Let R_{1} and R_{2} be the radii of two mercu | vedclass

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(D) When two mercury drops merge to form a single larger drop,the total volume of the mercury remains conserved.Let $R$ be the radius of the resultant

Vedclass · Accepted Answer

$R=\left(R_{1}^{3}+R_{2}^{3}\right)^{\frac{1}{3}}$

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(D) When two mercury drops merge to form a single larger drop,the total volume of the mercury remains conserved.Let $R$ be the radius of the resultant big drop.The volume of the first drop is $V_{1} = \frac{4}{3} \pi R_{1}^{3}$.The volume of the second drop is $V_{2} = \frac{4}{3} \pi R_{2}^{3}$.The volume of the resultant drop is $V = \frac{4}{3} \pi R^{3}$.Since the total volume is conserved,$V = V_{1} + V_{2}$.$\frac{4}{3} \pi R^{3} = \frac{4}{3} \pi R_{1}^{3} + \frac{4}{3} \pi R_{2}^{3}$.Canceling $\frac{4}{3} \pi$ from both sides,we get $R^{3} = R_{1}^{3} + R_{2}^{3}$.Therefore,$R = (R_{1}^{3} + R_{2}^{3})^{\frac{1}{3}}$.

Let $R_{1}$ and $R_{2}$ be the 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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