The density of a liquid at `100^(@)C` is `8.0g//cm^(3)` and at `0^(@)C` is `8.4g//cm^(3)`, the coefficient of cubical expansion of the liquid is

Options

A. `10^(-4)//^(@)C`
B. `5 xx 10^(-4)//^(@)C`
C. `8 xx 10^(-4)//^(@)C`
D. `4 xx 10^(-4)//^(@)C`

Explanation

To find the coefficient of cubical expansion of the liquid, we can use the relationship between density and temperature change. Here's a step-by-step solution:

### Step 1: Understand the given information
We have the densities of the liquid at two different temperatures:
- Density at \(0^\circ C\) (\(d_0\)) = \(8.4 \, g/cm^3\)
- Density at \(100^\circ C\) (\(d_{100}\)) = \(8.0 \, g/cm^3\)

### Step 2: Write the formula for the relationship between density and cubical expansion
The relationship between the densities at two temperatures and the coefficient of cubical expansion (\(\gamma\)) is given by the formula:
\[ d_{100} = \frac{d_0}{1 + \gamma \Delta T} \]
where \(\Delta T\) is the change in temperature.

### Step 3: Calculate the change in temperature
The change in temperature (\(\Delta T\)) from \(0^\circ C\) to \(100^\circ C\) is:
\[ \Delta T = 100 - 0 = 100^\circ C \]

### Step 4: Substitute the known values into the formula
Now, substituting the known values into the formula:
\[ 8.0 = \frac{8.4}{1 + \gamma \cdot 100} \]

### Step 5: Rearrange the equation to solve for \(\gamma\)
Rearranging the equation gives:
\[ 1 + \gamma \cdot 100 = \frac{8.4}{8.0} \]
Calculating the right side:
\[ \frac{8.4}{8.0} = 1.05 \]
So, we have:
\[ 1 + \gamma \cdot 100 = 1.05 \]

### Step 6: Isolate \(\gamma\)
Subtracting 1 from both sides:
\[ \gamma \cdot 100 = 1.05 - 1 = 0.05 \]
Now, divide both sides by 100:
\[ \gamma = \frac{0.05}{100} = 0.0005 \, \text{or} \, 5 \times 10^{-4} \, \text{per} \, ^\circ C \]

### Final Answer
The coefficient of cubical expansion of the liquid is:
\[ \gamma = 5 \times 10^{-4} \, \text{per} \, ^\circ C \]

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