Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has radius equal to

Question

Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has radius equal to (A) (r1 + r2/2) (B) (r1 r2/r1 + r2) (C) √r1 r2 (D) √r12 + r22

Tardigrade Admin · Accepted Answer

Excess of pressure, inside the first bubble

p_{1} = \frac{4 T}{r _{1}}

S imi 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 = r_{1}^{2} + r_{2}^{2}

Q. Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has radius equal to

$\frac{r _{1} + r _{2}}{2}$

$\frac{r _{1} r _{2}}{r _{1} + r _{2}}$

$r_{1} r_{2}$

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

Q. Two spherical soap bubbles of radii r1​ and r2​ in vacuum combine under isothermal conditions. The resulting bubble has radius equal to

2r1​+r2​​

r1​+r2​r1​r2​​

r1​r2​​

r12​+r22​​

Q. Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has radius equal to

$\frac{r _{1} + r _{2}}{2}$

$\frac{r _{1} r _{2}}{r _{1} + r _{2}}$

$r_{1} r_{2}$

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