A glass flask is filled up to a mark with 50 cc of mercury at `18^@C`. If the flask and contents are heated to `38^@C`, how much mercury will be above the mark (`alpha` for glass is `9xx10^(-6)//^@C` and coeffiecient of real expansion of mercury is `180xx10^(-6)//^@C`)?

Options

A. 0.85 cc
B. 0.46 cc
C. 0.153 cc
D. 0.05 cc

Explanation

To solve the problem, we need to calculate the expansion of both the glass flask and the mercury when the temperature is increased from \(18^\circ C\) to \(38^\circ C\). We will then find out how much mercury will be above the mark after heating.

### Step 1: Calculate the change in volume of the glass flask

The formula for the change in volume due to thermal expansion is given by:

\[ \Delta V_f = \gamma_f V \Delta T \]

Where:
- \(\Delta V_f\) = change in volume of the flask
- \(\gamma_f\) = coefficient of volume expansion of glass = \(3 \alpha\)
- \(V\) = initial volume of the flask = \(50 \, \text{cc}\)
- \(\Delta T\) = change in temperature = \(38^\circ C - 18^\circ C = 20^\circ C\)

Given \(\alpha = 9 \times 10^{-6} \, ^\circ C^{-1}\), we can calculate \(\gamma_f\):

\[ \gamma_f = 3 \times 9 \times 10^{-6} = 27 \times 10^{-6} \, ^\circ C^{-1} \]

Now substituting the values into the volume expansion formula:

\[ \Delta V_f = (27 \times 10^{-6}) \times 50 \times 20 \]

Calculating this gives:

\[ \Delta V_f = 27 \times 10^{-6} \times 1000 = 0.027 \, \text{cc} \]

### Step 2: Calculate the change in volume of mercury

The formula for the change in volume of mercury is:

\[ \Delta V_m = \gamma_m V \Delta T \]

Where:
- \(\Delta V_m\) = change in volume of mercury
- \(\gamma_m\) = coefficient of real expansion of mercury = \(180 \times 10^{-6} \, ^\circ C^{-1}\)

Substituting the values:

\[ \Delta V_m = (180 \times 10^{-6}) \times 50 \times 20 \]

Calculating this gives:

\[ \Delta V_m = 180 \times 10^{-6} \times 1000 = 0.18 \, \text{cc} \]

### Step 3: Determine the amount of mercury above the mark

To find out how much mercury will be above the mark, we subtract the change in volume of the flask from the change in volume of mercury:

\[ \text{Mercury above the mark} = \Delta V_m - \Delta V_f \]

Substituting the values we calculated:

\[ \text{Mercury above the mark} = 0.18 - 0.027 = 0.153 \, \text{cc} \]

### Final Answer

Thus, the amount of mercury that will be above the mark is:

\[ \boxed{0.153 \, \text{cc}} \]