A `U-`tube is made up of capillaries of bores `1 mm` and `2 mm` respectively. The tube is held vertically and partically filled with a liquid of surface tension `49 "dyne"//"cm"` and zero contact angle. Calculate the density of the liquid of the difference in the levels of the meniscus is `1.25 cm`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a U-tube with two capillaries of different diameters (1 mm and 2 mm). The surface tension of the liquid is given as 49 dyne/cm, and the contact angle is zero. The difference in the levels of the liquid meniscus is given as 1.25 cm. We need to calculate the density of the liquid.

### Step 2: Convert Units
Convert the diameters of the capillaries from mm to cm:
- Diameter of capillary 1 (d1) = 1 mm = 0.1 cm
- Diameter of capillary 2 (d2) = 2 mm = 0.2 cm

Now, calculate the radii:
- Radius of capillary 1 (r1) = d1/2 = 0.1 cm / 2 = 0.05 cm
- Radius of capillary 2 (r2) = d2/2 = 0.2 cm / 2 = 0.1 cm

### Step 3: Use the Formula for Height Difference
The height difference (h) in a U-tube due to surface tension can be calculated using the formula:
\[ h = \frac{2T}{\rho g} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]
Where:
- \( T \) = surface tension (49 dyne/cm)
- \( \rho \) = density of the liquid (unknown)
- \( g \) = acceleration due to gravity (approximately 980 cm/s²)
- \( r_1 \) and \( r_2 \) are the radii of the capillaries.

### Step 4: Rearranging the Formula
Rearranging the formula to find the density \( \rho \):
\[ \rho = \frac{2T}{g h} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)^{-1} \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:
- \( T = 49 \, \text{dyne/cm} = 49 \times 10^{-5} \, \text{N/m} \) (since 1 dyne = \( 10^{-5} \) N)
- \( g = 980 \, \text{cm/s}^2 \)
- \( h = 1.25 \, \text{cm} \)
- \( r_1 = 0.05 \, \text{cm} \)
- \( r_2 = 0.1 \, \text{cm} \)

Now, calculate:
\[ \rho = \frac{2 \times 49 \, \text{dyne/cm} \times 1 \, \text{cm}}{980 \, \text{cm/s}^2 \times 1.25 \, \text{cm}} \left( \frac{1}{0.05} - \frac{1}{0.1} \right)^{-1} \]

### Step 6: Calculate the Terms
Calculate \( \frac{1}{r_1} - \frac{1}{r_2} \):
\[ \frac{1}{0.05} - \frac{1}{0.1} = 20 - 10 = 10 \, \text{cm}^{-1} \]

Now substitute this back into the density equation:
\[ \rho = \frac{2 \times 49 \times 1}{980 \times 1.25} \times \frac{1}{10} \]

### Step 7: Final Calculation
Calculate the density:
\[ \rho = \frac{98}{1225} \approx 0.08 \, \text{g/cm}^3 \]

### Conclusion
Thus, the density of the liquid is approximately **0.08 g/cm³**.