The radii of two soap bubbles are r₁ and r₂. In isothermal conditions, two meet together in vacuum. Then the radius of the resultant bubble is given by

The correct option is R² = r₁² + r₂².

Step-by-Step Explanation:

When two soap bubbles of radii r₁ and r₂ coalesce in isothermal conditions to form a single bubble of radius R in a vacuum, we can apply the conservation of the number of air moles inside the bubbles (Boyle's Law).

1. Excess Pressure inside a Soap Bubble:
The excess pressure inside a soap bubble of radius r with surface tension T is given by:

$P=\frac{4T}{r}$
Since the process takes place in a vacuum, the external atmospheric pressure is zero. Therefore, the absolute pressure inside each bubble is equal to the excess pressure.

2. Volume of a Soap Bubble:
The volume V of a spherical soap bubble of radius r is:

$V=\frac{4}{3}\pi r^3$

3. Applying Boyle's Law under Isothermal Conditions:
Under isothermal conditions, the temperature remains constant. Since no gas escapes during the coalescence of the two bubbles, the total number of moles of gas is conserved, which gives:

$P_1{V}_1+P_2{V}_2=PV$
where P₁, V₁ and P₂, V₂ are the pressures and volumes of the two initial bubbles, and P, V are the pressure and volume of the resultant bubble.

4. Deriving the Relation:
Substitute the values of pressure and volume into the equation:

$\left(\frac{4T}{r_1}\right)\left(\frac{4}{3}\pi r_1^3\right)+\left(\frac{4T}{r_2}\right)\left(\frac{4}{3}\pi r_2^3\right)=\left(\frac{4T}{R}\right)\left(\frac{4}{3}\pi R^3\right)$
Simplifying each term:

$\frac{16}{3}\pi T{r}_1^2+\frac{16}{3}\pi T{r}_2^2=\frac{16}{3}\pi T{R}^2$
Dividing both sides by the common factor $\frac{16}{3}\pi T$ yields:

$r_1^2+r_2^2=R^2$
Therefore, the radius of the resultant bubble R is related by:

$R^2=r_1^2+r_2^2$