Question

Pressure inside the two soap bubbles are 1.01 atm and 1.02 atm. The ratio of their free surface area is​

Answer 1

Answer:

Pressure × Radius = Surface Tension

Here the pressure is Gauge Pressure, i.e the difference between absolute pressure and atmospheric pressure

Hence ratio of gauge pressures is 1:2

Accordingly ratio of radii is 2:1

Since Volume∝radius

3

Hence ratio of volumes is 8:1

Answer 2

Answer:

The ratio of the surface area of the two bubbles is 4 : 1.

Explanation:

The excess pressure for the bubble with surface tension T and radius R is given by:-

$\triangle P=\frac{4T}{R}$

Excess pressure is the difference between inside pressure and outside pressure of the bubble.

ΔP$=P_i-P_o$

$P_i-P_o=\frac{4T}{R}$

We know that outside pressure or atmospheric pressure is equal to 1 atm.

Therefore, $P_i=1+\frac{4T}{R}$

Given, the inside pressure of bubble with radius R₁ is, $P_{i,1}=1.01\;atm$

$1.01=1+\frac{4T}{R_1}$

$\frac{4T}{R_1}=0.01$                                                                                .............(2)

The inside pressure of bubble with radius R₂ is, $P_{i,2}=1.02\;atm$

$1.02=1+\frac{4T}{R_2}$

$\frac{4T}{R_2}=0.02$                                                                                ..............(3)

Divide the equation (3) by eq. (2):-

$\frac{4T/R_2}{4T/R_1}=\frac{0.02}{0.01}$

$\frac{R_1}{R_2} =2$

The surface area of the bubble is given by : S = 4πR²

The ratio of the surface area of the two bubbles is:

$\frac{S_1}{S_2}=\frac{4\pi R^2_1}{4\pi R^2_2}$

$\frac{S_1}{S_2}=(\frac{R_1}{R_2})^2$

$\frac{S_1}{S_2}=(2)^2$

$\frac{S_1}{S_2}=4$

Therefore, the ratio of the surface area of the two bubbles is 4 : 1.

To learn more about "Excess pressure for the bubble"

https://brainly.in/question/9549309

To learn more about "Excess pressure inside an air bubble released in mercury "

https://brainly.in/question/14541550