Question

A soap bubble of radius \[r\] is blown up to form a bubble radius of \[2r\] under isothermal conditions. If \[T\] is surface tension of soap solution, the energy spent in the blowing(in \[J\]) is:
A. \[3\pi T{r^2}\]
B. \[6\pi T{r^2}\]
C. \[12\pi T{r^2}\]
D. \[24\pi T{r^2}\]

Answer 1

Hint: As in the question we are given with isothermal condition that means temperature is constant and to calculate energy spent we will be using \[T \times \Delta A\]where \[\Delta A\]is change in area of soap bubble as radius is doubled.

Complete step by step answer:
As in the question we are given with a soap bubble with initial radius as,\[r\]
and final radius as,\[2r\]
And we are also given with surface tension as, \[T\]
So to calculate the energy spent we have formula,\[E = T \times \Delta A\]
Initial area is, \[{A_i} = 4\pi {r^2}\]
Final area is, \[{A_f} = 4\pi {\left( {2r} \right)^2}\]
As we have calculated both areas therefore\[\Delta A\]is, \[{A_f} - {A_i}\]
\[{A_f} - {A_i}\]
\[4\pi {\left( {2r} \right)^2} - 4\pi {r^2}\]
\[16\pi {r^2} - 4\pi {r^2}\]
\[12\pi {r^2}\]
So this is increase in surface area so a bubble has two surface therefore increase in surface area for bubble will be double of it, so we have increase in surface area as \[24\pi {r^2}\]
Therefore to calculate the energy spent substitute the value calculated above of\[\Delta A\] in the formula of energy spent as
\[E = T \times \Delta A\]
\[E = T \times 24\pi {r^2}\]
\[E = 24\pi {r^2}T\]
Therefore the correct option is D.

Additional Information: A cleanser bubble is an amazingly slender film of sudsy water encasing air that shapes an empty circle with a luminous surface. Cleanser bubbles normally keep going for just a short time before blasting, either all alone or on contact with another article. Amassing a few air pockets brings about froth. A soap bubble can also expand on charging because the air pocket will extend on the grounds that the charged particles consistently dispersed on it makes them repulse each other because of the electrostatic power. This will happen to both positive and adversely charged air pockets due to the charge on it.


Note: In this question we have used the formula as written but take care of two surfaces of soap bubble otherwise you will mark the C option which is present as \[12\pi {r^2}T\]and your answer will be wrong.