Let R_1, R_2 and R_3 be the radii of three me | vedclass

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(A) When three mercury drops of radii $R_1, R_2$ and $R_3$ combine to form a single large drop of radius $R$,the total volume remains conserved becaus

Vedclass · Accepted Answer

$(R_1^3+R_2^3+R_3^3)^{\frac{1}{3}}$

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(A) When three mercury drops of radii $R_1, R_2$ and $R_3$ combine to form a single large drop of radius $R$,the total volume remains conserved because the density of mercury is constant.The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.Equating the sum of the volumes of the three small drops to the volume of the large drop:$\frac{4}{3}\pi R_1^3 + \frac{4}{3}\pi R_2^3 + \frac{4}{3}\pi R_3^3 = \frac{4}{3}\pi R^3$Dividing both sides by $\frac{4}{3}\pi$,we get:$R_1^3 + R_2^3 + R_3^3 = R^3$Taking the cube root on both sides:$R = (R_1^3 + R_2^3 + R_3^3)^{\frac{1}{3}}$Thus,the correct option is $A$.

Let $R_1, R_2$ and $R_3$ be the radii of three mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

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