The coefficient of volumetric expansion of mercury is `18xx10^(-5)//^(@)C` . A thermometer bulb has a volume `10^(-6)m^(3)` and cross section of stem is `0.004 cm^(2)`. Assuming that bulb is filled with mercury at `0^(@)C` then the length of the mercury column at `100^(@)C` is

Options

A. `18.8 mm`
B. `9.2 mm`
C. `7.4 mm`
D. `4.5 cm`

Answer

To find the length of the mercury column at 100°C, we will follow these steps:

### Step 1: Understand the problem
We have a thermometer bulb filled with mercury at 0°C. The coefficient of volumetric expansion of mercury is given as \( \gamma = 18 \times 10^{-5} \, \text{°C}^{-1} \). The initial volume of the bulb is \( V_0 = 10^{-6} \, \text{m}^3 \) and the cross-sectional area of the stem is \( A = 0.004 \, \text{cm}^2 \).

### Step 2: Convert the cross-sectional area to square meters
To use the area in calculations, we need to convert it from cm² to m²:
\[ A = 0.004 \, \text{cm}^2 = 0.004 \times 10^{-4} \, \text{m}^2 = 4 \times 10^{-7} \, \text{m}^2 \]

### Step 3: Calculate the change in volume (\( \Delta V \))
The change in volume due to thermal expansion can be calculated using the formula:
\[ \Delta V = V_0 \cdot \gamma \cdot \Delta T \]
where \( \Delta T = 100 - 0 = 100 \, \text{°C} \). Substituting the values:
\[ \Delta V = 10^{-6} \, \text{m}^3 \cdot (18 \times 10^{-5} \, \text{°C}^{-1}) \cdot 100 \]
\[ = 10^{-6} \cdot 18 \cdot 10^{-5} \cdot 100 \]
\[ = 10^{-6} \cdot 18 \cdot 10^{-3} \]
\[ = 18 \times 10^{-9} \, \text{m}^3 \]

### Step 4: Relate the change in volume to the change in height
The change in volume can also be expressed in terms of the change in height (\( \Delta H \)) of the mercury column:
\[ \Delta V = A \cdot \Delta H \]
Thus, we can set the two expressions for \( \Delta V \) equal to each other:
\[ A \cdot \Delta H = 18 \times 10^{-9} \, \text{m}^3 \]
Substituting the area:
\[ (4 \times 10^{-7}) \cdot \Delta H = 18 \times 10^{-9} \]

### Step 5: Solve for \( \Delta H \)
Now, we can solve for \( \Delta H \):
\[ \Delta H = \frac{18 \times 10^{-9}}{4 \times 10^{-7}} \]
\[ = \frac{18}{4} \times 10^{-2} \]
\[ = 4.5 \times 10^{-2} \, \text{m} \]
Converting to centimeters:
\[ \Delta H = 4.5 \, \text{cm} \]

### Final Answer
The length of the mercury column at 100°C will be \( 4.5 \, \text{cm} \).

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