91Ó°ÊÓ

• 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 ${\mathrm{ÒÏ}}_w=1000kg/m^3$
  

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:

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

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

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

$\mathrm{V}=\frac{4}{3}{\mathrm{Ï€°ù}}^3$

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

$V_w=V_b$

$=\frac{4}{3}\mathrm{Ï€}{r}^3$

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,

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

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

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

Rearranging the equation,

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

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

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

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