Question

The excess pressure inside a soap bubble is three times that of other bubble, than the ratio of their volume will be
A. $1:3$
B. $1:10$
C. $1:17$
D. $1:27$

Answer 1

Hint: Study about the formation and concepts of soap bubble formation, find out what is excess pressure and obtain the mathematical expression for excess pressure. Then put the values according to the two given conditions where we will take a constant surface tension for both the bubbles. At the end we can take the ratio of the volumes of the two bubbles.

Complete Step-by-Step solution:
In a soap bubble we have two surfaces. Consider a soap bubble of radius R and surface tension T. now, due to the presence of surface tension the outer surface experiences a net inward force. So, we have excess pressure on the inside.
Excess pressure inside a soap bubble can be expressed as,
$P=\dfrac{4T}{R}$
Now, let the first soap bubble has a excess pressure ${{P}_{1}}$ and the second soap bubble have the excess pressure ${{P}_{2}}$
According to the question,
${{P}_{1}}=3{{P}_{2}}$
Let the first bubble have radius ${{R}_{1}}$ and the second bubble has a radius ${{R}_{2}}$.
Now,
$\begin{align}
  & {{P}_{1}}=3{{P}_{2}} \\
 & \dfrac{4T}{{{R}_{1}}}=3\times \dfrac{4T}{{{R}_{2}}} \\
 & \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{1}{3\infty } \\
\end{align}$
Now, taking the ratio of their volumes, we get
$\begin{align}
  & \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{\dfrac{4}{3}\pi R_{1}^{3}}{\dfrac{4}{3}\pi R_{2}^{3}}={{\left( \dfrac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}} \\
 & \dfrac{{{V}_{1}}}{{{V}_{2}}}={{\left( \dfrac{1}{3} \right)}^{3}} \\
 & \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{1}{27} \\
\end{align}$
So, the ratio of their volumes will be, $1:27$
The correct option is, (D).

Note: In a liquid the surface tension which holds the liquid molecules together is very strong and we can’t obtain bubbles from the liquids. When we add soap or detergents to the liquids it can lower the surface tension of the liquid by a certain amount and soap bubbles can form due to this low surface tension.
Due to surface tension the bubbles are spherical in shape.
When drops are formed, they are also spherical in shape because of the surface tension. In a drop we have only one surface and hence the excess pressure becomes, $P=\dfrac{2T}{R}$