Question

The work done in blowing a soap bubble of radius $0.2m$, given that the surface tension of soap solution is $60\times {10}^{-3}N/m$ is

• A. $192\pi \times {10}^{-4}J$
• B. $96\pi \times {10}^{-4}J$
• C. $8\pi \times {10}^{-4}J$
• D. $124\pi \times {10}^{-4}J$

Answer 1

The correct option is A

$192\pi \times {10}^{-4}J$

Step1: Given data

The radius of the soap bubble, $r=0.2m$

The surface tension of soap bubble, $T=60\times {10}^{-3}$

Step 2: Formula used

$Workdone=Surfacetension\times surfacearea$

Step 3: Calculation

The soap bubble has two surfaces, one inner and one outer.

Total surface area, $A=2\times 4\pi r^2$

$\Rightarrow A=8\pi r^2$

$$\begin{gathered} Workdone=Surfacetension\times surfacearea \\ =T\times 8\pi r^2 \\ =(60\times {10}^{-3})\times 8\pi \times {(0.2)}^2 \\ =192\pi \times {10}^{-4}J \end{gathered}$$

Hence, the work done in blowing a soap bubble of radius 0.2m, given that the surface tension of soap solution is $60\times {10}^{-3}N/m$ is $192\pi \times {10}^{-4}J$.

So, option (A) is correct option.