Let's tackle your questions one at a time, focusing on the thermal expansion properties of materials. Understanding how materials expand with temperature changes is crucial in many engineering applications. We'll break down each question step by step.
1. Temperature at Which the Steel Sphere Falls Through the Aluminium Plate
To determine the temperature at which the steel sphere will fall through the hole in the aluminium plate, we need to consider the thermal expansion of both materials. When the temperature increases, both the aluminium plate and the steel sphere will expand, but at different rates due to their differing coefficients of linear expansion.
Given Data
- Diameter of the hole in aluminium (D_aluminium) = 2 cm
- Diameter of the steel sphere (D_steel) = 2 cm (assumed for this example)
- Coefficient of linear expansion for steel (α_steel) = 11 x 10-6 °C-1
- Coefficient of linear expansion for aluminium (α_aluminium) = 23 x 10-6 °C-1
- Initial temperature (T_initial) = 10 °C
Calculating the Expansion
The change in diameter due to thermal expansion can be calculated using the formula:
ΔD = D_initial × α × ΔT
Where:
- ΔD = change in diameter
- D_initial = initial diameter
- α = coefficient of linear expansion
- ΔT = change in temperature
For the aluminium hole:
ΔD_aluminium = D_aluminium × α_aluminium × (T - T_initial)
For the steel sphere:
ΔD_steel = D_steel × α_steel × (T - T_initial)
We want to find the temperature T at which the diameter of the steel sphere exceeds the diameter of the hole in the aluminium plate:
D_steel + ΔD_steel > D_aluminium + ΔD_aluminium
Substituting the values:
2 + 2 × 11 x 10-6 × (T - 10) > 2 + 2 × 23 x 10-6 × (T - 10)
Solving this inequality will give us the temperature at which the sphere falls through the hole. Simplifying:
11 x 10-6 × (T - 10) > 23 x 10-6 × (T - 10)
Rearranging leads to:
(23 - 11) x 10-6 × (T - 10) > 0
Thus, we find:
12 x 10-6 × (T - 10) > 0
This implies that T must be greater than 10 °C. The exact temperature can be calculated by setting the two expansions equal and solving for T, which will yield a specific value.
2. Temperature Range for Accurate Measurements with a Steel Metre Scale
Next, let's address the second question regarding the accuracy of a steel metre scale at different temperatures. The scale is accurate at 20 °C, and we need to find the temperature range where the measurements remain within 0.055 mm over a 1 m length.
Understanding Measurement Accuracy
The change in length of the steel scale due to thermal expansion can be expressed as:
ΔL = L_initial × α × ΔT
Where:
- ΔL = change in length
- L_initial = initial length (1 m = 1000 mm)
- α = coefficient of linear expansion for steel (11 x 10-6 °C-1)
- ΔT = change in temperature
We want the change in length ΔL to be less than or equal to 0.055 mm:
ΔL ≤ 0.055 mm
Substituting the values:
1000 mm × 11 x 10-6 × (T - 20) ≤ 0.055
Solving for T:
11 x 10-3 × (T - 20) ≤ 0.055
This simplifies to:
T - 20 ≤ 5
Thus, we find:
T ≤ 25 °C
Now, we also need to consider the lower limit. If we assume the same expansion in the negative direction:
1000 mm × 11 x 10-6 × (T - 20) ≥ -0.055
Solving this gives:
T - 20 ≥ -5
Thus, we find:
T ≥ 15 °C
Final Temperature Range
Combining both results, the temperature range for accurate measurements with the steel metre scale is:
15 °C ≤ T ≤ 25 °C
This range ensures that the thermal expansion does not exceed the specified accuracy of 0.055 mm over a 1 m length. Understanding these principles of thermal expansion is essential for ensuring precision in measurements and applications involving temperature variations.
