Question

A drop of olive oil of radius $0.30mm$ spreads into a nearly circular film of radius $12cm$ on the water surface. Estimate the molecular size of a drop of olive oil.

Answer 1

Hint: In order to solve this question, firstly we will find the volume of a spherical olive oil drop which is calculated by $V = \dfrac{4}{3}\pi {r^3}$ where $r$ is the radius of sphere and then will calculate the area of circular film formed on water surface which is calculated as $A = \pi {R^2}$ where $R$ is the radius of circular film.

Complete step-by-step solution:
It’s given that radius of spherical oil drop of olive oil is $r = 0.30mm$
We will convert this millimetre into centimetres which is calculated as:
$1mm = 0.1cm$
$0.30mm = 0.03cm$
So, radius of olive oil drop is $r = 0.03cm$
Volume of this spherical oil drop is given by:
$V = \dfrac{4}{3}\pi {r^3}$
$V = \dfrac{4}{3}\pi {(0.03)^3} \to (i)$
Now, as this spherical drop turns into a circular film on the water surface so, the area formed on water surface by olive oil drop is calculated as:
$A = \pi {R^2}$ Where $R$ is the radius of film given as $R = 12cm$
$A = \pi {(12)^2} \to (ii)$
In order to know the diameter of olive oil drop we will divide its volume by the area formed on the water surface.
So, let olive oil diameter be $d$ then it’s calculated as
$d = \dfrac{{Volume}}{{Area}}$
So using equation $(i)and(ii)$ we get,
$d = \dfrac{{4\pi {{(0.03)}^3}}}{{3\pi {{(12)}^2}}}$
$d = \dfrac{{4{{(0.03)}^3}}}{{432}}$
$d = 2.5 \times {10^{ - 7}}cm$
Hence, the size of olive oil as a diameter is $d = 2.5 \times {10^{ - 7}}cm$ .

Note:It should be remembered that, volume of a body is nothing but the multiplication of its area in two dimensional surface and its height and the conversion of millimetres into centimetres area as $1mm = 0.1cm$ and if we need to convert this into metre then we will use $1mm = 0.001m$ .