Surface tension of water is $0.072N{m}^{-1}$ .The excess pressure inside a water drop of a diameter $1.2mm$ is:

The correct option is: D: $24N{m}^{-2}$

Given : Surface tension, $S=0.072N{m}^{-1}$

Diameter of water drop, $D=1.2mm=0.012m$ D=1.2mm=0.012D = 1.2mm = 0.012D=1.2mm=0.012mm

Thus radius of drop $r=\frac{0.012m}{2}$

Excess pressure inside water drop, $\Delta p=\frac{2S}{r}$ ΔP=2Sr\Delta P = \dfrac{2S}{r

$\Delta p=\frac{2\times 0.072}{\frac{0.012}{2}}=24{Nm}^{-2}$

∴\therefor ΔP=2×0.0720.006=24\Delta P = \dfrac{2\times 0.072}{0.006} = 24