Problem 24 - Master Resource Book in JEE Main Physics

Question

A wheel is \(80.3 \mathrm{~cm}\) in circumference. An iron tyre measures \(80.0 \mathrm{~cm}\) around its inner face. If the coefficient of linear expansion for iron is \(1.2 \times 0^{-5}{\underline{\phantom{xx}}}^{\circ} \mathrm{C}^{-1}\), the temperature of the tyre must be raised by (a) \(105^{\circ} \mathrm{C}\) (b) \(417^{\circ} \mathrm{C}\) (c) \(312^{+} \mathrm{C}\) (d) \(223^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The temperature of the tyre must be raised by 312°C (option c).

Step 1: Understanding the Problem

We are given a wheel with a circumference of 80.3 cm that needs to fit an iron tyre originally measuring 80.0 cm in circumference. The task is to determine the temperature increase needed for the tyre to expand and fit over the wheel using the linear expansion formula.

Step 2: Linear Expansion Formula

The formula for linear expansion is: \[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \] where \(\Delta L\) is the change in length, \(L_0\) is the original length (80.0 cm), \(\alpha\) is the coefficient of linear expansion (\(1.2 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1}\)), and \(\Delta T\) is the change in temperature (what we want to find).

Step 3: Calculate the Required Change in Length (\(\Delta L\))

The required \(\Delta L\) is the difference between the circumference of the wheel and the current size of the inner tyre: \[ \Delta L = 80.3 \, \mathrm{cm} - 80.0 \, \mathrm{cm} = 0.3 \, \mathrm{cm} \]

Step 4: Rearrange the Expansion Formula to Find\(\Delta T\)

Rearrange the linear expansion formula to solve for \(\Delta T\): \[ \Delta T = \frac{\Delta L}{L_0 \cdot \alpha} \]

Step 5: Substitute Values into the Formula

Substitute the known values into the rearranged formula: \[ \Delta T = \frac{0.3 \, \mathrm{cm}}{80.0 \, \mathrm{cm} \cdot 1.2 \times 10^{-5} \, ^{\circ}\mathrm{C}^{-1}} \] This calculation finds the temperature increase required.

Step 6: Perform the Calculation

Calculate \(\Delta T\): \[ \Delta T = \frac{0.3}{80.0 \times 1.2 \times 10^{-5}} = 312.5 \, ^{\circ} \mathrm{C} \] The necessary temperature increase is approximately 312.5 degrees Celsius.

Step 7: Choose the Closest Option

Looking at the options provided, the closest temperature increase required for the tyre is approximately 312 degrees Celsius, which matches option (c), 312\(^{+} \mathrm{C}\).

Key Concepts

Linear Expansion

Linear expansion is a principle explaining how materials expand upon heating. This is crucial when dealing with connections like fitting a metal tyre over a wheel. As temperature rises, the properties of materials change, causing an increase in length. For example, when the iron tyre heats up, its circumference expands to fit the wheel.

• The process depends on the type of material.
• It's important to know the original size and targeted expansion.

The linear expansion formula helps us calculate the change in length due to temperature change. This is essential for ensuring components like tires fit and function correctly, despite changes in environment or usage conditions.

Coefficient of Linear Expansion

The coefficient of linear expansion, denoted as \( \alpha \), is a material-specific value that quantifies how much a material expands per degree change in temperature. It’s measured in terms of \( \frac{1}{\text{°C}} \).

• This coefficient varies by material, highlighting different expansion rates.
• For iron in this scenario, \( \alpha \) is given as \( 1.2 \times 10^{-5} \, ^{\circ} \text{C}^{-1} \).

Understanding this coefficient is essential for predicting how a material behaves with thermal changes. It's used in the linear expansion formula to calculate how much a tyre, for example, needs to be heated to fit over a wheel precisely without damage.

Temperature Change Calculation

To determine the temperature change needed for expansion, we use the linear expansion formula rearranged to solve for temperature:\[ \Delta T = \frac{\Delta L}{L_0 \cdot \alpha} \]Where:

• \( \Delta L \) is the change in length required (0.3 cm in this problem).
• \( L_0 \) is the original length or initial measurement of the object (80.0 cm for the tyre).
• \( \alpha \) is the coefficient of linear expansion (\( 1.2 \times 10^{-5} \, ^{\circ} \text{C}^{-1} \)).

This formula provides a way to calculate exactly how much heat is necessary for the object—in this case, a metal tyre—to expand just the right amount. Through substitution and solving, you find \( \Delta T \) which tells you the exact temperature increase needed.

Circumference Increase

The problem specifically addresses a circumference increase, which is a practical application of linear expansion. For the metal tyre initially measuring 80.0 cm, it must expand to 80.3 cm to fit securely onto the wheel. By calculating the difference in these lengths, you estimate the required change (\\[ \Delta L = 0.3 \, \text{cm} \]).

• This comparison allows us to set the needed expansion during heating.
• Without precise calculations, the tyre might not fit or be too loose.

These changes are necessary whenever thermal fitting is required, highlighting the significance of understanding how even small temperature changes can influence material dimensions effectively.