What is the excess pressure inside a drop of mercury of radius 3.0 mm? (Surface tension of mercury is `4.65 xx 10^(-1) N m^(-1)`)

Options

A. 310 Pa
B. 410 Pa
C. 210 Pa
D. 510 Pa

Explanation

To find the excess pressure inside a drop of mercury, we can use the formula for excess pressure in a liquid drop, which is given by:

\[ \Delta P = \frac{2T}{R} \]

where:
- \(\Delta P\) is the excess pressure,
- \(T\) is the surface tension of the liquid,
- \(R\) is the radius of the drop.

**Step 1: Convert the radius from millimeters to meters.**
The radius given is \(3.0 \, \text{mm}\). We need to convert this to meters:

\[ R = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m} \]

**Step 2: Identify the surface tension of mercury.**
The surface tension of mercury is given as:

\[ T = 4.65 \times 10^{-1} \, \text{N/m} = 0.465 \, \text{N/m} \]

**Step 3: Substitute the values into the excess pressure formula.**
Now we can substitute the values of \(T\) and \(R\) into the formula for excess pressure:

\[ \Delta P = \frac{2T}{R} = \frac{2 \times 0.465 \, \text{N/m}}{3.0 \times 10^{-3} \, \text{m}} \]

**Step 4: Calculate the excess pressure.**
Perform the calculation:

\[ \Delta P = \frac{0.93}{3.0 \times 10^{-3}} = 310 \, \text{Pa} \]

Thus, the excess pressure inside the drop of mercury is:

\[ \Delta P = 310 \, \text{Pa} \]

### Final Answer:
The excess pressure inside a drop of mercury of radius 3.0 mm is **310 Pa**.

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