Question

What is the excess pressure inside a soap bubble with a radius of 1.5 cm and the surface tension of 0.03 N/m?
A.7 pa
B.3 pa
C.2 pa
D.8 pa

Answer 1

Hint: In this question we first write all the force component that are acting on the bubble and then equate them following the sign convention that is \[{F_{{P_1}}} + {F_T} = {F_{{P_2}}}\] then we get the expression for excess pressure as $\Delta P = {P_2} - {P_1} = \dfrac{{4T}}{R}$ then we substitute the values provided in the question and get the answer.

Complete step-by-step answer:
First will find the excess pressure for a spherical bubble in water and for that, we first draw the cross-section view of a spherical bubble with radius R equal to 1.5 cm and surface tension T equal to 0.03 N/m as can be shown in figure 1.
We know that a bubble has two liquid surfaces in contact with air, one is the inner surface of the bubble and the other one is the outer surface of the bubble. So, the force due to surface tension will become
\[2 \times \left( {2\pi RT} \right)\].

Figure 1
Now writing the all forces acting on the bubble that are
Force due to the external pressure that is \[{F_{{P_1}}} = {P_1}\pi {R^2}\] acting in the right direction
Force due to the internal pressure that is \[{F_{{P_2}}} = {P_2}\pi {R^2}\] acting in the left direction
Force due to the surface tension that is \[{F_T} = 4\pi RT\] acting in the right direction
Since the drop is in equilibrium we will get
\[{F_{{P_1}}} + {F_T} = {F_{{P_2}}}\]
\[ \Rightarrow {P_1}\pi {R^2} + 4\pi RT = {P_2}\pi {R^2}\]
\[ \Rightarrow ({P_2} - {P_1})\pi {R^2} = 4\pi RT\]
$ \Rightarrow \Delta P = {P_2} - {P_1} = \dfrac{{4T}}{R}$ ---------------------------- (1)
Now substituting $R = 1.5cm = 1.5 \times {10^{ - 2}}m$ and $T = 0.03N/m$ in equation (1) to find the excess pressure, we get
${P_2} - {P_1} = \dfrac{{4T}}{R} = \dfrac{{4 \times 0.03}}{{1.5 \times {{10}^{ - 2}}}}$
$ \Rightarrow \Delta P = {P_2} - {P_1} = \dfrac{{4 \times 3}}{{1.5}}$
$ \Rightarrow \Delta P = {P_2} - {P_1} = 8pa$
Hence the Excess pressure $\Delta P = 8pa$ inside the bubble,
So option D is correct.

Note:For these types of questions, we need the concept of surface tension and pressure due to the air-water interface. We also need to remember the excess pressure for different shapes of the air-water interface that we see in real life that is cylindrical drops and bubbles, spherical drops and bubbles, and meniscus. We need to know how excess pressure varies with the free surfaces and how to calculate numerically related to this topic.