A soap bubble has radius R and thickness \[d( < < R)\] as shown in the figure. It collapses into a spherical drop. The ratio of excess pressure in the spherical drop to the excess pressure inside the bubble is \[{\left( {\dfrac{R}{{xD}}} \right)^{\dfrac{1}{3}}}\]. Find the value of x.

Answer
301.2k+ views
Hint: For a soap bubble, the pressurized bubble of air is contained within a thin, elastic surface of the liquid having a large volume and surface area. When the bubble bursts, it will form a number of spherical drops with lesser volume and surface area, the difference in pressure causes an audible sound.
Formula Used:
The equation of volume of a soap bubble is given by,
\[4\pi d{R^2} = \dfrac{4}{3}\pi {r^3}\]…………. (1)
Where, \[R\] is the radius of the soap bubble, \[d\] is the diameter of the soap bubble and \[r\] is the radius of the spherical drop.
Complete step by step solution:
To find the value of x we need to find the ratio of excess pressure.
The formula to find the ratio of excess pressure is given by,
\[{P_1} = \dfrac{{4S}}{R}\]..........(Excess pressure of the soap bubble initially)
\[{P_2} = \dfrac{{2S}}{r}\]..........(Excess pressure of the drop finally)
Now, \[\dfrac{{{P_2}}}{{{P_1}}} = \left( {\dfrac{{\dfrac{{4S}}{R}}}{{\dfrac{{2S}}{r}}}} \right)\]
\[\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{R}{{2r}}\]
By rearranging the equation (1), the value of r will be written as,
\[r = {\left( {3{R^2}d} \right)^{\dfrac{1}{3}}}\]
Put the value of r in equation (2) then we get,
\[\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{R}{{2r}}\]
\[\Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{R}{{2{{\left( {3{R^2}d} \right)}^{\dfrac{1}{3}}}}}\\\]
\[\therefore \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{R}{{{{\left( {24d} \right)}^{\dfrac{1}{3}}}}}\]
Therefore, the value of x is 24.
Note:Surface tension is defined as the property of any liquid by virtue of which it tries to minimize its surface area. The surface tension of water provides the necessary wall tension for the formation of the water bubbles. This tendency to minimize the tension on the walls pulls the bubbles into spherical shapes.
Formula Used:
The equation of volume of a soap bubble is given by,
\[4\pi d{R^2} = \dfrac{4}{3}\pi {r^3}\]…………. (1)
Where, \[R\] is the radius of the soap bubble, \[d\] is the diameter of the soap bubble and \[r\] is the radius of the spherical drop.
Complete step by step solution:
To find the value of x we need to find the ratio of excess pressure.
The formula to find the ratio of excess pressure is given by,
\[{P_1} = \dfrac{{4S}}{R}\]..........(Excess pressure of the soap bubble initially)
\[{P_2} = \dfrac{{2S}}{r}\]..........(Excess pressure of the drop finally)
Now, \[\dfrac{{{P_2}}}{{{P_1}}} = \left( {\dfrac{{\dfrac{{4S}}{R}}}{{\dfrac{{2S}}{r}}}} \right)\]
\[\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{R}{{2r}}\]
By rearranging the equation (1), the value of r will be written as,
\[r = {\left( {3{R^2}d} \right)^{\dfrac{1}{3}}}\]
Put the value of r in equation (2) then we get,
\[\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{R}{{2r}}\]
\[\Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{R}{{2{{\left( {3{R^2}d} \right)}^{\dfrac{1}{3}}}}}\\\]
\[\therefore \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{R}{{{{\left( {24d} \right)}^{\dfrac{1}{3}}}}}\]
Therefore, the value of x is 24.
Note:Surface tension is defined as the property of any liquid by virtue of which it tries to minimize its surface area. The surface tension of water provides the necessary wall tension for the formation of the water bubbles. This tendency to minimize the tension on the walls pulls the bubbles into spherical shapes.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2026-27 Free PDF Download (Sign-in Required)

