Question

If T is surface tension of soap solution, the amount of work done is blowing a soap bubble from a diameter D to a diameter 2D.
A.) $2\pi {{D}^{2}}T$
B.) $4\pi {{D}^{2}}T$
C.) $6\pi {{D}^{2}}T$
D.) $8\pi {{D}^{2}}T$

Answer 1

Hint: Surface tension is nothing but stretchable membrane on liquid. Soap bubble has two surfaces, inner and outer. Use the formula of work done. Work done is nothing but change in area and tension in surface tension. Here displacement is replaced by area.

Complete step by step solution:

Initial diameter is D and the final diameter is 2D.

I.e Di=D and Df=2D

Now to find the amount of work done in blowing soap bubbles from diameter D to 2D is.
We know that surface tension is given by,

$T=\dfrac{wD}{A}$

And work done in blowing bubble is given by,

$\Delta w=T.dA$

We know that bubbles have two surfaces, one is inner and other one is outer.
Therefore our work down is given by,

$\Delta w=2T.dA$

Since our diameter is from D to 2D, therefore our area dA will also change.

$ \Delta w=2T\left[ 4\pi {{\left( \dfrac{2D}{2} \right)}^{2}}-4\pi {{\left( \dfrac{D}{2} \right)}^{2}} \right] $
By solving,

$\Delta w=2T\left( 4\pi \dfrac{3{{D}^{2}}}{4} \right)$
$\Delta w=6\pi {{D}^{2}}T$
Answer- (C)

Additional information:
Surface tension is the force per unit length acting at right angles to an imaginary line drawn on the surface of liquid. If the length of the line is l and F is force acting on it. S.I. unit of surface tension is N/m, surface tension is equal to potential energy per unit of liquid surface.

Note: While putting value in place of area, always look for surface. For soap bubbles, the number of surfaces are two. In the given question, diameter is given, not radius. Most students just forget to convert diameter into radius.