
If the surface tension of a soap solution is 0.03 MKS units, then the excess of pressure inside a soap bubble of diameter 6 mm over the atmospheric pressure will be
A. Less than 40 $N/{{m}^{2}}$
B. Greater than 40 $N/{{m}^{2}}$
C. less than 20 $N/{{m}^{2}}$
D. Greater than 20 $N/{{m}^{2}}$
Answer
164.7k+ views
Hint: In this question, we have given the surface tension and we have to find the excess pressure inside the soap bubble. We know the pressure above the atmospheric pressure in a soap bubble can be written in the form of a soap bubble and its diameter. By putting the values in the identity and solving them, we are able to get our desirable answer.
Formula used:
the formula for excess pressure inside the soap bubble is,
$\Delta P=\dfrac{4\sigma }{R}$
Where $\sigma $ is the surface tension and R is the radius of the bubble.
Complete step by step solution:
Given the diameter of the soap bubble is 6 mm.
Then the radius of soap bubble = $\dfrac{D}{2}$= $\dfrac{6}{2}$= $3\,mm$
The amount of pressure exists inside the bubble is $\Delta P=0.03\,MKS$
We know the formula for excess pressure inside the soap bubble is,
$\Delta P=\dfrac{4\sigma }{R} \\ $
Now we will substitute the value of R and $\sigma $ in the above equation, we get
$\Delta P=\dfrac{4\times 0.03}{3} \\ $
Then $\sigma =40\,N/{{m}^{2}}$
Hence the surface tension of the soap bubble is greater than $40\,N/{{m}^{2}}$.
Thus, option B is the correct answer.
Note: The formula for excess pressure is formulated using the condition of equilibrium for the forces acting on the bubble. Surface tension and the force due to pressure are the forces which are acting on the soap bubble.
Formula used:
the formula for excess pressure inside the soap bubble is,
$\Delta P=\dfrac{4\sigma }{R}$
Where $\sigma $ is the surface tension and R is the radius of the bubble.
Complete step by step solution:
Given the diameter of the soap bubble is 6 mm.
Then the radius of soap bubble = $\dfrac{D}{2}$= $\dfrac{6}{2}$= $3\,mm$
The amount of pressure exists inside the bubble is $\Delta P=0.03\,MKS$
We know the formula for excess pressure inside the soap bubble is,
$\Delta P=\dfrac{4\sigma }{R} \\ $
Now we will substitute the value of R and $\sigma $ in the above equation, we get
$\Delta P=\dfrac{4\times 0.03}{3} \\ $
Then $\sigma =40\,N/{{m}^{2}}$
Hence the surface tension of the soap bubble is greater than $40\,N/{{m}^{2}}$.
Thus, option B is the correct answer.
Note: The formula for excess pressure is formulated using the condition of equilibrium for the forces acting on the bubble. Surface tension and the force due to pressure are the forces which are acting on the soap bubble.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
