Question

The excess pressure inside a spherical drop of water is four times the other drop of water. Their respective mass ratio is

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

Hint:

Excess pressure varies inversely with radius, and mass varies as radius cubed.

Answer 1

The Correct Option is C

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

Step 2: Radius ratio
\(P_1=4P_2\) gives \(\dfrac{r_1}{r_2}=\dfrac{P_2}{P_1}=\dfrac{1}{4}\).

Step 3: Mass ratio
Mass \(m\propto r^{3}\), so \(\dfrac{m_1}{m_2}=\left(\dfrac{1}{4}\right)^{3}=\dfrac{1}{64}\).

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