An alumium rod when measured with a steel scale both being at 25 degree celcius appears to be 1 meter long. If the scale is correct at 0 degree celcius, what is the true length of the rod at 25 degree celcius? What wull be the length of rod at 0 degree celcius?


To solve the problem, we need to consider the concept of thermal expansion. When the temperature of a material changes, its length changes as well, depending on the material's coefficient of linear expansion. Here's how to solve it:

### Given Data:
- **Temperature at measurement:** \( T_1 = 25^\circ C \)
- **Temperature at correct scale:** \( T_2 = 0^\circ C \)
- **Measured length of the rod at \( T_1 \):** \( L_1 = 1 \text{ meter} \)
- **Coefficient of linear expansion for aluminum:** \( \alpha_{Al} = 23 \times 10^{-6} \, \text{per degree Celsius} \)
- **Coefficient of linear expansion for steel:** \( \alpha_{Steel} = 11 \times 10^{-6} \, \text{per degree Celsius} \)

### 1. **True Length of the Rod at \( 25^\circ C \):**
When the aluminum rod is measured with a steel scale at 25°C, both the rod and the scale expand. Since the scale expands, the measured length will be less than the actual length. 

Let's denote the true length of the rod at 25°C by \( L \).

The expansion of the steel scale causes the length measurement to appear shorter, so the relation can be given by:

\[
L = \text{Measured length} \times (1 + \alpha_{Steel} \times (T_1 - T_2))
\]

\[
L = 1 \times \left( 1 + 11 \times 10^{-6} \times (25 - 0) \right)
\]

\[
L = 1 \times \left( 1 + 11 \times 10^{-6} \times 25 \right)
\]

\[
L = 1 \times \left( 1 + 275 \times 10^{-6} \right)
\]

\[
L = 1 \times 1.000275 \, \text{meters}
\]

\[
L \approx 1.000275 \, \text{meters}
\]

### 2. **Length of the Rod at 0°C:**
To find the length of the rod at 0°C (denoted as \( L_0 \)), we must consider the true length of the rod at 25°C and then apply the contraction due to the temperature decrease.

\[
L_0 = L \times \left( 1 - \alpha_{Al} \times (T_1 - T_2) \right)
\]

\[
L_0 = 1.000275 \times \left( 1 - 23 \times 10^{-6} \times 25 \right)
\]

\[
L_0 = 1.000275 \times \left( 1 - 575 \times 10^{-6} \right)
\]

\[
L_0 = 1.000275 \times 0.999425
\]

\[
L_0 \approx 0.999996 \, \text{meters}
\]

### Final Results:
- The **true length of the rod at 25°C** is approximately **1.000275 meters**.
- The **length of the rod at 0°C** is approximately **0.999996 meters**.

Would you like more details or have any questions about this? 

Here are 5 related questions to explore further:
1. How would the results change if the coefficient of linear expansion for aluminum were different?
2. What if the temperature difference was greater, say 100°C instead of 25°C?
3. How does the concept of thermal expansion affect the accuracy of engineering measurements?
4. What happens if the steel scale's coefficient of linear expansion was higher than aluminum's?
5. How would you calculate the change in volume if the rod was not only expanding in length but in all dimensions?

**Tip:** Always remember to adjust measurements based on the material's thermal expansion when taking measurements in varying temperatures. This is crucial for precision engineering tasks.

An alumium rod when measured with a steel scale both being at 25 degree celcius appears to be 1 meter long. If the scale is correct at 0 degree celcius, what is the true length of the rod at 25 degree celcius? What will be the length of rod at 0 degree celcius?

Explore how to calculate the true length of an aluminum rod measured at 25°C using a steel scale, accounting for thermal expansion. This problem includes the expansion coefficients of aluminum and steel, solving for the true length at 25°C and the length at 0°C. Suitable for Grades 11-12 students.

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