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 analyze the volume expansion of both the beaker and the mercury as the temperature changes. The key point given in the problem is that the unfilled volume of the beaker remains constant when the temperature is varied.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Volume capacity of the beaker, \( V_b = 500 \, \text{cc} \)
- Coefficient of volume expansion of the beaker, \( \gamma_b = 6 \times 10^{-6} \, \text{°C}^{-1} \)
- Coefficient of volume expansion of mercury, \( \gamma_m = 1.5 \times 10^{-4} \, \text{°C}^{-1} \)
- Volume of mercury at \( 30 \, \text{°C} \), \( V_m \)

2. **Understanding Volume Changes:**
- When the temperature changes, the volume of the beaker and the volume of mercury will expand.
- The change in volume of the beaker, \( \Delta V_b \), can be expressed as:
\[ \Delta V_b = V_b \cdot \gamma_b \cdot \Delta T \]
- The change in volume of mercury, \( \Delta V_m \), can be expressed as:
\[ \Delta V_m = V_m \cdot \gamma_m \cdot \Delta T \]

3. **Setting Up the Equation:**
- Since the unfilled volume of the beaker remains constant, the change in volume of the mercury must equal the change in volume of the beaker:
\[ \Delta V_m = \Delta V_b \]
- Substituting the expressions for \( \Delta V_m \) and \( \Delta V_b \):
\[ V_m \cdot \gamma_m \cdot \Delta T = V_b \cdot \gamma_b \cdot \Delta T \]
- Since \( \Delta T \) is the same for both, we can cancel it out (assuming \( \Delta T \neq 0 \)):
\[ V_m \cdot \gamma_m = V_b \cdot \gamma_b \]

4. **Solving for \( V_m \):**
- Rearranging the equation to find \( V_m \):
\[ V_m = \frac{V_b \cdot \gamma_b}{\gamma_m} \]
- Substituting the known values:
\[ V_m = \frac{500 \, \text{cc} \cdot 6 \times 10^{-6} \, \text{°C}^{-1}}{1.5 \times 10^{-4} \, \text{°C}^{-1}} \]

5. **Calculating \( V_m \):**
- Simplifying the right-hand side:
\[ V_m = \frac{500 \cdot 6}{1.5} \times 10^{-6 + 4} \]
\[ V_m = \frac{3000}{1.5} \times 10^{-2} \]
\[ V_m = 2000 \times 10^{-2} = 20 \, \text{cc} \]

### Final Answer:
The volume of mercury \( V_m \) is approximately \( 20 \, \text{cc} \).

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