Question

The radius of a soup bubble is increased from $\dfrac{1}{{\sqrt \pi }}cm\,to\,\dfrac{2}{{\sqrt \pi }}cm$. If the surface tension of water is 30 dynes per cm, then the work done will be
(A) 180 erg
(B) 360 erg
(C) 720 erg
(D) 960 erg

Answer 1

Hint: A soap bubble is in a spherical shape and its assumed to be a perfect sphere and due to atmospheric pressure a surface tension acts on normally on the surface of the sphere and due tothis work done is needed to increase or decrease its size so using general relation of work between tension and surface area, we will solve for work done according to the given parameters value.


Complete answer:
Start with the formula of amount of work done in blowing a soup bubble. We know that, Work done is equal to tension in surface energy and surface energy is tension multiply to change in area as follows:
$W = T \times \Delta A$
Where,
$W = workdone$
$T = Surface\,tension$
$\Delta A = Area$

Now, in given case of soup bubble, work done will be;
$W = 8\pi T\left( {r_2^2 - r_1^1} \right)$

From the question, we know that;
${r_2} = \dfrac{2}{{\sqrt \pi }}\,and\,{r_1} = \dfrac{1}{{\sqrt \pi }}$

Putting these values in equation of work done, we get;
$W = 8\pi T\left[ {{{\left( {\dfrac{2}{{\sqrt \pi }}} \right)}^2} - {{\left( {\dfrac{1}{{\sqrt \pi }}} \right)}^2}} \right]$

By solving, we get;
$W = 8\pi \times 30 \times \dfrac{3}{\pi } = 720\,erg$

Hence the correct answer is Option(C).

Note: While solving these kind of questions, make sure the work done formula is applied only when there is change in radius of the soap bubble and the decrease in surface area is changed while surface tension remain same and surface tension spreads more densely on small spherical bubble than larger bubbles.