Question
10) The difference in length between a brass and an iron rod is 14 cm at 10
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C. What must be the length of the iron for this difference to remain at 14 cm when both rods are heated to 100
∘
C ? (linear expansivity of brass is
∘
C. What must be the length of the iron for this difference to remain at 14 cm when both rods are heated to 100
∘
C ? (linear expansivity of brass is
Answers
GPT-4o mini
To solve this problem, we need to use the concept of linear thermal expansion, which is given by the formula:
\[
\Delta L = L_0 \cdot \alpha \cdot \Delta T
\]
Where:
- \(\Delta L\) = change in length
- \(L_0\) = original length
- \(\alpha\) = coefficient of linear expansion
- \(\Delta T\) = change in temperature
Let:
- The length of the brass rod at \(10^\circ C\) be \(L_{b,0}\).
- The length of the iron rod at \(10^\circ C\) be \(L_{i,0}\).
From the problem statement, the difference between the two lengths at \(10^\circ C\) is given by:
\[
L_{b,0} - L_{i,0} = 14 \, \text{cm} \quad \text{(1)}
\]
When heated from \(10^\circ C\) to \(100^\circ C\), the change in temperature (\(\Delta T\)) is:
\[
\Delta T = 100 - 10 = 90^\circ C
\]
The change in length for each rod is given by linear thermal expansion:
\[
\Delta L_b = L_{b,0} \cdot \alpha_b \cdot 90
\]
\[
\Delta L_i = L_{i,0} \cdot \alpha_i \cdot 90
\]
Now, the new lengths at \(100^\circ C\) will be:
\[
L_{b} = L_{b,0} + \Delta L_b = L_{b,0} + L_{b,0} \cdot \alpha_b \cdot 90 = L_{b,0}(1 + \alpha_b \cdot 90)
\]
\[
L_{i} = L_{i,0} + \Delta L_i = L_{i,0} + L_{i,0} \cdot \alpha_i \cdot 90 = L_{i,0}(1 + \alpha_i \cdot 90)
\]
We want to maintain the length difference of 14 cm at \(100^\circ C\):
\[
L_b - L_i = 14 \, \text{cm} \quad \text{(2)}
\]
Substituting the expressions for \(L_b\) and \(L_i\) from the thermal expansion formulas into equation (2):
\[
L_{b,0}(1 + \alpha_b \cdot 90) - L_{i,0}(1 + \alpha_i \cdot 90) = 14
\]
Using equation (1) from above, \(L_{b,0} = L_{i,0} + 14\):
Substituting \(L_{b,0}\) into the equation gives:
\[
(L_{i,0} + 14)(1 + \alpha_b \cdot 90) - L_{i,0}(1 + \alpha_i \cdot 90) = 14
\]
Expanding this:
\[
L_{i,0} (1 + \alpha_b \cdot 90) + 14(1 + \alpha_b \cdot 90) - L_{i,0}(1 + \alpha_i \cdot 90) = 14
\]
Simplifying gives:
\[
L_{i,0} \alpha_b \cdot 90 + 14 + 14 \alpha_b \cdot 90 - L_{i,0} \alpha_i \cdot 90 = 14
\]
So now we can simplify and isolate \(L_{i,0}\):
\[
L_{i,0}(\alpha_b - \alpha_i) \cdot 90 + 14 \alpha_b \cdot 90 = 0
\]
To find \(L_{i,0}\):
\[
L_{i,0} = -\frac{14 \alpha_b}{(\alpha_b - \alpha_i)} \quad \text{(3)}
\]
We need the values of \(\alpha_b\) and \(\alpha_i\):
- Coefficient of linear expansion of brass, \(\alpha_b ≈ 19 \times 10^{-6} \, \text{°C}^{-1}\)
- Coefficient of linear expansion of iron, \(\alpha_i ≈ 11 \times 10^{-6} \, \text{°C}^{-1}\)
Now, substituting these values into equation (3):
\[
L_{i,0} = -\frac{14 \times 19 \times 10^{-6}}{(19 \times 10^{-6} - 11 \times 10^{-6})}
\]
Calculating this gives the original length of the iron rod \(L_{i,0}\).
Calculating:
\[
= -\frac{266 \times 10^{-6}}{(8 \times 10^{-6})} = L_{i,0} \approx 33.25 \, \text{cm}
\]
Thus, the length of the iron rod at \(10^\circ C\) should be approximately 33.25 cm for the difference to remain 14 cm when heated to \(100^\circ C\).
\[
\Delta L = L_0 \cdot \alpha \cdot \Delta T
\]
Where:
- \(\Delta L\) = change in length
- \(L_0\) = original length
- \(\alpha\) = coefficient of linear expansion
- \(\Delta T\) = change in temperature
Let:
- The length of the brass rod at \(10^\circ C\) be \(L_{b,0}\).
- The length of the iron rod at \(10^\circ C\) be \(L_{i,0}\).
From the problem statement, the difference between the two lengths at \(10^\circ C\) is given by:
\[
L_{b,0} - L_{i,0} = 14 \, \text{cm} \quad \text{(1)}
\]
When heated from \(10^\circ C\) to \(100^\circ C\), the change in temperature (\(\Delta T\)) is:
\[
\Delta T = 100 - 10 = 90^\circ C
\]
The change in length for each rod is given by linear thermal expansion:
\[
\Delta L_b = L_{b,0} \cdot \alpha_b \cdot 90
\]
\[
\Delta L_i = L_{i,0} \cdot \alpha_i \cdot 90
\]
Now, the new lengths at \(100^\circ C\) will be:
\[
L_{b} = L_{b,0} + \Delta L_b = L_{b,0} + L_{b,0} \cdot \alpha_b \cdot 90 = L_{b,0}(1 + \alpha_b \cdot 90)
\]
\[
L_{i} = L_{i,0} + \Delta L_i = L_{i,0} + L_{i,0} \cdot \alpha_i \cdot 90 = L_{i,0}(1 + \alpha_i \cdot 90)
\]
We want to maintain the length difference of 14 cm at \(100^\circ C\):
\[
L_b - L_i = 14 \, \text{cm} \quad \text{(2)}
\]
Substituting the expressions for \(L_b\) and \(L_i\) from the thermal expansion formulas into equation (2):
\[
L_{b,0}(1 + \alpha_b \cdot 90) - L_{i,0}(1 + \alpha_i \cdot 90) = 14
\]
Using equation (1) from above, \(L_{b,0} = L_{i,0} + 14\):
Substituting \(L_{b,0}\) into the equation gives:
\[
(L_{i,0} + 14)(1 + \alpha_b \cdot 90) - L_{i,0}(1 + \alpha_i \cdot 90) = 14
\]
Expanding this:
\[
L_{i,0} (1 + \alpha_b \cdot 90) + 14(1 + \alpha_b \cdot 90) - L_{i,0}(1 + \alpha_i \cdot 90) = 14
\]
Simplifying gives:
\[
L_{i,0} \alpha_b \cdot 90 + 14 + 14 \alpha_b \cdot 90 - L_{i,0} \alpha_i \cdot 90 = 14
\]
So now we can simplify and isolate \(L_{i,0}\):
\[
L_{i,0}(\alpha_b - \alpha_i) \cdot 90 + 14 \alpha_b \cdot 90 = 0
\]
To find \(L_{i,0}\):
\[
L_{i,0} = -\frac{14 \alpha_b}{(\alpha_b - \alpha_i)} \quad \text{(3)}
\]
We need the values of \(\alpha_b\) and \(\alpha_i\):
- Coefficient of linear expansion of brass, \(\alpha_b ≈ 19 \times 10^{-6} \, \text{°C}^{-1}\)
- Coefficient of linear expansion of iron, \(\alpha_i ≈ 11 \times 10^{-6} \, \text{°C}^{-1}\)
Now, substituting these values into equation (3):
\[
L_{i,0} = -\frac{14 \times 19 \times 10^{-6}}{(19 \times 10^{-6} - 11 \times 10^{-6})}
\]
Calculating this gives the original length of the iron rod \(L_{i,0}\).
Calculating:
\[
= -\frac{266 \times 10^{-6}}{(8 \times 10^{-6})} = L_{i,0} \approx 33.25 \, \text{cm}
\]
Thus, the length of the iron rod at \(10^\circ C\) should be approximately 33.25 cm for the difference to remain 14 cm when heated to \(100^\circ C\).