Question

If a bubble in sparkling water accelerates upward at the rate of$\mathbf{0}\mathbf{.}\mathbf{225}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{}$and has a radius of $\mathbf{0}\mathbf{.}\mathbf{500}\mathit{m}\mathit{m}$, what is its mass? Assume that the drag force on the bubble is negligible.

Short answer

The mass of the bubble in sparkling water is$5.11x{10}^{-7}kg.$

Step-by-step solution

Listing the given quantities.

• The radius of the bubble

$\left(r\right)=0.500mm$

$=0.500x{10}^{-3}m$

• The acceleration is, $a=0.225m/s^2.$
  
• The density of water is ${\rho }_w=1000kg/m^3$
  

Understanding the concept of buoyant force.

There are two forces acting on the bubble – buoyant force $\mathbf{\left(}{\mathbf{F}}_{\mathbf{b}}\mathbf{\right)}$and weight.As the bubble is accelerating, using Newton’s second law, we can find the mass of the bubble$\mathbf{\left(}{\mathbf{M}}_{\mathbf{b}}\mathbf{\right)}$.

Formula:

${\mathrm{F}}_{\mathrm{net}}=\mathrm{Ma}$

${\mathrm{F}}_{\mathrm{b}}={\mathrm{M}}_{\mathrm{f}}\mathrm{g}$

$\mathrm{W}={\mathrm{M}}_{\mathrm{b}}\mathrm{g}$

$\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3$

Calculation of mass of the bubble.

The volume of the water displaced by the bubble $\left(V_w\right)$= Volume of the bubble $\left(V_b\right)$

$V_w=V_b$

$=\frac{4}{3}\pi r^3$

Substitute the value in the above equation.

$=\frac{4}{3}x3.14x{(500x{10}^{-3}m)}^3$

role="math" localid="1657556016855" $=1.57x{10}^{-9}{m}^3\left(1\right)$

Using Newton’s second law,

${\mathrm{F}}_{\mathrm{b}}-\mathrm{W}={\mathrm{M}}_{\mathrm{b}}\mathrm{a}$

${\rho }_w{V}_{b}g-{\rho }_{b}g{V}_b={\rho }_b{V}_{b}a$

Since the volume is the same, it would be canceled.

Rearranging the equation,

${\rho }_b=\frac{{\rho }_{w}g}{g+a}$

Substitute the values in the above equation.

$=\frac{998{\mathrm{m}}^3\mathrm{x}9.8\mathrm{m}/{\mathrm{s}}^2}{9.8\mathrm{m}/{\mathrm{s}}^2+0.255\mathrm{m}/{\mathrm{s}}^2}$

role="math" localid="1657556387852" $=975.6\mathrm{kg}/{\mathrm{m}}^3\left(2\right)$

${\mathrm{M}}_{\mathrm{b}}={\mathrm{\rho }}_{\mathrm{b}}\mathrm{x}{\mathrm{V}}_{\mathrm{b}}$

Using the values from (1) and (2), we have

${\mathrm{M}}_{\mathrm{b}}=1.57\mathrm{x}{10}^{-9}\mathrm{kg}/{\mathrm{m}}^3\mathrm{x}975.6{\mathrm{m}}^3$

$=5.11x{10}^{-7}kg$

Mass of the bubble in sparkling water is$5.11x{10}^{-7}kg.$