Question

If the excess pressure inside a spherical mercury drop is \(1240\,Nm^{-2}\), then the radius of the mercury drop is:

• A. \(0.75\,mm\)
• B. \(1.5\,mm\)
• C. \(0.375\,mm\)
• D. \(2.25\,mm\)

Hint:

For a liquid drop: \[ \Delta P=\frac{2T}{R} \] whereas for a soap bubble: \[ \Delta P=\frac{4T}{R}. \]

Answer 1

The Correct Option is A

Concept: For a liquid drop, \[ \Delta P=\frac{2T}{R}. \] For mercury, \[ T=0.465\,N/m. \]

Step 1: Use excess pressure formula.
\[ 1240=\frac{2(0.465)}{R}. \] \[ R=\frac{0.93}{1240}. \] \[ R=7.5\times10^{-4}m. \]

Step 2: Convert into millimetres.
\[ R=0.75mm. \] Hence \[ \boxed{0.75\,mm}. \]