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. \( R = \sqrt{R_1^2 - R_2^2} \)
• B. \( R = \frac{R_1 + R_2}{2} \)
• C. \( R = \sqrt{R_1^2 + R_2^2} \)
• D. \( R = \left( R_1^3 + R_2^3 \right)^{1/3} \)

Hint:

When combining spherical drops of the same material under isothermal conditions, the total volume is conserved. Use the sum of the volumes to calculate the radius of the resulting drop.

Answer 1

The Correct Option is D

Step 1: Volume of mercury drops.
The volume of a sphere is given by \( V = \frac{4}{3} \pi R^3 \). Since the process is isothermal, the total volume is conserved when the drops combine. Therefore, the volume of the resulting large drop is the sum of the volumes of the individual drops: \[ \frac{4}{3} \pi R^3 = \frac{4}{3} \pi R_1^3 + \frac{4}{3} \pi R_2^3 \] Simplifying: \[ R^3 = R_1^3 + R_2^3 \] Step 2: Conclusion.
Thus, the radius of the resultant drop is \( R = \left( R_1^3 + R_2^3 \right)^{1/3} \), which is option (D).