If the excess pressure inside a soap bubble is balanced by an oil column of height 2 mm, then the surface tension of soap solution will be (r = 1cm, density of oil = 0.8g/cm3)
${\text{A}}{\text{. 3}}{\text{.9N/m}}$
${\text{B}}{\text{. 3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{N/m}}$
${\text{C}}{\text{. 3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{N/m}}$
${\text{D}}{\text{. 3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{ - 1}}{\text{N/m}}$

Hint: The excess pressure inside the bubble is given by, $P = \dfrac{{4T}}{r}$ and the pressure to height column of oil is given as, $P = \rho hg$ .

Complete step-by-step answer:

Formula used - $P = \dfrac{{4T}}{r}$ , $P = \rho hg$

Given that the excess pressure inside a soap bubble is balanced by an oil column of height 2 mm.
To find – surface tension of soap solution.

Now, as we know, the excess pressure inside the soap bubble is given by the formula, $P = \dfrac{{4T}}{r}$ and the pressure to height column of oil is given by the formula, $P = \rho hg$ .

Now, since the excess pressure inside the bubble is balanced by the height of the column and so we can write that-

$\dfrac{{4T}}{r} = \rho hg$
now, $r = 1cm = 1 \times {10^{ - 2}}m$
\[\rho = 0.8g/c{m^3} = \dfrac{{0.8g}}{{c{m^3}}} = \dfrac{{0.8 \times {{10}^{ - 3}}kg}}{{{{({{10}^{ - 2}})}^3}{m^3}}} = 0.8 \times {10^3}kg/{m^3}\]
$h = 2mm = 2 \times {10^{ - 3}}m$
So, now using $\dfrac{{4T}}{r} = \rho hg$ , we can write-

$T = \dfrac{{r\rho hg}}{4}$
Putting the value, we get-

$
  T = \dfrac{{1 \times {{10}^{ - 2}} \times 0.8 \times {{10}^3} \times 9.8 \times 2 \times {{10}^{ - 3}}}}{4} = \dfrac{{0.8 \times {{10}^{ - 2}} \times 9.8 \times 2}}{4} = \dfrac{{7.84 \times {{10}^{ - 2}}}}{2} \\
   \Rightarrow T = 3.9 \times {10^{ - 2}}N/m \\
$

So, the surface tension of soap solution is ${\text{3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{N/m}}$ .

Hence, the correct option is B.

Note: Whenever such types of questions appear then always write down the things given in question. Then, as mentioned in the solution, that the excess pressure inside a soap bubble is balanced by an oil column, so we have written, $\dfrac{{4T}}{r} = \rho hg$ and then after putting the values of density, r, h and g we found out the value of surface tension, T.