Question

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

Answer 1

Use the formula for excess pressure inside a soap bubble. This formula gives the relation between the excess pressure in the soap bubble, surface tension and radius of the soap bubble. Also, use the formula for volume of a spherical object. Determine the ratio of radii of the soap bubbles and then determine the ratio of volumes of the soap bubbles.

Formulae used:
The excess pressure $\Delta P$ inside a soap bubble is given by
$\Delta P = \dfrac{{4T}}{R}$ …… (1)
Here, $T$ is the surface tension on the soap bubble and $R$ is the radius of the soap bubble.
The volume $V$ of a spherical object is given by
$V = \dfrac{4}{3}\pi {R^3}$ …… (2)
Here, $R$ is the radius of the spherical object.

Complete step by step answer:
Let us consider that the excess pressure inside the first and second soap bubble is $\Delta {P_1}$ and $\Delta {P_2}$ respectively.
Let the radii and volumes of the first and second soap bubbles are ${V_1}$ and ${V_2}$ and ${V_1}$ and ${V_2}$ respectively.

We have given that the excess pressure inside the first soap bubble is three times the excess pressure in the second soap bubble.
$\Delta {P_1} = 3\Delta {P_2}$
The surface tension for both the soap bubbles is the same.

From equation (1), we can conclude that the excess pressure inside the soap bubble is inversely proportional to radius of the bubble.
$\Delta P \propto \dfrac{1}{R}$
Rewrite the above relation for two soap bubbles.
$\dfrac{{\Delta {P_1}}}{{\Delta {P_2}}} = \dfrac{{{R_2}}}{{{R_1}}}$
Substitute $3\Delta {P_2}$ for $\Delta {P_1}$ in the above equation.
$\dfrac{{3\Delta {P_2}}}{{\Delta {P_2}}} = \dfrac{{{R_2}}}{{{R_1}}}$
$\Rightarrow 3 = \dfrac{{{R_2}}}{{{R_1}}}$
$\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{3}$
The soap bubbles are spherical in shape.

Now from equation (2), we can conclude that the volume of the spherical soap bubble is directly proportional to the cube of radius of the soap bubble.
$V \propto {R^3}$
Rewrite the above relation for two soap bubbles.
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{R_1^3}}{{R_2^3}}$
$\Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = {\left( {\dfrac{{{R_1}}}{{{R_2}}}} \right)^3}$
Substitute $\dfrac{1}{3}$ for $\dfrac{{{R_1}}}{{{R_2}}}$ in the above equation.
$\Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = {\left( {\dfrac{1}{3}} \right)^3}$
$\therefore \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{1}{{27}}$
Therefore, the ratio of volumes of the soap bubbles is 1:27.

Hence, the correct option is D.

Note: The students may think that how we can take the shape of soap bubble spherical and use formula for volume of sphere if its shape is not mentioned in the question. But the surface tension and excess pressure acting on the soap bubble balances such that the shape of the soap bubble in steady state is always spherical.