What is the pressure inside a drop of mercury of radius 3.0mm at room temperature? Surface tension of mercury at that temperature `(20^(@)C)` is `4.65xx10^(-1)Nm^(-1)`. The atmospheric pressure is `1.01xx10^(5)Pa`. Also give the excess pressure inside the drop.

Explanation

To find the pressure inside a drop of mercury, we will follow these steps:

### Step 1: Identify the given values
- Radius of the mercury drop, \( r = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m} \)
- Surface tension of mercury, \( \sigma = 4.65 \times 10^{-1} \, \text{N/m} \)
- Atmospheric pressure, \( P_0 = 1.01 \times 10^5 \, \text{Pa} \)

### Step 2: Calculate the excess pressure
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