Problem 9 - Arihant AIEEE Physics

Question

At \(30^{\circ} \mathrm{C}\), the hole in a steel plate has diameter of \(0.99970\) \(\mathrm{cm}\). A cylinder of diameter exactly \(1 \mathrm{~cm}\) at \(30^{\circ} \mathrm{C}\) is to be slide into the hole. To what temperature the plate must be heated ? (Given : \(\alpha_{\text {steel }}=1.1 \times 10^{-5}{\underline{\phantom{xx}}}^{\circ} \mathrm{C}^{-1}\) ) (a) \(58^{\circ} \mathrm{C}\) (b) \(55^{\circ} \mathrm{C}\) (c) \(57.3^{\circ} \mathrm{C}\) (d) \(60^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The steel plate must be heated to approximately \(57.3^{\circ} \mathrm{C}\) (Option c).

Step 1: Understand the problem

We need to increase the diameter of a hole in a steel plate from 0.99970 cm to 1.00 cm by heating it. We are given the coefficient of linear expansion, \( \alpha_{\text{steel }} = 1.1 \times 10^{-5} {}^{\circ} \mathrm{C}^{-1} \), and the initial temperature as 30°C. Our task is to find the temperature to which we need to heat the plate so that the hole accommodates the cylinder.

Step 2: Set up the linear expansion formula

The linear expansion formula relates the change in length (or diameter, in this case) \( \Delta L \), the original length \( L_0 \), the coefficient of linear expansion \( \alpha \), and the change in temperature \( \Delta T \). It is given by: \[ \Delta L = L_0 \alpha \Delta T \] where \( \Delta L \) is the amount we need to increase the diameter of the hole.

Step 3: Calculate the change in diameter \( \Delta L \)

The change in diameter required, \( \Delta L \), is the difference between the final and initial diameters of the hole: \[ \Delta L = 1.00 \text{ cm} - 0.99970 \text{ cm} = 0.00030 \text{ cm} \]

Step 4: Substitute the known values into the formula

Using the values \( L_0 = 0.99970 \text{ cm} \), \( \Delta L = 0.00030 \text{ cm} \), and \( \alpha = 1.1 \times 10^{-5} {}^{\circ} \mathrm{C}^{-1} \), substitute them into the linear expansion formula: \[ 0.00030 = 0.99970 \times 1.1 \times 10^{-5} \times \Delta T \]

Step 5: Solve for \( \Delta T \)

Rearrange the equation to solve for \( \Delta T \): \[ \Delta T = \frac{0.00030}{0.99970 \times 1.1 \times 10^{-5}} \] Calculate \( \Delta T \): \[ \Delta T \approx 27.3^{\circ} \mathrm{C} \]

Step 6: Find the final temperature

The final temperature \( T_f \) is the initial temperature plus the change in temperature: \[ T_f = 30^{\circ} \mathrm{C} + 27.3^{\circ} \mathrm{C} = 57.3^{\circ} \mathrm{C} \]

Key Concepts

Coefficient of Linear Expansion

The concept of the Coefficient of Linear Expansion is key in understanding how materials expand when heated. This coefficient, often denoted by \( \alpha \), measures the degree to which a certain material expands per degree change in temperature. In our exercise, we consider steel with a coefficient of linear expansion \( \alpha_{\text{steel}} = 1.1 \times 10^{-5} \,{}^{\circ}\mathrm{C}^{-1} \). This value implies that for every degree Celsius increase in temperature, each unit length (or diameter, in our case) will expand by \( 1.1 \times 10^{-5} \times \text{original length} \).Understanding \( \alpha \) helps explain why different materials react differently to heat. For example, steel will expand at a specific rate compared to other materials like aluminum or copper.

Linear Expansion Formula

The Linear Expansion Formula helps calculate how much a material's length changes with temperature. It's expressed as \( \Delta L = L_0 \alpha \Delta T \), where:

• \( \Delta L \) is the change in length (or diameter in this context),
• \( L_0 \) is the initial length (initial diameter of the hole),
• \( \alpha \) is the coefficient of linear expansion.

In the exercise, \( L_0 \) is 0.99970 cm, and we need a \( \Delta L \) of 0.00030 cm, which is the amount by which the hole's diameter must increase to fit the cylinder. The formula allows easy calculation of how much a material needs to change. By rearranging the formula, you can solve for \( \Delta T \), the change in temperature required to achieve this expansion.

Temperature Change Calculation

The Temperature Change Calculation in our exercise involves finding \( \Delta T \) using the Linear Expansion Formula. We set up the equation:\[ 0.00030 = 0.99970 \times 1.1 \times 10^{-5} \times \Delta T \]Rearranging and solving the above gives:\[ \Delta T = \frac{0.00030}{0.99970 \times 1.1 \times 10^{-5}} \]After calculation, you find \( \Delta T \approx 27.3^{\circ} \mathrm{C} \).
Now, to find the final temperature \( T_f \), add this change to the initial temperature:
\[ T_f = 30^{\circ} \mathrm{C} + 27.3^{\circ} \mathrm{C} = 57.3^{\circ} \mathrm{C} \]Thus, the steel plate must be heated to approximately 57.3°C for the cylinder to fit snugly into the expanded hole. This calculation shows the relationship between temperature and physical changes in materials, demonstrating how knowledge of thermodynamics and material properties is practically applied.