Question

The pressure inside two soap Bubbles is 1.01 atm and 1.02 atm. find the ratio of their volume? ​

Answer 1

AnswEr :

• Ratio of their volume is 8 : 1, respectively.

GivEn :

• Pressure inside the 1st soap bubble = 1.01 atm.

• Pressure inside the 2nd soap bubble = 1.02 atm.

To find :

• Ratio of the volume of soap bubbles = ?

Formula usEd :

To solve the question you need to know the concept of Gauge Pressure !

In simple words, gauge pressure is the difference between the absolute pressure and atmospheric pressure .

• Gauge Pressure = Absolute Pressure - Atmospheric pressure

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SoluTion :

Excess pressure of the 2 bubbles :

• ΔP1 = 1.01 - 1 = 0.01

• ΔP2 = 1.02 - 1 = 0.02

Also we know that,

Excess pressure is inversely proportional to the radius (As bubble is spherical in shape).

$\implies$ ΔP ∝ 1 / r

$\implies$ r ∝ 1 / ΔP

which implies,

$\dashrightarrow \sf \: \: \: \dfrac{r_1}{r_2} = \dfrac{ \Delta P_1}{ \Delta P_2}$

$\dashrightarrow \sf \: \: \: \dfrac{r_1}{r_2} = \dfrac{ 0.02}{ 0.01}$

$\dashrightarrow \sf \: \: \: \dfrac{r_1}{r_2} = \dfrac{ 2}{ 1}$

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Also, we know the formula for Volume of the Sphere (Soap Bubbles),

$\star \: { \boxed{ \sf{ \purple{v = \dfrac{4}{3}\pi {r}^{3} }}}}$

⠀⠀✩ Ratio of their Volume :

$\dashrightarrow \sf \: \: \: \dfrac{V_1}{V_2} = \dfrac{4/3\pi {r}^{3}}{4/3\pi {r}^{3}}$

$\dashrightarrow \sf \: \: \: \dfrac{V_1}{V_2} = \bigg ({\dfrac{r_1}{r_2}}\bigg)^{3}$

$\dashrightarrow \sf \: \: \: \dfrac{V_1}{V_2} = {\dfrac{8}{1}}$

Therefore, ratio of their volume is 8 : 1 .

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