How much work will be done in making a soap bubble of diameter \[2.0\;{\rm{cm}}\].
A. \[8.54 \times {10^{ - 5}}\]
B. \[7.54 \times {10^{ - 5}}\]
C. \[9.54 \times {10^{ - 5}}\]
D. \[10.54 \times {10^{ - 5}}\]
Answer
638.4k+ views
Hint: The above problem can be resolved by applying the concept and mathematical formula of the work done to make the soap bubble. The mathematical relation is given by taking the product of surface tension and area of the bubble. The standard value for the surface tension of the soap bubble can be taken to make the solution more reliable.
Complete step by step answer:
Given:
The diameter of the soap bubble is, \[d = 2.0\;{\rm{cm}}\].
The mathematical formula for the work done to make the soap bubble is given as,
\[W = \sigma \times A\]
Here, \[\sigma \] is the surface tension of the soap bubble and its standard value is \[3 \times {10^{ - 2}}\;{\rm{N/m}}\].And A is the area of the soap bubble and its value is,
\[A = 4\pi {r^2}\]
Here, r denoted the radius of the soap bubble and its value is, \[r = d/2\].
Solve by substituting the values as,
\[\begin{array}{l}
W = \sigma \times A\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {r^2}} \right)\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {{\left( {\dfrac{d}{2}} \right)}^2}} \right)\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {{\left( {\dfrac{{20\;{\rm{cm}} \times \dfrac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}}{2}} \right)}^2}} \right)\\
W = 7.54 \times {10^{ - 5}}\;{\rm{J}}
\end{array}\]
Therefore, the required work to make the soap bubble is \[7.54 \times {10^{ - 5}}\;{\rm{J}}\].And option B is correct.
Note:In order to resolve the given problem, one must be aware of the concepts and applications of the surface tension, along with the work required to increase or decrease the surface tension of any object. Moreover, the mathematical relation of work and surface tension is to be remembered. The surface tension plays a vital role in the formation and the analysis of the soap bubbles, along with the involvement of the surface energy.
Complete step by step answer:
Given:
The diameter of the soap bubble is, \[d = 2.0\;{\rm{cm}}\].
The mathematical formula for the work done to make the soap bubble is given as,
\[W = \sigma \times A\]
Here, \[\sigma \] is the surface tension of the soap bubble and its standard value is \[3 \times {10^{ - 2}}\;{\rm{N/m}}\].And A is the area of the soap bubble and its value is,
\[A = 4\pi {r^2}\]
Here, r denoted the radius of the soap bubble and its value is, \[r = d/2\].
Solve by substituting the values as,
\[\begin{array}{l}
W = \sigma \times A\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {r^2}} \right)\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {{\left( {\dfrac{d}{2}} \right)}^2}} \right)\\
W = 3 \times {10^{ - 2}}\;{\rm{N/m}} \times \left( {4\pi {{\left( {\dfrac{{20\;{\rm{cm}} \times \dfrac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}}{2}} \right)}^2}} \right)\\
W = 7.54 \times {10^{ - 5}}\;{\rm{J}}
\end{array}\]
Therefore, the required work to make the soap bubble is \[7.54 \times {10^{ - 5}}\;{\rm{J}}\].And option B is correct.
Note:In order to resolve the given problem, one must be aware of the concepts and applications of the surface tension, along with the work required to increase or decrease the surface tension of any object. Moreover, the mathematical relation of work and surface tension is to be remembered. The surface tension plays a vital role in the formation and the analysis of the soap bubbles, along with the involvement of the surface energy.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

