A rod is heated from `0^@` to `10^@`, its length is changed by `0.02%`. By what `%` mass density is changed?

Options

A. `0.06`
B. `0.008`
C. `2.3`
D. `0.8`

Explanation

To solve the problem of how much the mass density of a rod changes when it is heated from \(0^\circ\) to \(10^\circ\) Celsius and its length changes by \(0.02\%\), we can follow these steps:

### Step 1: Understand the relationship between length change and temperature change
The change in length of the rod due to temperature change is given by the formula for linear expansion:
\[ \alpha = \frac{\Delta L}{L \Delta T} \]
where:
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta L\) is the change in length,
- \(L\) is the original length,
- \(\Delta T\) is the change in temperature.

### Step 2: Calculate the coefficient of linear expansion
Given that the length change \(\Delta L\) is \(0.02\%\), we can express this as:
\[ \Delta L = 0.02 \times 10^{-2} L \]
The change in temperature \(\Delta T\) is \(10^\circ\) Celsius. Plugging these values into the formula gives:
\[ \alpha = \frac{0.02 \times 10^{-2} L}{L \times 10} = \frac{0.02 \times 10^{-2}}{10} = 2 \times 10^{-5} \, \text{per degree Celsius} \]

### Step 3: Relate linear expansion to volumetric expansion
The volumetric expansion coefficient \(\gamma\) is related to the linear expansion coefficient \(\alpha\) by:
\[ \gamma = 3\alpha \]
Thus, substituting the value of \(\alpha\):
\[ \gamma = 3 \times (2 \times 10^{-5}) = 6 \times 10^{-5} \, \text{per degree Celsius} \]

### Step 4: Calculate the relative change in volume
The relative change in volume \(\frac{\Delta V}{V}\) can be calculated using:
\[ \frac{\Delta V}{V} = \gamma \Delta T \]
Substituting the values:
\[ \frac{\Delta V}{V} = 6 \times 10^{-5} \times 10 = 6 \times 10^{-4} \]

### Step 5: Relate density change to volume change
The density \(\rho\) is defined as mass per unit volume:
\[ \rho = \frac{m}{V} \]
The relative change in density can be expressed as:
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} - \frac{\Delta V}{V} \]
Since the mass remains constant during expansion (\(\Delta m = 0\)), we have:
\[ \frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V} \]
Substituting the value of \(\frac{\Delta V}{V}\):
\[ \frac{\Delta \rho}{\rho} = -6 \times 10^{-4} \]

### Step 6: Convert to percentage
To express the change in density as a percentage:
\[ \frac{\Delta \rho}{\rho} = -6 \times 10^{-4} = -0.06\% \]

### Final Answer
The percentage change in mass density is \(-0.06\%\).