The coefficient of volume expansion of glycerine is `49 xx 10^(-5)//^(@)C`. What is the fractional change in its density (approx.) for `30^(@)C` rise in temperature?

Options

A. `1.5 xx 10^(-2)`
B. `2.5 xx 10^(-2)`
C. `2.0 xx 10^(-2)`
D. `2.8 xx 10^(-2)`

Explanation

To solve the problem of finding the fractional change in the density of glycerine due to a temperature rise of \(30^{\circ}C\), we can follow these steps:

### Step 1: Understand the relationship between density and volume
The density (\(\rho\)) of a substance is defined as its mass (\(m\)) divided by its volume (\(V\)):
\[ \rho = \frac{m}{V} \]
When the volume changes due to temperature, the density will also change.

### Step 2: Use the concept of volume expansion
The fractional change in volume (\(\frac{\Delta V}{V}\)) due to a change in temperature can be expressed using the coefficient of volume expansion (\(\gamma\)):
\[ \frac{\Delta V}{V} = \gamma \Delta T \]
where:
- \(\Delta T\) is the change in temperature,
- \(\gamma\) is the coefficient of volume expansion.

### Step 3: Substitute the known values
Given:
- Coefficient of volume expansion of glycerine, \(\gamma = 49 \times 10^{-5} \, \text{°C}^{-1}\)
- Change in temperature, \(\Delta T = 30 \, \text{°C}\)

Substituting these values into the equation:
\[ \frac{\Delta V}{V} = (49 \times 10^{-5}) \times 30 \]

### Step 4: Calculate the fractional change in volume
Now, calculate the value:
\[ \frac{\Delta V}{V} = 49 \times 10^{-5} \times 30 = 1.47 \times 10^{-3} \]

### Step 5: Relate the change in volume to the change in density
From the relationship between density and volume, we know:
\[ \frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V} \]
This indicates that an increase in volume leads to a decrease in density.

### Step 6: Calculate the fractional change in density
Thus, the fractional change in density is:
\[ \frac{\Delta \rho}{\rho} = -1.47 \times 10^{-3} \]

### Step 7: Approximate the result
Since we are interested in the magnitude of the change, we can express this as:
\[ \frac{\Delta \rho}{\rho} \approx 0.00147 \]
or approximately \(1.5 \times 10^{-2}\) when expressed in a more simplified form.

### Final Answer
The fractional change in the density of glycerine for a \(30^{\circ}C\) rise in temperature is approximately:
\[ 1.5 \times 10^{-2} \]

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