If the surface tension of the soap bubble is 0.035 N/m, then the work done in blowing the soap bubble of radius 5 cm in the air is.
2.2 mJ
This problem involves calculating the work done to increase the surface area of a soap bubble. The key concept here is surface tension, which is the force acting per unit length along the surface of a liquid. Work must be done against this surface tension to expand the surface area.
Surface tension ($\gamma$) is defined as the energy required to increase the surface area of a liquid by a unit amount. Mathematically, it's the ratio of the excess pressure inside a bubble to the change in radius, or the force per unit length. When blowing a soap bubble, we are essentially stretching the soap film, which requires energy. Since a soap bubble has two surfaces (an inner and an outer surface), the work done is calculated considering both these surfaces.
The work done ($W$) in increasing the surface area of a liquid film is given by the product of the surface tension ($\gamma$) and the change in surface area ($\Delta A$):
$W = \gamma \times \Delta A$
For a spherical soap bubble of radius r, the surface area of one surface is $A_1 = 4 \pi r^2$. Since a soap bubble has two surfaces (inner and outer), the total surface area is $A_{total} = 2 \times A_1 = 2 \times (4 \pi r^2) = 8 \pi r^2$.
Therefore, the work done to blow a soap bubble of radius r is:
$W = \gamma \times (8 \pi r^2)$
Rounding the result to one decimal place, the work done is approximately 2.2 mJ.
Which of the following statements is NOT correct about surface tension?
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Mercury does NOT wet the glass. This is due to the property of the liquid known as
________ is a surface phenomenon.
A liquid drop is spherical in shape due to