An air bubble of radius `r` in water is at a depth `h` below the water surface at some instant. If `P` is atmospheric pressure, `d` and `T` are density and surface tension of water respectively . the pressure inside the bubble will be :

Options

A. `p + hdg - (4T)/(r)`
B. `p + hdg + (2T)/(r)`
C. `p + hdg - (2T)/(r)`
D. `p + hdg + (4T)/(r)`

Answer

To find the pressure inside an air bubble of radius `r` at a depth `h` in water, we can follow these steps:

### Step 1: Understand the pressures acting on the bubble
The pressure inside the bubble is influenced by two main factors:
1. The atmospheric pressure above the water surface.
2. The pressure exerted by the water column above the bubble.

### Step 2: Calculate the pressure due to the water column
The pressure exerted by the water column at depth `h` can be calculated using the formula:
\[ P_{\text{water}} = dgh \]
where:
- \( d \) is the density of water,
- \( g \) is the acceleration due to gravity,
- \( h \) is the depth of the bubble.

### Step 3: Consider the effect of surface tension
The surface tension \( T \) of the water creates an excess pressure inside the bubble. The excess pressure due to surface tension for a spherical bubble is given by:
\[ P_{\text{excess}} = \frac{2T}{r} \]
where:
- \( T \) is the surface tension,
- \( r \) is the radius of the bubble.

### Step 4: Combine the pressures
The total pressure inside the bubble \( P_{\text{inside}} \) can be expressed as the sum of the atmospheric pressure, the pressure due to the water column, and the excess pressure due to surface tension:
\[ P_{\text{inside}} = P + dgh + \frac{2T}{r} \]

### Final Answer
Thus, the pressure inside the bubble is:
\[ P_{\text{inside}} = P + dgh + \frac{2T}{r} \]