A liquid of specific gravity `1.5` is observed to rise `3.0 cm` in a capillary tube of diameter `0.50 mm` and the liquid wets the surface of the tube. Calculate the excess pressure inside a spherical bubble of `1.0 cm` diameter blown from the same liquid.

Explanation

To solve the problem of calculating the excess pressure inside a spherical bubble of a liquid with a specific gravity of 1.5, we will follow these steps:

### Step 1: Understand the relationship between capillary rise and surface tension
The height of the liquid column (h) in a capillary tube is related to the surface tension (T) of the liquid, the radius of the tube (r), and the density (ρ) of the liquid using the formula:
\[ h = \frac{2T \cos \theta}{r \rho g} \]
Since the liquid wets the surface of the tube, the contact angle (θ) is 0 degrees, and thus \( \cos(0) = 1 \).

### Step 2: Calculate the radius of the capillary tube
The diameter of the capillary tube is given as 0.50 mm. Therefore, the radius (r) is:
\[ r = \frac{0.50 \, \text{mm}}{2} = 0.25 \, \text{mm} = 0.25 \times 10^{-3} \, \text{m} = 0.00025 \, \text{m} \]

### Step 3: Convert specific gravity to density
The specific gravity (SG) of the liquid is given as 1.5. The density of water (ρ_water) is approximately \( 1000 \, \text{kg/m}^3 \). Therefore, the density (ρ) of the liquid is:
\[ \rho = SG \times \rho_{\text{water}} = 1.5 \times 1000 \, \text{kg/m}^3 = 1500 \, \text{kg/m}^3 \]

### Step 4: Calculate the surface tension (T)
Rearranging the capillary rise equation to solve for surface tension (T):
\[ T = \frac{h \cdot r \cdot \rho \cdot g}{2} \]
Where:
- \( h = 3.0 \, \text{cm} = 0.03 \, \text{m} \)
- \( g = 9.81 \, \text{m/s}^2 \)

Substituting the values:
\[ T = \frac{0.03 \, \text{m} \cdot 0.00025 \, \text{m} \cdot 1500 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2}{2} \]
\[ T = \frac{0.03 \cdot 0.00025 \cdot 1500 \cdot 9.81}{2} \]
\[ T = \frac{0.0001171875}{2} \]
\[ T = 0.00005859375 \, \text{N/m} \]

### Step 5: Calculate the excess pressure inside the bubble
The excess pressure (P) inside a spherical bubble is given by the formula:
\[ P = \frac{4T}{R} \]
Where R is the radius of the bubble. The diameter of the bubble is given as 1.0 cm, so:
\[ R = \frac{1.0 \, \text{cm}}{2} = 0.5 \, \text{cm} = 0.005 \, \text{m} \]

Substituting the values into the pressure formula:
\[ P = \frac{4 \times 0.00005859375 \, \text{N/m}}{0.005 \, \text{m}} \]
\[ P = \frac{0.000234375}{0.005} \]
\[ P = 0.046875 \, \text{N/m}^2 \]

### Step 6: Convert to appropriate units
To express the pressure in Pascals (Pa), we note that \( 1 \, \text{N/m}^2 = 1 \, \text{Pa} \):
\[ P = 0.046875 \, \text{Pa} \]

### Final Answer
The excess pressure inside the spherical bubble is approximately:
\[ P \approx 0.0469 \, \text{Pa} \]