Question

The excess pressure inside a spherical drop of water is four times that of another drop. Then, their respective mass ratio is

• A. \(1:16\)
• B. \(8:1\)
• C. \(1:4\)
• D. \(1:64\)

Hint:

Excess pressure varies inversely with radius, and mass varies as the cube of radius.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Excess pressure inside a drop is \(\Delta P=\dfrac{2T}{r}\), so \(\Delta P\propto\dfrac{1}{r}\).

Step 2: Radius Ratio
\(\dfrac{\Delta P_{1}}{\Delta P_{2}}=4\Rightarrow\dfrac{r_{2}}{r_{1}}=4\), so \(\dfrac{r_{1}}{r_{2}}=\dfrac{1}{4}\).

Step 3: Mass Ratio
Mass \(m=\dfrac{4}{3}\pi r^{3}\rho\propto r^{3}\), so \(\dfrac{m_{1}}{m_{2}}=\left(\dfrac{1}{4}\right)^{3}=\dfrac{1}{64}\).

Final Answer:
The mass ratio of the two drops is 1:64. \[ \boxed{\text{(D) }1:64} \]