The radii of two mercury drops are R_1 and R_ | vedclass

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(D) Since the process occurs under isothermal conditions and the total mass of the mercury remains constant,the total volume of the two drops must be

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$R^3=R_1^3+R_2^3$

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(D) Since the process occurs under isothermal conditions and the total mass of the mercury remains constant,the total volume of the two drops must be equal to the volume of the single large drop formed.Let the volume of the first drop be $V_1 = \frac{4}{3} \pi R_1^3$ and the volume of the second drop be $V_2 = \frac{4}{3} \pi R_2^3$.The volume of the resulting large drop is $V = \frac{4}{3} \pi R^3$.By the principle of conservation of volume: $V = V_1 + V_2$.Substituting the expressions: $\frac{4}{3} \pi R^3 = \frac{4}{3} \pi R_1^3 + \frac{4}{3} \pi R_2^3$.Dividing both sides by $\frac{4}{3} \pi$,we get: $R^3 = R_1^3 + R_2^3$.

The radii of two mercury drops are $R_1$ and $R_2$. Under isothermal conditions,a single drop of radius $R$ is formed from them. The relation between $R, R_1$ and $R_2$ is

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