Question

The excess pressure inside a soap bubble is equal to an oil column of height 3mm. The total surface area of the soap bubble in contact with air will be what? (given, density of oil = 0.9g/cubic cm , surface tension of soap solution = 0.03 N/m and g=10m/square sec). "PLS GIVE ANSWER WITH DETAILED EXPLANATION".

Answer 1

Explanation:

Hope it help you

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Answer 2

Given:

Height of the column $=3$ $mm$

Density of the oil $=0.9$ $g/cm^{3}$

Surface tension of the soap solution $=0.03$ $N/m$

Gravity acceleration $=10$ $m/s^{2}$

To find:

Total surface area of the soap bubble.

Explanation:

We have been given the soap bubble is in contact with the air and the pressure inside the liquid column is equal to the pressure inside the soap bubble and, we know The pressure inside the soap bubble is greater than the atmospheric pressure.

The pressure inside the soap bubble is given by $\frac{4T}{R}$ where, $T$ is the surface tension and $R$ is the radius of the soap bubble.

In a liquid column of height $h$, where density of liquid is ρ the pressure at the bottom is given by ρgh where g is the acceleration due to gravity.

Solution:

Step 1

According to the question,

The pressure inside the capillary tube = The pressure inside the soap bubble

ρgh $=\frac{4T}{R}$

We have,

ρ $=0.9g/cm^{3} =0.9$ × $10^{3}kg/m^{3}$

$g=10$ $m/s^{2}$

$h=3$ $mm=3$ × $10^{-3}$ $m$

$T=0.03N/m$

Substituting the known values, we get

$0.9*10^{3}*(10)*3*10^{-3} =\frac{4(0.03)}{R}$

$27=\frac{0.12}{R}$

$R=0.0045m$     $;$ $or$

$R=4.5$ $mm$

Step 2

Now,

The area of the soap bubble $=4\pi R^{2}$

We have, $R=0.0045m$

Hence,

Area of the soap bubble will be,

$A=4(\frac{22}{7} )(4.5*10^{-3} )^{2}$

$A=\frac{1782*10^{-6} }{7}$

$A={254.6}$ × $10^{-6}$ $m^{2}$

Final answer:

The total surface area of the soap bubble is 254.6 × 10⁻⁶ m².