Question

Let \( R_1 \) and \( R_2 \) be 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. \( \left( R_1^3 + R_2^3 \right)^{\frac{1}{3}} \)
• C. \( \frac{R_1 + R_2}{2} \)
• D. \( \sqrt{R_1^2 + R_2^2} \)

Hint:

For isothermal processes involving spherical drops, volume is conserved and the radius of the resultant drop is the cube root of the sum of the cubes of individual radii.

Answer 1

The Correct Option is B

Step 1: Understanding the concept of volume conservation.
Since the process is isothermal, the volume of the two smaller drops is conserved when they combine into a larger drop. The volume of a spherical drop is given by \( V = \frac{4}{3} \pi R^3 \). Therefore, the total volume before and after the merging of the drops is: \[ \frac{4}{3} \pi R_1^3 + \frac{4}{3} \pi R_2^3 = \frac{4}{3} \pi R^3 \] Simplifying the equation, we get the relation for the radius \( R \) of the resultant drop: \[ R = \left( R_1^3 + R_2^3 \right)^{\frac{1}{3}} \] Step 2: Conclusion.
Thus, the radius of the resultant drop is \( \left( R_1^3 + R_2^3 \right)^{\frac{1}{3}} \), corresponding to option (B).