
: A soap bubble of radius ${r_1}$ is placed on another soap bubble of radius ${r_2}$.$\left( {{r_1} < {r_2}} \right)$. The radius $R$ of the soapy film separating the two bubbles is:
A. ${r_1} + {r_2}$
B. $\sqrt {{r_1}^2 + {r_2}^2} $
C. ${r_1}^2 + {r_2}^2$
D. $\dfrac{{{r_2}{r_1}}}{{{r_2} - {r_1}}}$
Answer
590.1k+ views
Hint: To find the radius $R$ of the soapy film, we will use the expression of the pressure exerted by the soap bubble. We will write the expression for the pressure exerted by the soap bubble of ${r_1}$ and ${r_2}$. Then we write the expression for the pressure exerted by the soap bubble of radius $R$. Now, we will equate the difference of the pressure exerted by the radius ${r_1}$ and ${r_2}$with the pressure exerted by the soap bubble of radius $R$.
Formula Used:
The pressure exerted by the soap bubble can be expressed as:
$\Rightarrow P = \dfrac{{4T}}{r}$
Where $P$is the pressure, $T$is the surface tension and $r$is the radius of the bubble.
Complete step by step answer:
Given:
The radius of the soap bubble is ${r_1}$.
The radius of another soap bubble is ${r_2}$.
The radius ${r_1}$is less than radius ${r_2}$.
We will assume ${P_1}$ and ${P_2}$ as the pressure exerted by the bubble of radius ${r_1}$ and ${r_2}$ respectively.
We will express the relation for the pressure exerted by the bubble of radius ${r_1}$.
$\Rightarrow {P_1} = \dfrac{{4T}}{{{r_1}}}$……(i)
Where $T$is the surface tension in the soap bubble.
We will express the relation for the pressure exerted by the bubble of radius ${r_2}$.
$\Rightarrow {P_2} = \dfrac{{4T}}{{{r_2}}}$……(ii)
It is given in the question that the radius ${r_1}$ is less than radius ${r_2}$. Therefore $\Rightarrow {P_2}$ is less than ${P_1}$.
The radius of the bubble separating the two bubbles will be $R$.
We will write the expression for the pressure in the new bubble.
$\Rightarrow {P_R} = \dfrac{{4T}}{R}$……(iii)
We also know that this pressure will be equivalent to the difference of the pressure exerted by the bubbles of radius ${r_1}$ and ${r_2}$. This can be expressed as:
$\Rightarrow {P_R} = {P_1} - {P_2}$
We will substitute the values of ${P_1}$,${P_2}$ and ${P_R}$ from the equation (i), (ii) and (iii) respectively in the above expression.
$
\Rightarrow \dfrac{{4T}}{R} = \dfrac{{4T}}{{{r_1}}} - \dfrac{{4T}}{{{r_2}}}\\
\Rightarrow \dfrac{1}{R} = \dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}\\
\Rightarrow \dfrac{1}{R} = \dfrac{{{r_2} - {r_1}}}{{{r_1}{r_2}}}
$
On reciprocating the above expression, we will get
$\Rightarrow R = \dfrac{{{r_2}{r_1}}}{{{r_2} - {r_1}}}$
Hence, the radius $R$ of the soap bubble is $\dfrac{{{r_2}{r_1}}}{{{r_2} - {r_1}}}$.
Therefore, option D is the correct answer.
Note: The value of surface tension for an interface say liquid air interface is always a constant. In this question, we are assuming $T$ as the value of surface tension. This is applicable to all the three bubbles of different radius as the surface tension will have the same value $T$ for a particular interface.
Formula Used:
The pressure exerted by the soap bubble can be expressed as:
$\Rightarrow P = \dfrac{{4T}}{r}$
Where $P$is the pressure, $T$is the surface tension and $r$is the radius of the bubble.
Complete step by step answer:
Given:
The radius of the soap bubble is ${r_1}$.
The radius of another soap bubble is ${r_2}$.
The radius ${r_1}$is less than radius ${r_2}$.
We will assume ${P_1}$ and ${P_2}$ as the pressure exerted by the bubble of radius ${r_1}$ and ${r_2}$ respectively.
We will express the relation for the pressure exerted by the bubble of radius ${r_1}$.
$\Rightarrow {P_1} = \dfrac{{4T}}{{{r_1}}}$……(i)
Where $T$is the surface tension in the soap bubble.
We will express the relation for the pressure exerted by the bubble of radius ${r_2}$.
$\Rightarrow {P_2} = \dfrac{{4T}}{{{r_2}}}$……(ii)
It is given in the question that the radius ${r_1}$ is less than radius ${r_2}$. Therefore $\Rightarrow {P_2}$ is less than ${P_1}$.
The radius of the bubble separating the two bubbles will be $R$.
We will write the expression for the pressure in the new bubble.
$\Rightarrow {P_R} = \dfrac{{4T}}{R}$……(iii)
We also know that this pressure will be equivalent to the difference of the pressure exerted by the bubbles of radius ${r_1}$ and ${r_2}$. This can be expressed as:
$\Rightarrow {P_R} = {P_1} - {P_2}$
We will substitute the values of ${P_1}$,${P_2}$ and ${P_R}$ from the equation (i), (ii) and (iii) respectively in the above expression.
$
\Rightarrow \dfrac{{4T}}{R} = \dfrac{{4T}}{{{r_1}}} - \dfrac{{4T}}{{{r_2}}}\\
\Rightarrow \dfrac{1}{R} = \dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}\\
\Rightarrow \dfrac{1}{R} = \dfrac{{{r_2} - {r_1}}}{{{r_1}{r_2}}}
$
On reciprocating the above expression, we will get
$\Rightarrow R = \dfrac{{{r_2}{r_1}}}{{{r_2} - {r_1}}}$
Hence, the radius $R$ of the soap bubble is $\dfrac{{{r_2}{r_1}}}{{{r_2} - {r_1}}}$.
Therefore, option D is the correct answer.
Note: The value of surface tension for an interface say liquid air interface is always a constant. In this question, we are assuming $T$ as the value of surface tension. This is applicable to all the three bubbles of different radius as the surface tension will have the same value $T$ for a particular interface.
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