Problem 41 - Master Resource Book in JEE Main Physics

Question

An iron tyre is to be fitted on a wooden wheel \(1 \mathrm{~m}\) in diameter. The diameter of tyre is \(6 \mathrm{~mm}\) smaller than that of wheel. The tyre should be heated so that its temperature increases by a minimum of (the coefficient of cubical expansion of iron is \(3.6 \times 10^{-5} /{ }^{\circ} \mathrm{C}\) (a) \(167^{\circ} \mathrm{C}\) (b) \(334^{\circ} \mathrm{C}\) (c) \(500^{\circ} \mathrm{C}\) (d) \(1000^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The tyre should be heated to \(500^{\circ} \mathrm{C}\) to fit the wheel.

Step 1: Understanding the Problem

We have a wooden wheel with a diameter of \(1\) meter and an iron tyre that is \(6\, \text{mm}\) smaller in diameter. We need to find how much we need to increase the temperature of the iron tyre to make its diameter equal to the wheel. The coefficient of cubical expansion for iron is given as \(3.6 \times 10^{-5} /{}^{\circ}\mathrm{C}\).

Step 2: Calculate Initial Diameter and Required Expansion

The initial diameter of the tyre is \(1 - 0.006 = 0.994\) meters. Therefore, the required change in diameter to fit the wheel is \(0.006\) meters.

Step 3: Understanding Linear Expansion Relation

Thermal expansion for linear dimensions is given by the formula \(\Delta L = \alpha L_0 \Delta T\), where \(\alpha\) is the coefficient of linear expansion, \(L_0\) is the original length (diameter in this case), and \(\Delta T\) is the change in temperature. Note the coefficient of linear expansion \(\alpha = \frac{\text{cubical coefficient}}{3}\).

Step 4: Calculate Coefficient of Linear Expansion

Given the coefficient of cubical expansion \(3.6 \times 10^{-5} /{}^{\circ}\mathrm{C}\), the coefficient of linear expansion is \( \alpha = \frac{3.6 \times 10^{-5}}{3} = 1.2 \times 10^{-5}/{}^{\circ}\mathrm{C}\).

Step 5: Apply the Linear Expansion Formula

Substituting the known values into the formula: \[ 0.006 = 1.2 \times 10^{-5} \times 0.994 \times \Delta T \] Solving for \(\Delta T\):\[ \Delta T = \frac{0.006}{1.2 \times 10^{-5} \times 0.994} \]

Step 6: Calculate Temperature Increase

Calculate \(\Delta T\):\[ \Delta T \approx \frac{0.006}{1.1928 \times 10^{-5}} \approx 502.68 \] Rounding to the nearest available option, the temperature increase required is approximately \(500^{\circ} \mathrm{C}\).

Key Concepts

Coefficient of Cubical Expansion

The coefficient of cubical expansion is a measure of how a material's volume changes with temperature. When a substance is heated, its molecules move faster and tend to occupy more space, leading to an increase in volume. The coefficient, often represented by the Greek letter beta (β), quantifies this change and is expressed as a fractional change in volume per degree of temperature change. This value is crucial in scenarios where precise volume changes are necessary, such as engineering applications.For solids like iron, which was mentioned in the exercise, cubical expansion is particularly relevant when estimating temperature-related volume changes. The coefficient of cubical expansion tells us how much a material will expand in three-dimensional space given a temperature change. In the case presented, iron has a coefficient of cubical expansion of \(3.6 \times 10^{-5} /{}^{\circ}\mathrm{C}\). This value helps us understand the degree to which the iron tyre's volume must increase to fit the wheel perfectly.

Linear Expansion

Linear expansion focuses on changes in one dimension, usually length. For the problem with the iron tyre, we look at how an increase in temperature affects the tyre's diameter. Similar to cubical expansion, linear expansion is governed by a coefficient, usually denoted by alpha (α), which represents the fractional change in length per degree of temperature change.To convert cubical expansion to linear, we divide the cubical coefficient by three. In this case, from \(3.6 \times 10^{-5}\) to \(1.2 \times 10^{-5}/{^{\circ}\mathrm{C}}\). This equation: \(\Delta L = \alpha L_0 \Delta T\) is then used to calculate the expansion needed for a desired change in length, \(\Delta L\). This calculation indicates how much we need to heat the tyre to achieve a specific increase in diameter so it fits snugly over the wheel. By understanding these expansion principles, we can ensure structures and components maintain their functionality under temperature variations.

Temperature Change Calculation

Calculating the temperature change needed for linear expansion involves simple algebraic manipulation. Given the initial measurements and desired changes, we rearrange the linear expansion formula \(\Delta L = \alpha L_0 \Delta T\) to solve for \(\Delta T\), the temperature change.In the tyre problem, you start with the desired increase in diameter, \(0.006\ \text{meters}\), over the original diameter \(0.994\ \text{meters}\) and apply the calculated linear expansion coefficient \(1.2 \times 10^{-5}/{^{\circ}\mathrm{C}}\). By substituting these values into the formula, we calculate \(\Delta T\). The result is a temperature change of approximately \(502.68^{\circ}\mathrm{C}\), which rounds to \(500^{\circ}\mathrm{C}\).This calculation highlights the importance of precise measurements in engineering. Accurate understanding of how temperature affects materials ensures the safety and functionality of assembled structures, like our iron tyre and wooden wheel.