Question

A water drop is divided into 8 equal droplets. The pressure difference between the inner and outer side of the big drop will be

• A. same as for smaller droplet
• B. half of that for smaller droplet
• C. \(\left(\frac{1}{4}\right)^{th}\) of that for smaller droplet
• D. twice that for smaller droplet

Hint:

Use volume conservation to relate the radii of the big drop and the small droplets, then compare excess pressures given by 2T over R.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The excess pressure inside a drop of radius \(R\) is \(P=\frac{2T}{R}\), so \(P\propto\frac{1}{R}\).

Step 2: Volume Conservation
\(\frac{4}{3}\pi R^{3}=8\times\frac{4}{3}\pi r^{3}\), so \(R^{3}=8r^{3}\) and \(R=2r\).

Step 3: Compare Pressures
\(\dfrac{P_{big}}{P_{small}}=\dfrac{r}{R}=\dfrac{1}{2}\).

Step 4: Compare the options
The big drop has half the excess pressure of a smaller droplet.

Final Answer:
The pressure difference for the big drop is half of that for a smaller droplet. \[ \boxed{\text{(B) }\text{half of that for smaller droplet}} \]