Question 7 of 20intermediate🔧 ApplyNumerical3 marks

A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel=1.20×10−5 K−1\alpha_{\text{steel}} = 1.20 \times 10^{-5} \text{ K}^{-1}.

Correct Answer

The shaft must be cooled to approximately -69 °C for the wheel to slip on.

Exercise: EXERCISES | Q: 10.7 | (Page 23)
For More Understanding

Explanation

This numerical problem applies the concept of thermal contraction. Since the shaft diameter (8.70 cm) is larger than the wheel hole (8.69 cm) at room temperature, the shaft must be cooled to contract. The diameter must decrease by 0.01 cm. Using the linear expansion formula, the required temperature change is calculated, and the final temperature is found by subtracting this change from the initial temperature.

Solution Steps

  1. Step 1: Identify given values - Initial temperature T₁ = 27 °C, Shaft diameter D = 8.70 cm, Hole diameter = 8.69 cm, αsteel = 1.20×101.20 \times 10⁻⁵ K⁻¹

  2. Step 2: Calculate required change in diameter - ΔD = 8.69 - 8.70 = -0.01 cm (shaft must contract by 0.01 cm)

  3. Step 3: Apply linear expansion formula - ΔD = D × α × ΔT

  4. Step 4: Solve for temperature change - ΔT = ΔD/(D × α) = -0.01/(8.70×1.208.70 \times 1.20 × 10⁻⁵) = -0.01/(10.44×1010.44 \times 10⁻⁵) = -95.79 K

  5. Step 5: Calculate final temperature - T₂ = T₁ + ΔT = 27 - 95.79 = -68.79 °C ≈\approx -69 °C

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