A glass vessel just holds `50gm` of a liquid at `0^(0)C`. If the coefficient of linear expansion is `8 xx 10^(-6) // ^(0)C` The mass of the liquid it holds at `80^(0)C` is [coefficient of absolute expansion of liquid `= 5 xx 10(-4) //^(0)C` (nearly)

Options

A. `46 g`
B. `48 g`
C. `51 g`
D. `42 g`

Explanation

To solve the problem, we need to determine the mass of the liquid held by the glass vessel at `80°C` after accounting for the expansion of both the glass vessel and the liquid. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the change in volume of the glass vessel
The formula for the change in volume due to linear expansion is given by:
\[ \Delta V = 3 \alpha \Delta T V_0 \]
Where:
- \(\alpha\) = coefficient of linear expansion of glass = \(8 \times 10^{-6} \, ^\circ C^{-1}\)
- \(\Delta T\) = change in temperature = \(80 - 0 = 80 \, ^\circ C\)
- \(V_0\) = initial volume of the glass vessel

First, we need to express the initial volume \(V_0\) in terms of the mass of the liquid it holds:
\[ V_0 = \frac{m}{\rho} \]
Assuming the density of the liquid is \(\rho\), we can denote the initial volume as \(V_0\).

Now, substituting the values into the change in volume formula:
\[ \Delta V_{glass} = 3 \times (8 \times 10^{-6}) \times 80 \times V_0 \]
Calculating this:
\[ \Delta V_{glass} = 1920 \times 10^{-6} V_0 = 1.92 \times 10^{-3} V_0 \]

### Step 2: Calculate the change in volume of the liquid
The formula for the change in volume due to absolute expansion is given by:
\[ \Delta V = \gamma \Delta T V_0 \]
Where:
- \(\gamma\) = coefficient of absolute expansion of liquid = \(5 \times 10^{-4} \, ^\circ C^{-1}\)

Now substituting the values:
\[ \Delta V_{liquid} = (5 \times 10^{-4}) \times 80 \times V_0 \]
Calculating this:
\[ \Delta V_{liquid} = 40 \times 10^{-4} V_0 = 4 \times 10^{-3} V_0 \]

### Step 3: Calculate the net change in volume
The net change in volume is given by:
\[ \Delta V_{net} = \Delta V_{liquid} - \Delta V_{glass} \]
Substituting the values we calculated:
\[ \Delta V_{net} = (4 \times 10^{-3} V_0) - (1.92 \times 10^{-3} V_0) \]
Calculating this:
\[ \Delta V_{net} = (4 - 1.92) \times 10^{-3} V_0 = 2.08 \times 10^{-3} V_0 \]

### Step 4: Calculate the new volume of the liquid
The new volume of the liquid at `80°C` is:
\[ V_{new} = V_0 + \Delta V_{net} = V_0 + 2.08 \times 10^{-3} V_0 = (1 + 2.08 \times 10^{-3}) V_0 \]
This means the new volume is:
\[ V_{new} = 1.00208 V_0 \]

### Step 5: Calculate the new mass of the liquid
Since the volume of the liquid has increased, but the vessel can only hold a certain mass, we need to find the mass that corresponds to the new volume:
\[ m_{new} = \text{Density} \times V_{new} \]
Assuming the density remains constant, we can express the new mass as:
\[ m_{new} = m_{initial} - \text{mass lost} \]
Where the mass lost can be calculated as:
\[ \text{mass lost} = \Delta V_{glass} \times \text{Density} \]
Substituting the values:
\[ \text{mass lost} = 1.92 \times 10^{-3} \times \rho \]
Given that the initial mass \(m_{initial} = 50 \, gm\), we can find the new mass:
\[ m_{new} = 50 - \text{mass lost} \]

### Final Calculation
Substituting the values:
\[ \text{mass lost} = 1.92 \times 10^{-3} \times \rho \]
If we assume the density of the liquid is approximately \(1 \, gm/cm^3\) (for water), then:
\[ \text{mass lost} \approx 1.92 \, gm \]
Finally:
\[ m_{new} = 50 - 1.92 = 48.08 \approx 48 \, gm \]

### Conclusion
The mass of the liquid it holds at `80°C` is approximately **48 grams**.