First slide

Surface tension and surface energy

Question

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

Moderate

A

$R = (r_{1} + r_{2}) / 2$
B

$R = r_{1} (r_{1} r_{2} + r_{2})$
C

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

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

Solution

The radii of two bubbles are $r_{1}$ and $r_{2}$ . In isothermal condition, two meet together in a vacuum. then

Under isothermal condition.
$P_{1} V_{1} + P_{2} V_{2} = P V$
$P_{1} = \frac{4 T}{r_{1}}, P_{2} = \frac{4 T}{r_{2}}, P = \frac{4 T}{R}$
& volume of bubble = $\frac{4}{3} {πR}^{3}$

Hence $\frac{4 T}{r_{1}} \times (\frac{4}{3} {πr}_{1}^{3}) + \frac{4 T}{r_{2}} \times \frac{4}{3} {πr}_{2}^{3} = \frac{4 T}{R} \times \frac{4}{3} {πR}^{3}$
$R^{2} = r_{1}^{2} + r_{2}^{2}$

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