Question

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

Answer 1

$T=0.072$N/m

$d=4$mm

$\therefore r=\frac{4}{2}$mm$=2\times {10}^{-3}$m

$P_0=1.013\times {10}^5$N/$m^2$

To find :Pressure inside the raindrop $\left(P_i\right)$

Formula:$P_i=P_0+\frac{2T}{r}$

$=1.013\times {10}^5+\frac{2\times 0.072}{2\times {10}^{-3}}$

$=1.013\times {10}^5+0.00072\times {10}^5$

$=1.01372\times {10}^5$Pa

The pressure inside the raindrop is $1.01372\times {10}^5$Pa