Question

A soap bubble of diameter $8mm$ is formed in air. If the surface tension of liquid is $30dyne.c{m^{ - 1}}$, then excess pressure inside the soap bubble is
(A) $150~dyne.c{m^{ - 2}}$
(B) $300~dyne.c{m^{ - 2}}$
(C) $3 \times {10^{ - 3}}~dyne.c{m^{ - 2}}$
(D) $12~dyne.c{m^{ - 2}}$

Answer 1

Hint: In order to solve this question, we will first find the radius of the soap bubble, and then using the general formula of pressure inside the soap bubble using tension and radius, we will solve for the change in pressure inside the soap bubble.

Formula Used:
If T is the surface tension of the liquid of the soap bubble and r is the radius of the soap bubble and $\Delta P$ is the excess pressure inside the soap bubble then, we have
$\Delta P = \dfrac{{4T}}{r}$

Complete answer:
We have given that, A soap bubble of diameter $8mm$ is formed in air and the surface tension of the liquid is $30~dyne.c{m^{ - 1}}$ now, the radius of the soap bubble will be
$r = \dfrac{8}{2} = 4mm$
$r = 0.4~cm $ and
surface tension is given by $T = 30~dyne.c{m^{ - 1}}$

Now using the formula $\Delta P = \dfrac{{4T}}{r}$ for the excess pressure inside the soap bubble and putting the values of required parameters we get,
$\Delta P = \dfrac{{4(30)}}{{0.4}}$
$ \Delta P = 300~dyne.c{m^{ - 2}}$
So, the excess pressure inside the soap bubble will be $300~dyne.c{m^{ - 2}}$

Hence, the correct option is (B) $300~dyne.c{m^{ - 2}}$

Note:It should be remembered that the surface tension is given in the units of dyne and centimetres, so always convert all the physical quantities into the same unit such as radius is converted from millimetres to centimetres with conversion formula $1mm = 0.1cm$.