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Consider a spherical liquid and let the outside pressure be `P_(0)` and inside presure be `P_(i)`, such that the excess pressure is `P_(i)-P_(0).` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/GRK_10Y_SP_HSC_PHY_FEB_18_E01_016_S01.png" width="80%"> <br> Let, the radius of the drop increase from r to `r=Delta r`, where `Delta r` is very small, so that the pressure inside the drop remains almost constant. <br> Initial surface area `(A_(1))=4pi r^(2)` <br> Final surface area `(A_(2))=4pi (r+Delta r)^(2)` <br> `=4pi (r^(2)+ 2r Delta r +Delta r^(2))` <br> `=4pi r^(2)+8 pi r Delta r +4 pi Delta r^(2)` <br> As `Delta r` is very small, `Delta r^(2)` is neglected (i.e., `4 pi Delta r^(2) =0`) <br> Increase in surface area (dA) <br> `=A_(2)-A_(1)=4pi r^(2)+8pi r Delta r-4pi r^(2)` <br> Increase in surface area (dA) <br> `=8 pi r Delta r` <br> Work done to increase the surface area `8pi r Delta r` is extra surface energy. <br> `therefore dW=TdA ` <br> `therefore dW=Txx 8 pi r Delta r " " ` ...(i) <br> This work done is equal to the product of the force and the distace `Delta r. ` <br> Excess `" " ` Force = Excess Pressure `xx` Area <br> `dF =(P_(1)-P_(o))4pi r^(2)` <br> The increase in the radius of the bubble is `Delta r`. <br> `dW=dF Delta r=(P_(i)-P_(o))4pi r^(2) xx Delta r " " `...(ii) <br> Comparing equations (i) and (ii), we get <br> `(P_(i)-P_(o)) 4pi r^(2)xx Delta r =T xx 8 pi r Delta r` <br> `therefore (P_(i)-P_(o)) =2T//r` <br> This is called the Laplace's law of spherical membrane.