Problem 43 - Master Resource Book in JEE Main Physics

Question

A steel scale measures the length of a copper wire as \(80.0 \mathrm{~cm}\), when both are at \(20^{\circ} \mathrm{C}\) (the calibration temperature for scale). What would be the scale read for the length of the wire when both are at \(40^{\circ} \mathrm{C}\) ? (Given \(\alpha_{\text {nteel }}=11 \times 10^{-6}\) per \(^{\circ} \mathrm{C}\) and \(\alpha_{\text {copper }}=17 \times 10^{-6}\) per \(^{\circ} \mathrm{C}\) ) (a) \(80.0096 \mathrm{~cm}\) (b) \(80.0272 \mathrm{~cm}\) (c) \(1 \mathrm{~cm}\) (d) \(25.2 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The scale reads 80.0096 cm at 40°C.

Step 1: Understand Thermal Expansion Concepts

When a material is heated, it expands. The change in length due to temperature change is given by the formula: \( \Delta L = L_0 \alpha \Delta T \), where \( \Delta L \) is the change in length, \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.

Step 2: Calculate Temperature Change

The change in temperature, \( \Delta T \), is the difference between the final temperature and the initial temperature: \( \Delta T = 40^{\circ} C - 20^{\circ} C = 20^{\circ} C \).

Step 3: Calculate Expansion of Steel Scale

Using the formula \( \Delta L = L_0 \alpha \Delta T \) for the steel scale: \( \Delta L_{steel} = 80.0 \text{ cm} \times 11 \times 10^{-6} \times 20^{\circ} C \). Calculate \( \Delta L_{steel} \).

Step 4: Calculate Expansion of Copper Wire

Using the formula \( \Delta L = L_0 \alpha \Delta T \) for the copper wire: \( \Delta L_{copper} = 80.0 \text{ cm} \times 17 \times 10^{-6} \times 20^{\circ} C \). Calculate \( \Delta L_{copper} \).

Step 5: Calculate New Scale Reading

The new length of the wire, according to the scale, is \( (L_{0_wire} + \Delta L_{copper}) - \Delta L_{steel} \). Calculate this using the values obtained in Steps 3 and 4. Once computed, find the actual new length measured by the scale and compare to the options given.

Key Concepts

Linear Expansion

When materials are subjected to changes in temperature, they tend to expand or contract; this phenomenon is known as thermal expansion. A crucial aspect of thermal expansion is linear expansion, which specifically deals with the change in length of materials as they experience temperature variations.
Linear expansion is particularly important in engineering and construction where the precise fit and alignment of components are critical. To mathematically express this, the change in length (\( \Delta L \)) of a material is given by the formula:\[ \Delta L = L_0 \alpha \Delta T \]where:

• \( \Delta L \): Change in length
• \( L_0 \): Original length of the material
• \( \alpha \): Coefficient of linear expansion
• \( \Delta T \): Change in temperature

Understanding this formula helps us predict how materials will behave under different thermal conditions.

Coefficient of Linear Expansion

The coefficient of linear expansion, denoted as \( \alpha \), is a material-specific value that indicates how much a material expands per degree of temperature change. This value is generally expressed in units per degree Celsius (\(^{\circ}C^{-1}\)), and it varies greatly between different materials.
The importance of the coefficient of linear expansion cannot be overstated in applications where precise measurements are crucial. For instance:

• A higher \( \alpha \) means the material will expand more for each degree of temperature increase.
• A lower \( \alpha \) indicates that the material's length will change only slightly with temperature variations.

This coefficient is instrumental when considering material choices for construction and manufacturing, ensuring that temperature-induced dimensional changes do not compromise integrity and functionality.

Temperature Change

Temperature change, represented as \( \Delta T \), is the difference between the initial and final temperatures of a system. This is calculated using the simple formula:\[ \Delta T = T_{final} - T_{initial} \]Understanding the concept of temperature change is vital for predicting how materials will expand or contract.
For example, in the given exercise, the steel scale and copper wire were initially at 20°C and were then subjected to a temperature increase to 40°C. Consequently, this resulted in:

• An increase in temperature of \(20^{\circ} C\)
• This increase caused both materials to expand, but by different amounts due to their distinct coefficients of linear expansion.

Accurately calculating \( \Delta T \) ensures that we can precisely apply the formula of linear expansion to predict changes in material dimensions.