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Step-by-Step Solution
1. Identify the Known Data
• Temperature increase, $ \Delta T = 10^\circ \text{C} $
• Fractional (percentage) increase in length, $ \frac{\Delta L}{L} = 0.02\% $
2. Relate Linear Expansion Coefficient to Length Change
By definition of linear expansion,
$ \frac{\Delta L}{L} = \alpha \, \Delta T $
Given that $ \frac{\Delta L}{L} = 0.02\% $, we have:
$ \alpha \, \Delta T = 0.02\% \quad \Longrightarrow \quad \alpha = \frac{0.02\%}{\Delta T} = \frac{0.02\%}{10} = 0.002\% \text{ per }^\circ\text{C}. $
3. Determine Area Expansion Coefficient
For a metal wire, its cross-sectional area $A$ expands approximately with the areal expansion coefficient $ \beta $, which is related to the linear expansion coefficient by:
$ \beta = 2\,\alpha. $
Hence:
$ \beta = 2 \times 0.002\% = 0.004\% \text{ per }^\circ\text{C}. $
Therefore, the total fractional change in area for a $10^\circ \text{C}$ rise is:
$ \beta \, \Delta T = 0.004\% \times 10 = 0.04\%. $
4. Express Density and Its Change
Density $ \rho $ of the wire can be expressed as:
$ \rho = \frac{M}{A \, L}, $
where $M$ is the mass (which stays constant), $A$ is the cross-sectional area, and $L$ is the length. The percentage change in density can be found using differentials (or fractional changes):
$ \frac{\Delta \rho}{\rho}
= \frac{\Delta M}{M} - \frac{\Delta A}{A} - \frac{\Delta L}{L}. $
Since the mass $M$ remains constant,
$ \frac{\Delta M}{M} = 0, $
so:
$ \frac{\Delta \rho}{\rho}
= -\left(\frac{\Delta A}{A} + \frac{\Delta L}{L}\right). $
5. Substituting the Changes
We already calculated:
• $ \frac{\Delta L}{L} = 0.02\% $
• $ \frac{\Delta A}{A} = 0.04\% $
Hence,
$ \frac{\Delta \rho}{\rho}
= -\bigl(0.04\% + 0.02\%\bigr)
= -0.06\%. $
The negative sign shows that density decreases as temperature increases. Often, problems ask for the magnitude of this percentage change, so we quote:
$ \left|\frac{\Delta \rho}{\rho}\right| = 0.06\%. $
6. Final Answer
The closest value to the decrease in mass density of the wire is
0.06%.
