Question

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 1

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.

& dW=F\times s \\\ & \Rightarrow dW=p\times 4\pi {{R}^{2}}\times dR \\\ \end{aligned}$$…… (1) In the above equation, we have substituted the surface area of the sphere. The increase in the potential energy is calculated as follows, $$\begin{aligned} & dU=T\times {{A}_{inc}} \\\ & dU=T[2\\{4\pi {{(R+dR)}^{2}}-4\pi {{R}^{2}}\\}] \\\ \end{aligned}$$ Continue the further calculation. $$\begin{aligned} & 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{aligned}$$ The first within the bracket can be equated to zero because of the negligible value. Thus, we get, $$\begin{aligned} & dU=T[2\\{0+4\pi \times 2R\times dR\\}] \\\ & dU=T[2\\{4\pi (2RdR)\\}] \\\ \end{aligned}$$ Therefore, we have, $$dU=T[2\\{4\pi (2RdR)\\}]$$ …… (2) From the equations (1) and (2), we get, $$\begin{aligned} & p\times 4\pi {{R}^{2}}\times dR=T[2\\{4\pi (2RdR)\\}] \\\ & \Rightarrow p=\dfrac{4T}{R} \\\ \end{aligned}$$ Where T is the surface tension and R is the radius of the soap bubble. Here, the surface tension of the soap bubble can be considered to be a constant value. Thus, the proportionality constant can be taken as follows. $$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.