Question 10 of 20intermediate🔧 ApplyNumerical3 marks

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0×10−5 K−12.0 \times 10^{-5} \text{ K}^{-1}, steel = 1.2×10−5 K−11.2 \times 10^{-5} \text{ K}^{-1}).

Correct Answer

Step 1: Identify the change in temperature. The initial temperature is 40.0 °C and the final temperature is 250 °C. Change in temperature, ΔT=250−40=210 °C\Delta T = 250 - 40 = 210 \text{ °C}.

Step 2: Calculate the change in length for each rod. Using the relation ΔL=LαΔT\Delta L = L \alpha \Delta T: For brass rod (L=50 cmL = 50 \text{ cm}, α=2.0×10−5 K−1\alpha = 2.0 \times 10^{-5} \text{ K}^{-1}): ΔLbrass=50×2.0×10−5×210=0.21 cm\Delta L_{\text{brass}} = 50 \times 2.0 \times 10^{-5} \times 210 = \mathbf{0.21 \text{ cm}}.

For steel rod (L=50 cmL = 50 \text{ cm}, α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5} \text{ K}^{-1}): ΔLsteel=50×1.2×10−5×210=0.126 cm\Delta L_{\text{steel}} = 50 \times 1.2 \times 10^{-5} \times 210 = \mathbf{0.126 \text{ cm}}.

Step 3: Determine total change in length. Total change in length of the combined rod = ΔLbrass+ΔLsteel=0.21+0.126=0.336 cm\Delta L_{\text{brass}} + \Delta L_{\text{steel}} = 0.21 + 0.126 = \mathbf{0.336 \text{ cm}}.

Step 4: Determine if thermal stress is developed. No, there is no thermal stress developed at the junction. The text states that thermal stress is developed when thermal expansion is prevented by fixing the ends rigidly. Since the ends of the rod are free to expand, there is no external force preventing the expansion, and thus no thermal stress is developed.

Exercise: EXERCISES | Q: 10.10 | (Page 23)
For More Understanding

Explanation

The question asks for a numerical calculation of thermal expansion and a conceptual determination of thermal stress. The solution calculates the expansion for both brass and steel rods separately using the linear expansion formula derived from the context (strain = αΔT\alpha \Delta T) and sums them for the total change. For the stress component, the answer relies on the specific definition in the provided text: thermal stress arises only if expansion is prevented by rigid supports. Since the problem states the ends are free, no stress develops.

Solution Steps

  1. Step 1: Calculate temperature difference: ΔT=250−40=210 °C\Delta T = 250 - 40 = 210 \text{ °C}.

  2. Step 2: Calculate expansion of brass: ΔLbrass=50×2.0×10−5×210=0.21 cm\Delta L_{\text{brass}} = 50 \times 2.0 \times 10^{-5} \times 210 = 0.21 \text{ cm}.

  3. Step 3: Calculate expansion of steel: ΔLsteel=50×1.2×10−5×210=0.126 cm\Delta L_{\text{steel}} = 50 \times 1.2 \times 10^{-5} \times 210 = 0.126 \text{ cm}.

  4. Step 4: Sum the expansions: 0.21+0.126=0.336 cm0.21 + 0.126 = 0.336 \text{ cm}.

  5. Step 5: Conclude on thermal stress: No stress developed because ends are free to expand (expansion not prevented).

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