Question

A rain drop of diameter 4 mm is about to fall on the ground Calculate the pressure inside the raindrop.
[Surface Tension =$T=0.072N/m$, atmospheric pressure=${{P}_{a}}=1.013\times {{10}^{5}}N/{{m}^{2}}$]

Answer 1

Before proceeding with finding the solution, we must be thorough with a couple of terms Firstly, Surface tension which could be defined as the property of the surface of a liquid that allows resisting an external force due to the cohesive nature of the water molecules and symbolized as T.
And secondly, atmospheric pressure which is the pressure within the atmosphere of earth and is given to us here is${{P}_{a}}=1.013\times {{10}^{5}}N/{{m}^{2}}$.

Complete step by step answer:
Things given to us are:
Surface Tension =$T=0.072N/m$;
Atmospheric pressure=${{P}_{a}}=1.013\times {{10}^{5}}N/{{m}^{2}}$;
And diameter $d=4mm$
$\therefore r=\dfrac{4}{2}mm=2\times {{10}^{-3}}m$
And we are asked to find the Pressure inside the raindrop $\left( {{P}_{i}} \right)$
So the formula being used is ${{P}_{i}}={{P}_{a}}+\dfrac{2T}{r}$
Putting all the value in the formula
$=1.013\times {{10}^{5}}+\dfrac{2\times 0.072}{2\times {{10}^{-3}}}$
$=1.013\times {{10}^{5}}+0.00072\times {{10}^{5}}$
$=1.01372\times {{10}^{5}}Pa$

Hence, the pressure inside the raindrop is $1.01372\times {{10}^{5}}Pa$.

Note:
One must notice and keep in mind that in order to get the correct answer, you need to be accurate with the conversion of SI units just like we did while calculating the radius of the raindrop where we converted the radius given in millimeters to meters .