A steel tape measures that length of a copper rod as `90.0` cm when both are at `10^(@)C`, the calibration temperature, for the tape. What would the tape read for the length of the rod when both are at `30^(@)C`. Given `alpha_("steel")=1.2xx10^(-5)" per".^(@)Cand alpha_(Cu)=1.7xx10^(-5)per .^(@)C`

Options

A. `89.00cm`
B. `90.21cm`
C. `89.80cm`
D. `90.01cm`

Answer

To solve the problem of measuring the length of a copper rod with a steel tape at different temperatures, we will use the concept of linear thermal expansion. The formula for linear expansion is given by:

\[ \Delta L = L_0 \alpha \Delta T \]

Where:
- \(\Delta L\) is the change in length,
- \(L_0\) is the original length,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

### Step 1: Identify the given values
- Original length of the copper rod, \(L_0 = 90.0 \, \text{cm}\)
- Initial temperature, \(T_1 = 10^\circ C\)
- Final temperature, \(T_2 = 30^\circ C\)
- Coefficient of linear expansion for steel, \(\alpha_{\text{steel}} = 1.2 \times 10^{-5} \, \text{per} \, ^\circ C\)
- Coefficient of linear expansion for copper, \(\alpha_{\text{Cu}} = 1.7 \times 10^{-5} \, \text{per} \, ^\circ C\)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_2 - T_1 = 30^\circ C - 10^\circ C = 20^\circ C \]

### Step 3: Calculate the change in length for the copper rod
Using the formula for linear expansion:
\[ \Delta L_{\text{Cu}} = L_0 \alpha_{\text{Cu}} \Delta T \]
Substituting the known values:
\[ \Delta L_{\text{Cu}} = 90.0 \, \text{cm} \times (1.7 \times 10^{-5} \, \text{per} \, ^\circ C) \times (20 \, ^\circ C) \]
\[ \Delta L_{\text{Cu}} = 90.0 \times 1.7 \times 10^{-5} \times 20 \]
\[ \Delta L_{\text{Cu}} = 90.0 \times 0.000034 = 0.00306 \, \text{cm} \]

### Step 4: Calculate the change in length for the steel tape
Using the same formula:
\[ \Delta L_{\text{steel}} = L_0 \alpha_{\text{steel}} \Delta T \]
Substituting the known values:
\[ \Delta L_{\text{steel}} = 90.0 \, \text{cm} \times (1.2 \times 10^{-5} \, \text{per} \, ^\circ C) \times (20 \, ^\circ C) \]
\[ \Delta L_{\text{steel}} = 90.0 \times 1.2 \times 10^{-5} \times 20 \]
\[ \Delta L_{\text{steel}} = 90.0 \times 0.000024 = 0.00216 \, \text{cm} \]

### Step 5: Calculate the effective change in length
The effective change in length that the tape measures is the difference between the changes in length of the copper rod and the steel tape:
\[ \Delta L_{\text{effective}} = \Delta L_{\text{Cu}} - \Delta L_{\text{steel}} \]
\[ \Delta L_{\text{effective}} = 0.00306 \, \text{cm} - 0.00216 \, \text{cm} = 0.00090 \, \text{cm} \]

### Step 6: Calculate the final length measured by the tape
The final length measured by the tape at \(30^\circ C\) is:
\[ L_{\text{final}} = L_0 + \Delta L_{\text{effective}} \]
\[ L_{\text{final}} = 90.0 \, \text{cm} + 0.00090 \, \text{cm} = 90.00090 \, \text{cm} \]

### Step 7: Round the final answer
Rounding to two decimal places, the tape would read:
\[ L_{\text{final}} \approx 90.01 \, \text{cm} \]

### Final Answer
The tape would read approximately **90.01 cm** for the length of the copper rod when both are at \(30^\circ C\).

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