Question

Work done in increasing the size of a soap bubble from a radius of \[3cm\text{ to 5}cm\]is nearly (Surface tension of a soap solution=$0.03N{{m}^{-1}}$)
$\begin{align}
  & \text{A}\text{. 4}\pi mJ \\
 & \text{B}\text{. 0}\text{.2}\pi mJ \\
 & \text{C}\text{. 2}\pi mJ \\
 & \text{D}\text{. 0}\text{.4}\pi mJ \\
\end{align}$

Answer 1

Hint: Work done in increasing the surface area of the soap bubble is the work done against surface tension. Surface tension is the tendency of a liquid surface to contract to occupy minimum area. This work done against the surface tension is the product of surface tension and the change in volume. First convert the radius to SI unit and using formula calculate the work done against the surface tension to increase the surface area.

Formula used:
Increase in surface energy or work done in increasing the surface area is
$W=2\times \text{Surface tension }\times \text{ increase in area of liquid surface}$

Complete answer:
According to the question the soap solution has a surface tension $0.03N{{m}^{-1}}$.
Initial radius of soap bubble , ${{r}_{1}}=3cm=3\times {{10}^{-2}}m$
Initial surface area of soap bubble,${{A}_{1}}=4\pi {{r}_{1}}^{2}=36\pi \times {{10}^{-4}}{{m}^{2}}$
Final radius of soap bubble, ${{r}_{2}}=5cm=5\times {{10}^{-2}}m$
Final surface area of soap bubble,${{A}_{2}}=4\pi r_{2}^{2}=100\pi \times {{10}^{-4}}{{m}^{2}}$
Change in surface area $\Delta A={{A}_{2}}-{{A}_{1}}=100\pi \times {{10}^{-4}}{{m}^{2}}-36\pi \times {{10}^{-4}}{{m}^{2}}=64\pi \times {{10}^{-4}}{{m}^{2}}$

The work done in increasing the area
$\begin{align}
  & W=2\times \text{Surface tension}\times \text{ increase in surface area} \\
 & \Rightarrow \text{W=2}\times \text{0}\text{.03N}{{\text{m}}^{-1}}\times 64\pi \times {{10}^{-4}}{{m}^{2}}=3.84\times \pi \times {{10}^{-4}}Nm=3.84\times \pi \times {{10}^{-4}}J \\
 & \Rightarrow W=.384\times \pi \times {{10}^{-3}}J\simeq 0.4\pi mJ \\
\end{align}$

So, the correct answer is “Option D”.

Additional Information:
Surface tension is the surface of a liquid by virtue of which the free surface of a liquid at rest behaves like an elastic stretched membrane tending to contract so as to occupy the minimum surface area.
Surface tension is measured as the force acting per unit length of an imaginary line drawn on the liquid surface, the direction of force being perpendicular to this line and tangential to the liquid surface.

Note:
The unit of surface tension is $N{{m}^{-1}}$ and its dimension is $\left[ {{\text{M}}^{1}}{{\text{L}}^{0}}{{\text{T}}^{-2}} \right]$. Some examples of surface tension are,
(i) A needle supported on the water surface.
(ii) Raindrops are spherical due to surface tension.
(iii) Small mercury drops are spherical but larger one tend to flattened.
(iv) The hair of a painting brush clings together due to surface tension.