Question

The excess pressure due to surface tension inside a spherical drop is $6units$. If eight such drops coalesce, the excess pressure inside the new drop is:
A) $3units$
B) $6units$
C) $12units$
D) $48units$

Answer 1

In the question, it is given that eight drops coalesce to form a single drop, this implies that the volume of the single drop must be equal to the volume of eight drops combined. This will give us the relation between the radiuses of the bigger drop with respect to the small drop. Using this relation, in the formula for excess pressure, we can obtain the required value.

Complete step by step solution:
We know that drops take the most surface area available which is why they are spherical in shape.
Therefore, while evaluating the volume of the drop, we use the formula for the volume of a sphere.
Let $r$ be the radius of the small drops and $R$ be the radius of the bigger drop, formed as eight small drops coalesce.
Now, volume of sphere, we know is $= \dfrac{4}{3}\pi {r^3}$
Therefore, volume of smaller drop $= \dfrac{4}{3}\pi {r^3}$ and
The volume of bigger drop $= \dfrac{4}{3}\pi {R^3}$
Since eight small drops coalesce to form a big drop, the volume remains the same.
Therefore, we can write:
$\Rightarrow 8(\dfrac{4}{3}\pi {r^3}) = \dfrac{4}{3}\pi {R^3}$
On cancelling the common terms, we get:
$\Rightarrow R = 2r$
We know, excess pressure due to surface tension is basically the difference between output and input pressure.
Now, ${P_{OUT}} - {P_{IN}} = \dfrac{{4T}}{R}$
Where, $T$is the surface tension
Therefore, the excess pressure of the bigger drop is $\dfrac{{4T}}{R}$
Putting$R = 2r$, we get:
$\Rightarrow {P_{OUT}} - {P_{IN}} = \dfrac{{4T}}{{2r}}$
In the question, we now the excess pressure of the smaller drop is $6units$
Or, putting the values we can write:
$\Rightarrow {P_{OUT}} - {P_{IN}} = \dfrac{1}{2}(6)$
$\Rightarrow {P_{OUT}} - {P_{IN}} = 3units$
This is the required solution.

Hence, Option (A) is correct.

Note: The phenomenon of the surface tension is responsible for the shape of droplets. Due to surface tension, water droplets tend to be pulled from all the sides due to the cohesive forces and take up the most surface area. As the radius of the drop increases, the excess pressure of the drop decreases. Therefore, as excess pressure increases, the surface tension of the drop decreases.