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Let ${ }^{\prime} \mathrm{R}_{1}$ ' and ${ }^{\prime} \mathrm{R}_{2}$ ' are radii of two mercury drops. A big mercury drop is formed from
them under isothermal conditions. The radius of the resultant drop is
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them under isothermal conditions. The radius of the resultant drop is
Solution:
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Verified Answer
The correct answer is:
$\mathrm{R}=\left(\mathrm{R}_{1}^{3}+\mathrm{R}_{2}^{3}\right)^{\frac{1}{3}}$
The Volume of the bigger drop is equal to the sum of the volumes of the smaller drops.
$\begin{aligned}
& \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R_{1}^{3}+\frac{4}{3} \pi R_{2}^{3} \\
\therefore \quad & R=\sqrt[3]{R_{1}^{3}+R_{2}^{3}}
\end{aligned}$
$\begin{aligned}
& \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R_{1}^{3}+\frac{4}{3} \pi R_{2}^{3} \\
\therefore \quad & R=\sqrt[3]{R_{1}^{3}+R_{2}^{3}}
\end{aligned}$
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