Question

Two soap bubbles of radii$R_1$ and $R_2$ kept in atmosphere are combined isothermally to form a big bubble of radius $R$. The expression for surface tension will be

• A. $\frac{P_0(R^3+R_1^3-R_2^3)}{4(R^2-R_1^2-R_2^2)}$
• B. $\frac{P_0(R_1^3+R_2^3+R^3)}{4(R^2-R_1^2-R_2^2)}$
• C. $P_0(R_1^3+R_2^3-R^3)$
• D. $4P_0(R_1^3+R_2^3-R^3)$

Answer 1

The correct option is A $\frac{P_0(R^3+R_1^3-R_2^3)}{4(R^2-R_1^2-R_2^2)}$

$$\begin{array}{c}Accordingtoquestion................\\ P.V=nRT\\ \Rightarrow P_1{V}_1+P_2{V}_2=PV\\ but,\\ \Rightarrow P_{in}-P_{ext}=\frac{4T}{r},\left(P_{ext}=P\right)\\ \Rightarrow \left(P+\frac{4T}{R_1}\right)\frac{4}{3}\pi R_1^3+\left(P+\frac{4T}{R_2}\right)\frac{4}{3}\pi R_2^3=\left(P+\frac{4T}{R}\right)\frac{4}{3}\pi R^3\\ \Rightarrow P\left(R_1^3+R_2^3-R^3\right)=4T\left(R^2-R_1^2-R_2^2\right)\\ \therefore T=\frac{P_0\left(R^3+{R_1}^3-{R_2}^3\right)}{4\left(R^2-{R_1}^2-{R_2}^2\right)}\end{array}$$so the correct option is A