined as the maximum velocity attainable by an object as it falls through a fluid. We will use the equation of terminal velocity in terms of radius of the bubble, specific gravity, density of air and viscosity to calculate the problem. We also know that viscosity is a measure of a fluid's resistance to flow.
Formula used:
Vt=2r2(δσ)g9η{V_t} = \dfrac{{2{r^2}(\delta - \sigma )g}}{{9\eta }}
Where,
δ\delta is the density of particle,
σ\sigma is the density of fluid,
η\eta is the coefficient of viscosity,
rr is the radius of particle and
gg is the acceleration due to gravity.

Complete step-by-step solution:
According to the question it is given that the
Diameter of bubble is 0.8mm0.8mm
Density of liquid is 900kg m3900kg{\text{ }}{m^{ - 3}} and its viscosity is 0.15N sm20.15N{\text{ }}s{m^{ - 2}}
We know that terminal velocity is given by
Vt=2r2(δσ)g9η{V_t} = \dfrac{{2{r^2}(\delta - \sigma )g}}{{9\eta }}
\because no density of air is given δ=0\delta = 0
Now, putting all the values in above equation:
Vt=2(0.4×103)2×900×9.89×0.15 =0.0020m s1 Vt=0.20cm s1{V_t} = \dfrac{{2{{(0.4 \times {{10}^{ - 3}})}^2} \times 900 \times 9.8}}{{9 \times 0.15}} \\\ = 0.0020m{\text{ }}{s^{ - 1}} \\\ \therefore {V_t} = 0.20cm{\text{ }}{s^{ - 1}}
And now for water,
(VT)liq(VT)(H20)=σ1σ2 (VT)(H20)=(VT)liq×σ1σ2 =0.002×900997 (VT)(H20)=0.18cm s1 \dfrac{{{{({V_T})}_{liq}}}}{{{{({V_T})}_{({H_2}0)}}}} = \dfrac{{{\sigma _1}}}{{{\sigma _2}}} \\\ \therefore {({V_T})_{({H_2}0)}} = {({V_T})_{liq}} \times \dfrac{{{\sigma _1}}}{{{\sigma _2}}} \\\ = 0.002 \times \dfrac{{900}}{{997}} \\\ \Rightarrow {({V_T})_{({H_2}0)}} = 0.18cm{\text{ }}{s^{ - 1}}
So, the terminal velocity of air bubble in air is 0.20cm s10.20cm{\text{ }}{s^{ - 1}} and terminal velocity of air bubble in water is 0.18cm s10.18cm{\text{ }}{s^{ - 1}}.

Note: We should use the formula correctly without any confusion and also the units. Terminal velocity occurs when the sum of the drag force and the buoyancy is equal to the downward force of gravity acting on the object. Since the net force on the object is zero, the object has zero acceleration and drag depends on the projected area.