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

Options

A. 0.02
B. 0.04
C. 0.06
D. 0.08

Answer

To solve the problem of how much the mass density of a rod changes when it is heated from \(0^\circ\) to \(10^\circ\) 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 (\(\Delta L\)) of a rod due to a change in temperature (\(\Delta T\)) can be expressed using the formula:
\[ \frac{\Delta L}{L} = \alpha \Delta T \]
where \(\alpha\) is the coefficient of linear expansion.

### Step 2: Substitute the given values
We know that:
- \(\frac{\Delta L}{L} = 0.02\% = \frac{0.02}{100} = 0.0002\)
- \(\Delta T = 10^\circ - 0^\circ = 10^\circ\)

Substituting these values into the formula gives:
\[ 0.0002 = \alpha \times 10 \]

### Step 3: Solve for \(\alpha\)
Rearranging the equation to solve for \(\alpha\):
\[ \alpha = \frac{0.0002}{10} = 0.00002 \, \text{per degree} \]
This can also be expressed as:
\[ \alpha = 2 \times 10^{-5} \, \text{per degree} \]

### Step 4: Relate linear expansion to volumetric expansion
The volumetric coefficient of expansion (\(\gamma\)) is related to the linear coefficient of expansion (\(\alpha\)) by the formula:
\[ \gamma = 3\alpha \]
Substituting the value of \(\alpha\):
\[ \gamma = 3 \times (2 \times 10^{-5}) = 6 \times 10^{-5} \, \text{per degree} \]

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

### Step 6: Relate density change to volume change
The density (\(\rho\)) is defined as:
\[ \rho = \frac{m}{V} \]
The change in density can be expressed as:
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} - \frac{\Delta V}{V} \]
Since the mass (\(m\)) remains constant, \(\frac{\Delta m}{m} = 0\). Thus:
\[ \frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V} \]

### Step 7: Substitute the change in volume into the density change equation
Substituting the value of \(\frac{\Delta V}{V}\):
\[ \frac{\Delta \rho}{\rho} = -6 \times 10^{-4} \]

### Step 8: Convert to percentage change
To find the percentage change in density:
\[ \frac{\Delta \rho}{\rho} \times 100 = -6 \times 10^{-4} \times 100 = -0.06\% \]
The negative sign indicates a decrease in density.

### Final Answer
The mass density of the rod changes by \(0.06\%\).

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