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Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is
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The correct answer is:
$\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$
Under isothermal conditions the surface tension remains constant. If $r$ is the radius of the digger drop, then Final surface energy $=$ Initial surface energy $\therefore 8 \pi r^{2} T=8 \pi r_{1}^{2} T+8 \pi r_{2}^{2} T$
$\therefore r^{2}=r_{1}^{2}+r_{2}^{2}$
$\therefore r=\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$
$\therefore r^{2}=r_{1}^{2}+r_{2}^{2}$
$\therefore r=\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$
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