Question

The surface tension of soap solution is $T=0.03{\text{Nm}}^{-1}$. If the size of a soap bubble is increased from a radius of $3\text{cm}$ to $5\text{cm}$, the work done in increasing the size of the soap bubble is

• A. $4\pi \text{mJ}$
• B. $0.2\pi \text{mJ}$
• C. $2\pi \text{mJ}$
• D. $0.4\pi \text{mJ}$

Answer 1

The correct option is D $0.4\pi \text{mJ}$

Given,
Surface tension of soap solution $\left(T\right)=0.03{\text{Nm}}^{-1}$
Initial radius of soap bubble $\left(r_1\right)=3\text{cm}$
Final radius of soap bubble $\left(r_2\right)=5\text{cm}$
Soap bubble has two surface areas
Hence,
$W=T\Delta A=T\times [2\times 4\pi (r_2^2-r_1^2\left)\right]$
$\Rightarrow$ $0.03[2\times 4\pi (5^2-3^2\left)\right]\times {10}^{-4}$
$=0.384\pi \times {10}^{-3}\text{J}\cong 0.4\pi \text{mJ}$
Option (d) is the correct answer