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: Understanding the Concept
A water drop has one free surface, so the excess pressure inside it is \(\Delta P=\dfrac{2T}{r}\).

Step 2: Key Formula or Approach
\(\Delta P\propto\dfrac1r\) and mass \(m\propto r^3\).

Step 3: Detailed Explanation
\(\Delta P_A=4\Delta P_B\) gives \(r_A=\dfrac{r_B}{4}\).
\[ \frac{m_A}{m_B}=\left(\frac{r_A}{r_B}\right)^3=\left(\frac14\right)^3=\frac{1}{64} \]

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