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Difficulty: HardThermal Expansion of Solids (Linear, Area, and Volume Expansivity)

A steel rod and a brass rod are arranged such that the difference between their lengths remains constant at 10 cm10\text{ cm} at all temperatures. If the linear expansivity of steel is 1.2×10−5 K−11.2 \times 10^{-5}\text{ K}^{-1} and that of brass is 1.8×10−5 K−11.8 \times 10^{-5}\text{ K}^{-1}, what is the initial length of the steel rod in centimetres?

Answer: 30 cm

Answer

The initial length of the steel rod is 30 cm30\text{ cm}.
For the length difference between two rods to remain constant regardless of temperature change, both rods must undergo equal absolute expansion (\(\Delta L_1 = \Delta L_2\)). Since \(\Delta L = L_0 \alpha \Delta T\), this requires \(L_1 \alpha_1 = L_2 \alpha_2\). Substituting \(L_{\text{brass}} = L_{\text{steel}} - 10\text{ cm}\) and the given expansivity values gives \(1.2 \times 10^{-5} L_{\text{steel}} = 1.8 \times 10^{-5} (L_{\text{steel}} - 10)\), which simplifies to \(0.6 L_{\text{steel}} = 18\), giving \(L_{\text{steel}} = 30\text{ cm}\).

Step-by-Step Solution

1
Relate the condition for a constant difference in length to individual expansions
\(\Delta L_{\text{steel}} = \Delta L_{\text{brass}}\)
If the difference between the two lengths is constant across temperature changes, both rods must increase in length by the exact same amount for any given temperature change.
2
Apply the linear thermal expansion formula to both rods
\(L_{\text{steel}} \alpha_{\text{steel}} = L_{\text{brass}} \alpha_{\text{brass}}\)
Since \(\Delta L = L_0 \alpha \Delta T\), setting \(\Delta L_{\text{steel}} = \Delta L_{\text{brass}}\) gives \(L_{\text{steel}} \alpha_{\text{steel}} \Delta T = L_{\text{brass}} \alpha_{\text{brass}} \Delta T\). Cancelling \(\Delta T\) yields \(L_{\text{steel}} \alpha_{\text{steel}} = L_{\text{brass}} \alpha_{\text{brass}}\).
3
Substitute the length relationship into the equation
\(L_{\text{steel}} (1.2 \times 10^{-5}) = (L_{\text{steel}} - 10) (1.8 \times 10^{-5})\)
Because brass has a larger linear expansivity than steel, the brass rod must be shorter than the steel rod so that their products of length and expansivity remain equal, hence \(L_{\text{brass}} = L_{\text{steel}} - 10\text{ cm}\).
4
Solve for the length of the steel rod
\(L_{\text{steel}} = 30\text{ cm}\)
Dividing both sides by \(10^{-5}\) gives \(1.2 L_{\text{steel}} = 1.8 L_{\text{steel}} - 18\). Rearranging gives \(0.6 L_{\text{steel}} = 18\), which yields \(L_{\text{steel}} = \frac{18}{0.6} = 30\text{ cm}\).

Key Concept

Equal absolute linear expansion for constant length difference
Estimated Time:2m 0s
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Topics

Thermal Expansion of Solids (Linear, Area, and Volume Expansivity)

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