Surface tension of mercury is `0.465 N m^(-1)`. The excess pressure inside a mercury drop of diameter 6mm is

Options

A. 310 pa
B. 410 pa
C. 510 pa
D. 610 pa

Explanation

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

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

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

### Step-by-step Solution:

1. **Identify the Given Data:**
- Surface tension of mercury, \(S = 0.465 \, \text{N/m}\)
- Diameter of the mercury drop, \(d = 6 \, \text{mm}\)

2. **Convert Diameter to Radius:**
- The radius \(R\) is half of the diameter.
\[ R = \frac{d}{2} = \frac{6 \, \text{mm}}{2} = 3 \, \text{mm} \]
- Convert the radius from millimeters to meters:
\[ R = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \]

3. **Substitute Values into the Excess Pressure Formula:**
- Now, substitute the values of \(S\) and \(R\) into the formula:
\[ \Delta P = \frac{2S}{R} = \frac{2 \times 0.465 \, \text{N/m}}{3 \times 10^{-3} \, \text{m}} \]

4. **Calculate the Excess Pressure:**
- Perform the calculation:
\[ \Delta P = \frac{0.93 \, \text{N/m}}{3 \times 10^{-3} \, \text{m}} = 310 \, \text{Pa} \]

5. **Final Result:**
- The excess pressure inside the mercury drop is:
\[ \Delta P = 310 \, \text{Pa} \]

### Summary:
The excess pressure inside a mercury drop of diameter 6 mm is \(310 \, \text{Pa}\).

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