Solveeit Logo
Login

Question

Question: With what terminal velocity will an air bubble 8mm in diameter rise in a liquid of viscosity \(0.15N...

With what terminal velocity will an air bubble 8mm in diameter rise in a liquid of viscosity 0.15Nsm−20.15Ns{{m}^{-2}} and specific gravity 0.90.9. Density of air is 1.293kgm−31.293kg{{m}^{-3}}.

Explanation

Solution

We know that terminal velocity is defined as the steady speed achieved by an object falling freely through a fluid. We will use the relation 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 of the liquid is a measure of a fluid’s resistance to flow.

Formula used:
We are going to use the following formula to get the correct answer:-
vT=29×r2(ρ−ρs)gη{{v}_{T}}=\dfrac{2}{9}\times \dfrac{{{r}^{2}}\left( \rho -{{\rho }_{s}} \right)g}{\eta } .

Complete step by step solution:
Diameter of the bubble, d=8mm=8×10−3md=8mm=8\times {{10}^{-3}}m
Therefore, radius, r=4×10−3mr=4\times {{10}^{-3}}m
Specific gravity, ρs=0.9{{\rho }_{s}}=0.9 .
Density of air, ρ=1.293kgm−3\rho =1.293kg{{m}^{-3}} .
Viscosity, η=0.15Nsm−2\eta =0.15Ns{{m}^{-2}} .
We will use the following formula to solve this problem:-
vT=29×r2(ρ−ρs)gη{{v}_{T}}=\dfrac{2}{9}\times \dfrac{{{r}^{2}}\left( \rho -{{\rho }_{s}} \right)g}{\eta } …………….. (i)(i)
Putting values of the parameters in (i)(i) we get
vT=29×(4×10−3)2×(1.293−0.9)0.15{{v}_{T}}=\dfrac{2}{9}\times \dfrac{{{\left( 4\times {{10}^{-3}} \right)}^{2}}\times \left( 1.293-0.9 \right)}{0.15}
⇒vT=29×16×10−5×0.3930.15\Rightarrow {{v}_{T}}=\dfrac{2}{9}\times \dfrac{16\times {{10}^{-5}}\times 0.393}{0.15}
Solving further we get
vT=9.31×10−5ms−1{{v}_{T}}=9.31\times {{10}^{-5}}m{{s}^{-1}}
Therefore the terminal velocity is equal to vT=9.31×10−5ms−1{{v}_{T}}=9.31\times {{10}^{-5}}m{{s}^{-1}} .

Additional Information:
We know that the terminal velocity is defined as the maximum velocity attained by an object as it falls in a fluid. When the sum of drag force and buoyancy is equal to the force of gravity then the condition of terminal velocity occurs. There is a net zero acceleration during terminal velocity. Drag force increases with the speed of an object.

Note:
It should be noted that when an object is moving with the terminal velocity then its speed is uniform due to the restraining force exerted by the fluid in which it is moving. It should also be noted that drag depends on the projected area. We should use the formula correctly without any confusion and also the units.