Question

The excess pressure inside a spherical water drop A is four times that of another water drop B. Then, the ratio of the mass of water drop A to that of drop B is

• A. \(1:8\)
• B. \(1:16\)
• C. \(1:32\)
• D. \(1:64\)

Hint:

Excess pressure in a drop is \(\dfrac{2T}{r}\) and mass is proportional to \(r^3\).

Answer 1

The Correct Option is D

Step 1: Find the radius ratio
A has four times the pressure, so a radius four times smaller.

Step 2: Cube it
Volume ratio $=\left(\tfrac14\right)^3=\tfrac1{64}$, and same density gives the same mass ratio, so $1:64$, option (D).

Final Answer:
The ratio is $1:64$, option (D). \[ \boxed{1:64} \]