The tube of a mercury barometer is 4.5 mm in diameter . What error does surface tension introduce in the reading? Given angle of contact of mercury and glass is `140^(@)`. Surface tension of mercury `54xx 10^(-2) Nm^(-1)` and density 13.6 g m `//` cc .

Explanation

To find the error introduced by surface tension in the reading of a mercury barometer, we can follow these steps:

### Step 1: Gather the given data
- Diameter of the tube, \( D = 4.5 \, \text{mm} = 4.5 \times 10^{-3} \, \text{m} \)
- Radius of the tube, \( r = \frac{D}{2} = \frac{4.5 \times 10^{-3}}{2} = 2.25 \times 10^{-3} \, \text{m} \)
- Angle of contact, \( \theta = 140^\circ \)
- Surface tension of mercury, \( T = 54 \times 10^{-2} \, \text{N/m} \)
- Density of mercury, \( \rho = 13.6 \, \text{g/cm}^3 = 13.6 \times 10^{3} \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Convert the angle to radians
Since the cosine function typically uses radians, we can convert the angle:
\[ \theta = 140^\circ = \frac{140 \times \pi}{180} \, \text{radians} \]

### Step 3: Calculate the height \( h \) using the formula
The formula for the height \( h \) due to surface tension is given by:
\[ h = \frac{2T \cos \theta}{r \rho g} \]

### Step 4: Calculate \( \cos \theta \)
Calculate \( \cos(140^\circ) \):
\[ \cos(140^\circ) = \cos\left(\frac{140 \times \pi}{180}\right) = -\cos(40^\circ) \approx -0.766 \]

### Step 5: Substitute the values into the formula
Now substitute the values into the height equation:
\[ h = \frac{2 \times (54 \times 10^{-2}) \times (-0.766)}{(2.25 \times 10^{-3}) \times (13.6 \times 10^{3}) \times (9.8)} \]

### Step 6: Calculate the denominator
Calculate the denominator:
\[ (2.25 \times 10^{-3}) \times (13.6 \times 10^{3}) \times (9.8) \approx 2.25 \times 13.6 \times 9.8 \approx 303.6 \]

### Step 7: Calculate the height \( h \)
Now calculate \( h \):
\[ h = \frac{2 \times (54 \times 10^{-2}) \times (-0.766)}{303.6} \]
\[ h \approx \frac{-82.752 \times 10^{-2}}{303.6} \approx -0.272 \times 10^{-2} \, \text{m} = -0.275 \, \text{cm} \]

### Final Result
The error introduced by surface tension in the reading is approximately:
\[ h \approx -0.275 \, \text{cm} \]