Two liquid drop have diameters of 1 cm and 1.5 cm. The ratio of excess pressures inside them is

Options

A. `1:1`
B. `5 : 3`
C. `2:3`
D. `3:2`

Answer

To solve the problem of finding the ratio of excess pressures inside two liquid drops with diameters of 1 cm and 1.5 cm, we can follow these steps:

### Step 1: Understand the formula for excess pressure
The excess pressure (\(\Delta P\)) inside a liquid drop is given by the formula:
\[ \Delta P = \frac{2S}{r} \]
where \(S\) is the surface tension of the liquid and \(r\) is the radius of the drop.

### Step 2: Calculate the radii of the drops
Given the diameters of the drops:
- For the first drop (diameter = 1 cm), the radius \(r_1\) is:
\[ r_1 = \frac{1 \text{ cm}}{2} = 0.5 \text{ cm} \]
- For the second drop (diameter = 1.5 cm), the radius \(r_2\) is:
\[ r_2 = \frac{1.5 \text{ cm}}{2} = 0.75 \text{ cm} \]

### Step 3: Write the expressions for excess pressures
Using the formula for excess pressure:
- For the first drop:
\[ \Delta P_1 = \frac{2S}{r_1} = \frac{2S}{0.5} = 4S \]
- For the second drop:
\[ \Delta P_2 = \frac{2S}{r_2} = \frac{2S}{0.75} = \frac{8S}{3} \]

### Step 4: Find the ratio of excess pressures
Now, we can find the ratio of the excess pressures:
\[ \frac{\Delta P_1}{\Delta P_2} = \frac{4S}{\frac{8S}{3}} = 4S \times \frac{3}{8S} = \frac{12}{8} = \frac{3}{2} \]

### Conclusion
The ratio of the excess pressures inside the two liquid drops is:
\[ \frac{3}{2} \]