Which of the following is correct ?

Options

A. The excess pressure inside a spherical drop of liquid is 2T/R
B. The excess pressure inside a spherical soap bubble is 4T/R
C. The excess pressure inside a cylindrical drop of liquid is T/R
D. Only (A) and (B)

Answer

To solve the question regarding the correct statements about excess pressure in liquid drops and bubbles, we will analyze each option provided in the question step by step.

### Step 1: Excess Pressure in a Spherical Drop of Liquid
1. **Understanding the Concept**: The excess pressure inside a spherical drop of liquid is due to surface tension acting on the surface of the drop.
2. **Formula Derivation**:
- For a spherical drop, the excess pressure \( \Delta P \) is given by the formula:
\[ \Delta P = \frac{2T}{R} \]
- Here, \( T \) is the surface tension, and \( R \) is the radius of the drop.
3. **Conclusion**: The statement that "excess pressure inside a spherical drop of liquid is \( \frac{2T}{R} \)" is **correct**.

### Step 2: Excess Pressure in a Spherical Soap Bubble
1. **Understanding the Concept**: A soap bubble has two surfaces (inner and outer), which contributes to the excess pressure.
2. **Formula Derivation**:
- For a soap bubble, the excess pressure \( \Delta P \) is given by:
\[ \Delta P = \frac{4T}{R} \]
- This is because the surface tension acts on both surfaces of the bubble.
3. **Conclusion**: The statement that "excess pressure inside a spherical soap bubble is \( \frac{4T}{R} \)" is **correct**.

### Step 3: Excess Pressure in a Cylindrical Drop of Liquid
1. **Understanding the Concept**: A cylindrical drop has one curved surface and one flat surface.
2. **Formula Derivation**:
- For a cylindrical drop, the excess pressure \( \Delta P \) can be expressed as:
\[ \Delta P = T \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]
- For a cylinder, one radius of curvature \( R_1 \) is equal to \( R \) (the radius of the curved surface), and \( R_2 \) is infinite (the flat surface).
- Thus, the formula simplifies to:
\[ \Delta P = T \left( \frac{1}{R} + 0 \right) = \frac{T}{R} \]
3. **Conclusion**: The statement that "excess pressure inside a cylindrical drop of liquid is \( \frac{T}{R} \)" is **correct**.

### Final Conclusion
All three statements regarding excess pressure in a spherical drop, a soap bubble, and a cylindrical drop of liquid are correct.