The excess pressure inside a liquid drop is 500 Nm^-2. if the radius of the drop is 2mm, the surface tension of liquid is X $\times$ 10^-3Nm^-1. The value of X is______

The excess pressure inside a liquid drop is given by the formula: ΔP=2T​/r ​ 

Where:  ΔP is the excess pressure (500 Nm^−2) 

T is the surface tension of the liquid 

r is the radius of the drop (2 mm or 0.002 m) 

Rearranging the formula to find the surface tension ( 𝑇 ):

$\begin{array}{l}\mathrm{p}={\mathrm{p}}_{\circ }+\frac{2\mathrm{T}}{\mathrm{R}}\Rightarrow \mathrm{p}-{\mathrm{p}}_{\circ }=\frac{2\mathrm{T}}{\mathrm{R}}\\ 500=\frac{2\times \mathrm{T}}{2\times {10}^{-3}}\\ \mathrm{T}=500\times {10}^{-3}\\ \mathrm{SO},\mathrm{X}=500\end{array}$