Question

Pressure inside two soap bubbles is 1.01 and 1.02 atmosphere. The ratio between their volume is:
A. $102 : 101$
B. ${{\left( 102 \right)}^{2}}:{{\left( 101 \right)}^{2}}$
C. $8 : 1$
D. $2 : 1$

Answer 1

Hint: Use the formula for the excess pressure inside a bubble and find the ratio of the radii of the two bubbles with the help of the given data. To find the ratio of the volumes of the bubbles use the ratio of the radii of the two.

Formula used:
The excess the pressure for bubble of radius R with surface tension T is given as,
$\Delta P=\dfrac{4T}{R}$
$\Rightarrow\Delta P={{P}_{i}}-{{P}_{o}}$

Complete step by step answer:
When a bubble is formed, the pressure inside the bubble is more than the pressure outside the bubble. This means that the pressure exerted on the inner surface of the bubble is greater than the pressure exerted on its outer surface. Therefore, there is some difference between the pressure inside the bubble and the pressure outside the bubble. In other words, we can say that there is excess pressure inside the bubble. The excess pressure for the bubble of radius R with surface tension T is given as $\Delta P=\dfrac{4T}{R}$ ….. (i)

And we discuss that the excess pressure is the difference between the pressures inside and outside the bubble. Therefore, the excess can also be written as,
$\Delta P={{P}_{i}}-{{P}_{o}}$ ….. (ii)
From (i) and (ii) we get,
$\dfrac{4T}{R}={{P}_{i}}-{{P}_{o}}$
The pressure exerted on a bubble from outside is the atmospheric pressure. The atmospheric pressure is equal to $1atm$.
${{P}_{o}}=1atm$
Now let us write the excess pressure for both the bubbles.
Let the radii of the bubbles with inside pressure equal to ${{P}_{i,1}}=1.01atm$ and ${{P}_{i,2}}=1.02atm$ be ${{R}_{1}}$ and ${{R}_{2}}$, respectively.
Since both are soap bubbles, both will have equal surface tension (T).
Therefore,
${{P}_{i,1}}-{{P}_{o}}=\dfrac{4T}{{{R}_{1}}}$
$\Rightarrow {{P}_{i,2}}-{{P}_{o}}=\dfrac{4T}{{{R}_{2}}}$
Now substitute the known values.
$1.01-1=\dfrac{4T}{{{R}_{1}}}$
$\Rightarrow \dfrac{4T}{{{R}_{1}}}=0.01$ ….. (iii)
And
$1.02-1=\dfrac{4T}{{{R}_{2}}}$
$\Rightarrow \dfrac{4T}{{{R}_{2}}}=0.02$ ….. (iv).
Now, divide (iv) by (iii).
$\dfrac{\dfrac{4T}{{{R}_{2}}}}{\dfrac{4T}{{{R}_{1}}}}=\dfrac{0.02}{0.01}$
$\Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=2$.
The volume of a sphere is given as $V=\dfrac{4}{3}\pi {{R}^{3}}$, where R is the radius of the sphere.Therefore,
${{V}_{1}}=\dfrac{4}{3}\pi R_{1}^{3}$ …. (v)
And ${{V}_{2}}=\dfrac{4}{3}\pi R_{2}^{3}$ …. (vi), w
here ${{V}_{1}}$ and ${{V}_{2}}$ are the values of the bubbles.
Divide (v) by (vi).
$ \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{\dfrac{4}{3}\pi R_{1}^{3}}{\dfrac{4}{3}\pi R_{2}^{3}}\\
\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}} =\dfrac{R_{1}^{3}}{R_{2}^{3}}={{\left( \dfrac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}$
Substitute the value of $\dfrac{{{R}_{1}}}{{{R}_{2}}}$ that we found.
$\dfrac{{{V}_{1}}}{{{V}_{2}}}={{\left( 2 \right)}^{3}}\\
\therefore \dfrac{{{V}_{1}}}{{{V}_{2}}} =8$
This means that the ratio of the volumes of the bubbles is 8 : 1.

Hence, the correct option is C.

Note:The shape of a bubble will always be spherical. This is due to the surface tension of the bubble. Surface tension is the force per unit length exerted by the molecules on each other. Due to the surface tension, the bubble tends to occupy minimum surface area and sphere has least surface area. Hence, a bubble is formed in a spherical shape.