Question

A soap bubble of radius $1.0$ cm is formed inside another soap bubble of radius $2.0$ cm. The radius of an another soap bubble which has the same pressure difference as that between the inside of the smaller and outside of the large soap bubble , in meters is
A. $6.67 \times {10^{ - 3}}$ m
B. $3.34 \times {10^{ - 3}}$ m
C. $2.23 \times {10^{ - 3}}$ m
D. $4.5 \times {10^{ - 3}}$ m

Answer 1

To solve this problem we have to know the formula of the pressure difference. We know that, when two distinct objects are present then, both of them will face some kinds of pressures. The difference between those pressures is the pressure difference. The formula of this pressure difference of a circular object is $\dfrac{{2T}}{R}$. In this question, we are going to use this formula.

Complete step by step answer:
We have to assume that the change is pressure of inside the smaller bubble is $\Delta {P_1}$ which is equal to $\dfrac{{2T}}{{{r_1}}}$. Similarly, we can say, the pressure difference of the outside of the large soap bubble is $\Delta {P_2}$ which is equal to $\dfrac{{2T}}{{{r_2}}}$. Here we are assuming that ${r_1}$ is equal to the radius of the smaller inner bubble and ${r_2}$ is the radius of the large bubble. So, we can say the total pressure difference is equal to
$\Delta {P_2} + \Delta {P_1} = 2T(\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}}) = 3T$
$\Rightarrow{r_1} = 1.0cm \\\ \Rightarrow{r_2} = 2.0cm \\\$

$$\Rightarrow R = \dfrac{2}{3}\\\ \Rightarrow R = 0.667\,cm\\\ \therefore R = 6.67 \times {10^{ - 3}}\,m\\\$$

Hence, option A is the correct answer.

Note: We have to be careful about putting the values in the right place and to the right step to minimize the calculation part. We also can forget to transfer the unit centimeter from the meter. Because in this question we have to convert this at the end.