A bakelite beacker has volume capacity of `500 c c at 30^@C`. When it is partially filled with `V_m` volumne (at `30^@C)` of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If `gamma_("beaker") = 6 xx 10^(-6).^@C^(-1)` and `gamma_("mercury") = 1.5 xx 10^(-4).^@C^(-1)` , where `gamma` is the coefficient of volume expansion , then `V_(m)` (in cc) is close to __________.

Explanation

To solve the problem, we need to understand the relationship between the volume expansion of the beaker and the mercury as the temperature changes.

1. **Understanding Volume Expansion**:
The volume expansion of a substance can be expressed using the formula:
\[ \Delta V = V_0 \cdot \gamma \cdot \Delta T \]
where \( \Delta V \) is the change in volume, \( V_0 \) is the initial volume, \( \gamma \) is the coefficient of volume expansion, and \( \Delta T \) is the change in temperature.

2. **Initial Conditions**:
The beaker has a total volume of \( V_{beaker} = 500 \, \text{cc} \) at \( 30^\circ C \). Let the volume of mercury be \( V_m \) at \( 30^\circ C \).

3. **Volume Expansion of the Beaker**:
As the temperature changes, the volume of the beaker will expand. The change in volume of the beaker when the temperature changes by \( \Delta T \) is:
\[ \Delta V_{beaker} = 500 \cdot (6 \times 10^{-6}) \cdot \Delta T \]

4. **Volume Expansion of Mercury**:
The volume of mercury will also expand as the temperature changes. The change in volume of the mercury is:
\[ \Delta V_{mercury} = V_m \cdot (1.5 \times 10^{-4}) \cdot \Delta T \]

5. **Condition for Constant Unfilled Volume**:
The problem states that the unfilled volume of the beaker remains constant as temperature varies. This means that the expansion of the mercury must equal the expansion of the beaker:
\[ \Delta V_{mercury} = \Delta V_{beaker} \]
Substituting the expressions we derived:
\[ V_m \cdot (1.5 \times 10^{-4}) \cdot \Delta T = 500 \cdot (6 \times 10^{-6}) \cdot \Delta T \]

6. **Cancelling \( \Delta T \)**:
Since \( \Delta T \) is common on both sides and is not zero, we can cancel it out:
\[ V_m \cdot (1.5 \times 10^{-4}) = 500 \cdot (6 \times 10^{-6}) \]

7. **Solving for \( V_m \)**:
Now we can solve for \( V_m \):
\[ V_m = \frac{500 \cdot (6 \times 10^{-6})}{1.5 \times 10^{-4}} \]
\[ V_m = \frac{500 \cdot 6}{1.5} \cdot \frac{10^{-6}}{10^{-4}} = \frac{3000}{1.5} \cdot 10^{-2} \]
\[ V_m = 2000 \cdot 10^{-2} = 20 \, \text{cc} \]

Thus, the volume of mercury \( V_m \) is approximately **20 cc**.