Two spherical soap bubbles of radii r 1 and r 2 in vacuum combine under isothermal conditions. The resulting bubble has radius equal to

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TestHub · Accepted Answer

Excess of pressure, inside the first bubble $p_{1} = \frac{4 T}{r_{1}}$

$S i m i l a r l y, p_{2} = \frac{4 T}{r_{2}}$

Let the radius of the large bubble be R. then, excess of pressure inside the large bubble $p = \frac{4 T}{R}$

Under isothermal condition, temperature remains constant.

$S o, p V = p_{1} V_{1} + p_{2} V_{2}$

$\frac{4 T}{R} (\frac{4}{3} π r^{3}) = \frac{4 T}{r_{1}} (\frac{4}{3} π r_{1}^{3}) + \frac{4 T}{r_{2}} (\frac{4}{3} π r_{2}^{3})$

$R^{2} = r_{1}^{2} + r_{2}^{2}$

$⟹ R = \sqrt{r_{1}^{2} + r_{2}^{2}}$

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