Question

Let '$R_1$' and '$R_2$' are radii of two mercury drops. A big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

• A. $\sqrt{R_1^2 + R_2^2}$
• B. $(R_1^3 + R_2^3)^{1/3}$
• C. $\sqrt{R_1^2 - R_2^2}$
• D. $\frac{R_1 + R_2}{2}$

Hint:

Whenever spherical droplets fuse together without material loss, remember that volume conservation rules the math. Since volume scales with the cube of the radius ($V \propto R^3$), the cubical dimensions simply add up directly: $R^3 = R_1^3 + R_2^3$!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Two smaller liquid mercury drops of radii $R_1$ and $R_2$ coalesce together to form a single large drop. We need to express the radius $R$ of this combined drop in terms of the initial radii.

Step 2: Detailed Explanation:
Since fluid matter is incompressible, when multiple smaller droplets coalesce under isothermal conditions, the total collective volume of the fluid is preserved perfectly: $$ \text{Volume of large drop} = \text{Volume of drop 1} + \text{Volume of drop 2} $$ Substituting the standard volume formula for a geometric sphere ($V = \frac{4}{3}\pi r^3$): $$ \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_1^3 + \frac{4}{3}\pi R_2^3 $$ Canceling out the common structural multiplier $\frac{4}{3}\pi$ from both sides of the equation: $$ R^3 = R_1^3 + R_2^3 $$ Taking the cube root on both sides isolates our target radius parameter: $$ R = (R_1^3 + R_2^3)^{1/3} $$

Step 3: Final Answer:
The radius of the resultant mercury drop is $(R_1^3 + R_2^3)^{1/3}$, which corresponds to option (B).