In a U-tube diamter of two limbs are 0.5 cm and 1 cm respectively and tube has filled with water `(T=72"dyne"//cm`) then liquid level difference between two limbs will be

Options

A. 0.5 cm
B. 0.25 cm
C. 0.293 cm
D. none of these

Answer

To solve the problem of finding the liquid level difference between the two limbs of a U-tube, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Diameter of limb 1 (d1) = 0.5 cm
- Diameter of limb 2 (d2) = 1 cm
- Surface tension (T) = 72 dyne/cm

2. **Calculate the Cross-Sectional Areas:**
- The cross-sectional area (A) of a circular tube is given by the formula:
\[ A = \frac{\pi d^2}{4} \]
- For limb 1 (A1):
\[ A1 = \frac{\pi (0.5)^2}{4} = \frac{\pi \times 0.25}{4} = \frac{\pi}{16} \text{ cm}^2 \]
- For limb 2 (A2):
\[ A2 = \frac{\pi (1)^2}{4} = \frac{\pi}{4} \text{ cm}^2 \]

3. **Set Up the Equations for Each Limb:**
- The weight of the liquid column in limb 1 is balanced by the surface tension:
\[ T \cdot \pi d1 = A1 \cdot x \cdot \rho \cdot g \]
- The weight of the liquid column in limb 2 is also balanced by the surface tension:
\[ T \cdot \pi d2 = A2 \cdot y \cdot \rho \cdot g \]

4. **Substituting the Values into the Equations:**
- For limb 1:
\[ 72 \cdot \pi \cdot 0.5 = \frac{\pi}{16} \cdot x \cdot 1 \cdot 980 \]
Simplifying gives:
\[ 36\pi = \frac{\pi}{16} \cdot 980x \]
Canceling \(\pi\) and solving for \(x\):
\[ 36 = \frac{980x}{16} \implies x = \frac{36 \cdot 16}{980} \approx 0.5877 \text{ cm} \]

- For limb 2:
\[ 72 \cdot \pi \cdot 1 = \frac{\pi}{4} \cdot y \cdot 1 \cdot 980 \]
Simplifying gives:
\[ 72\pi = \frac{\pi}{4} \cdot 980y \]
Canceling \(\pi\) and solving for \(y\):
\[ 72 = \frac{980y}{4} \implies y = \frac{72 \cdot 4}{980} \approx 0.2938 \text{ cm} \]

5. **Calculate the Liquid Level Difference:**
- The difference in liquid levels (x - y) is:
\[ x - y = 0.5877 - 0.2938 \approx 0.2939 \text{ cm} \]

### Final Answer:
The liquid level difference between the two limbs is approximately **0.2939 cm**.

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