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Two liquid drops of radii $R_{1} a n d R_{2}$ are combined to form a liquid drop of radius R. The value of R in terms of $R_{1} a n d R_{2}$ is

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$\sqrt{R_{1}^{2} + R_{2}^{2}}$

$\frac{R_{1} + R_{2}}{2}$

$\sqrt{R_{1} R_{2}}$

$\sqrt[3]{R_{1}^{3} + R_{2}^{3}}$

answer is D.

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Detailed Solution

$V = V_{1} + V_{2}$

$\frac{4}{3} π R^{3} = \frac{4}{2} π R_{1}^{3} + \frac{4}{3} π R_{2}^{3}$

$\sqrt[3]{R_{1}^{3} + R_{2}^{3}}$

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