Two air bubbles of radii 0.002 m and 0.004 m of same liquid come together to form a single bubble under isothermal condition. Find the radius of the buble formed. Given surface tension of liquid is `0.072 Nm^(-1)`

Answer

`4.47 xx 10^(-3)m`

Explanation

Here, `r_1 = 0.02m , r_2 = 0.004m` Volume of the two bubbles `V_1 = (4)/(3) pi r_1^3 and V_2 = (4)/(3)pi r_2^3` If S is the surface tension of liquid and `P_1 and P_2` be the excess of pressure inside the two bubbles , then `P_1 = (4S)/(r_1) and P_2 =(4S)/(r_2)` If r is the radius of the new soap bubbles formed when two soap bubbles coalesce under isothermal conditions. Let V and P be the volume and excess of pressure indside this new bubbles, then `V = (4)/(3)pi r^3 and P = (4S)/(r )` Assuming the exterbnal pressure is negligible as compred to interanl pressure and new bubble is formed under isothermal condition so Boyle's Law is obeyed `:. P_1 V_1 + P_2V_2 = PV`
`or (4S)/(r_1)xx(4)/(3)pi r_1^3 + (4S)/(r_2)xx(4)/(3)pi r_2^3 = (4S)/(r )xx(4)/(3)pi r^3`
or `r_1^2 + r_2^2 = r^2 ` or `r = sqrt(r_1^2+ r_2^2)`
`:. r = sqrt((0.002)^2+ (0.004)^2) = 4.47 xx 10^(-3)m`