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`

Explanation

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-by-Step Solution:

1. **Identify the Given Data:**
- Diameter of the first drop (D1) = 1 cm
- Diameter of the second drop (D2) = 1.5 cm

2. **Calculate the Radii of the Drops:**
- The radius (R) is half of the diameter.
- For the first drop:
\[ R_1 = \frac{D_1}{2} = \frac{1 \text{ cm}}{2} = 0.5 \text{ cm} \]
- For the second drop:
\[ R_2 = \frac{D_2}{2} = \frac{1.5 \text{ cm}}{2} = 0.75 \text{ cm} \]

3. **Understand the Formula for Excess Pressure:**
- The excess pressure (ΔP) 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.

4. **Set Up the Ratio of Excess Pressures:**
- The ratio of excess pressures for the two drops can be expressed as:
\[ \frac{\Delta P_1}{\Delta P_2} = \frac{2S/R_1}{2S/R_2} = \frac{R_2}{R_1} \]
- The surface tension (S) cancels out since it is the same for both drops.

5. **Substitute the Values of Radii:**
- Now substituting the values of R1 and R2:
\[ \frac{\Delta P_1}{\Delta P_2} = \frac{R_2}{R_1} = \frac{0.75 \text{ cm}}{0.5 \text{ cm}} = \frac{3}{2} \]

6. **Conclusion:**
- Therefore, the ratio of excess pressures inside the two drops is:
\[ \frac{\Delta P_1}{\Delta P_2} = \frac{3}{2} \]

### Final Answer:
The ratio of excess pressures inside the two liquid drops is \( \frac{3}{2} \).

---