
The excess pressure inside a soap bubble is:
A. Inversely proportional to the surface tension.
B. Inversely proportional to its radius
C. Directly proportional to the square of its radius
D. Directly proportional to its radius
Answer
484.8k+ views
Hint: This problem is a direct question. Using the formula for computing the excess pressure inside a soap bubble, this question can be solved. We will remove the constant terms and instead of the equality sign, we will use the proportionality symbol to determine the quantity that is either direct or indirect proportionality value of the excess pressure inside a soap bubble.
Complete step by step answer:
Firstly, we will derive the expression for the excess pressure inside a soap bubble.
The work done by the excess pressure inside a soap bubble in displacing the surface is given as follows.
\[\begin{align}
& dW=F\times s \\
& \Rightarrow dW=p\times 4\pi {{R}^{2}}\times dR \\
\end{align}\]…… (1)
In the above equation, we have substituted the surface area of the sphere.
The increase in the potential energy is calculated as follows,
\[\begin{align}
& dU=T\times {{A}_{inc}} \\
& dU=T[2\{4\pi {{(R+dR)}^{2}}-4\pi {{R}^{2}}\}] \\
\end{align}\]
Continue the further calculation.
\[\begin{align}
& dU=T[2\{4\pi ({{R}^{2}}+d{{R}^{2}}+2R\times dR)-4\pi {{R}^{2}}\}] \\
& dU=T[2\{4\pi {{R}^{2}}+4\pi d{{R}^{2}}+4\pi \times 2R\times dR-4\pi {{R}^{2}}\}] \\
& dU=T[2\{4\pi d{{R}^{2}}+4\pi \times 2R\times dR\}] \\
\end{align}\]
The first within the bracket can be equated to zero because of the negligible value. Thus, we get,
\[\begin{align}
& dU=T[2\{0+4\pi \times 2R\times dR\}] \\
& dU=T[2\{4\pi (2RdR)\}] \\
\end{align}\]
Therefore, we have,
\[dU=T[2\{4\pi (2RdR)\}]\] …… (2)
From the equations (1) and (2), we get,
\[\begin{align}
& p\times 4\pi {{R}^{2}}\times dR=T[2\{4\pi (2RdR)\}] \\
& \Rightarrow p=\dfrac{4T}{R} \\
\end{align}\]
Where T is the surface tension and R is the radius of the soap bubble.
Here, the surface tension of the soap bubble can be considered to be a constant value. Thus, the proportionality constant can be taken as follows.
\[p\propto \dfrac{1}{R}\]
Therefore, the excess pressure inside a soap bubble is inversely proportional to its radius.
So, the correct answer is “Option B”.
Note: If we know the formula for computing the excess pressure inside a soap bubble, then, we can solve this problem, otherwise, we need to find the formula first. The derivation for the excess pressure inside a soap bubble is discussed above to make the student know how we obtain the formula for computing the excess pressure inside a soap bubble.
Complete step by step answer:
Firstly, we will derive the expression for the excess pressure inside a soap bubble.
The work done by the excess pressure inside a soap bubble in displacing the surface is given as follows.
\[\begin{align}
& dW=F\times s \\
& \Rightarrow dW=p\times 4\pi {{R}^{2}}\times dR \\
\end{align}\]…… (1)
In the above equation, we have substituted the surface area of the sphere.
The increase in the potential energy is calculated as follows,
\[\begin{align}
& dU=T\times {{A}_{inc}} \\
& dU=T[2\{4\pi {{(R+dR)}^{2}}-4\pi {{R}^{2}}\}] \\
\end{align}\]
Continue the further calculation.
\[\begin{align}
& dU=T[2\{4\pi ({{R}^{2}}+d{{R}^{2}}+2R\times dR)-4\pi {{R}^{2}}\}] \\
& dU=T[2\{4\pi {{R}^{2}}+4\pi d{{R}^{2}}+4\pi \times 2R\times dR-4\pi {{R}^{2}}\}] \\
& dU=T[2\{4\pi d{{R}^{2}}+4\pi \times 2R\times dR\}] \\
\end{align}\]
The first within the bracket can be equated to zero because of the negligible value. Thus, we get,
\[\begin{align}
& dU=T[2\{0+4\pi \times 2R\times dR\}] \\
& dU=T[2\{4\pi (2RdR)\}] \\
\end{align}\]
Therefore, we have,
\[dU=T[2\{4\pi (2RdR)\}]\] …… (2)
From the equations (1) and (2), we get,
\[\begin{align}
& p\times 4\pi {{R}^{2}}\times dR=T[2\{4\pi (2RdR)\}] \\
& \Rightarrow p=\dfrac{4T}{R} \\
\end{align}\]
Where T is the surface tension and R is the radius of the soap bubble.
Here, the surface tension of the soap bubble can be considered to be a constant value. Thus, the proportionality constant can be taken as follows.
\[p\propto \dfrac{1}{R}\]
Therefore, the excess pressure inside a soap bubble is inversely proportional to its radius.
So, the correct answer is “Option B”.
Note: If we know the formula for computing the excess pressure inside a soap bubble, then, we can solve this problem, otherwise, we need to find the formula first. The derivation for the excess pressure inside a soap bubble is discussed above to make the student know how we obtain the formula for computing the excess pressure inside a soap bubble.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Earth rotates from West to east ATrue BFalse class 6 social science CBSE

The easternmost longitude of India is A 97circ 25E class 6 social science CBSE

Write the given sentence in the passive voice Ann cant class 6 CBSE

Convert 1 foot into meters A030 meter B03048 meter-class-6-maths-CBSE

What is the LCM of 30 and 40 class 6 maths CBSE

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

What is the difference between superposition and e class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

State the laws of reflection of light

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