Problem 44 - 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 will read 80.0096 cm at 40°C.

Step 1: Understand the Problem

We need to determine the reading on the steel scale for the length of a copper wire initially measured at 80.0 cm at 20°C, when both the scale and wire are heated to 40°C. We'll be using thermal expansion formulas to calculate how both the wire and the scale expand with temperature change.

Step 2: Calculate the Change in Length for Steel Scale

Using the formula for linear expansion \( \Delta L = L_0 \alpha \Delta T \), where \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the temperature change:For the steel scale:\[ \Delta L_{\text{steel}} = 80.0 \times 11 \times 10^{-6} \times (40 - 20) \]\[ \Delta L_{\text{steel}} = 80.0 \times 11 \times 10^{-6} \times 20 \]\[ \Delta L_{\text{steel}} = 17.6 \times 10^{-3} \text{ cm} \]The steel scale expands by 0.0176 cm.

Step 3: Calculate the Actual Length of Copper Wire at 40°C

For the copper wire, also using the linear expansion formula:\[ \Delta L_{\text{copper}} = 80.0 \times 17 \times 10^{-6} \times (40 - 20) \]\[ \Delta L_{\text{copper}} = 80.0 \times 17 \times 10^{-6} \times 20 \]\[ \Delta L_{\text{copper}} = 27.2 \times 10^{-3} \text{ cm} \]The copper wire's new length at 40°C is:\[ L_{\text{copper}} = 80.0 + 0.0272 = 80.0272 \text{ cm} \]

Step 4: Calculate the Scale Reading

Since the steel scale expanded, its reading will not reflect the actual length of the wire directly. The scale's reading is updated using its expanded length:Taking the original measurement as \( 80 \text{ cm} \times \frac{80.0272}{80.0176} \), consider only the expansion impact:The length changes relative to the original precision, further adjustments hint at simple reading:\[ 80.0 + 0.0096 = 80.0096 \text{ cm} \]Here's why the slight deviation directly guides the accurate perception.

Key Concepts

Coefficient of Linear Expansion

The Coefficient of Linear Expansion describes how the size of an object changes with a change in temperature. This concept is crucial when dealing with different materials that expand at different rates as they heat up. It is represented by the symbol \( \alpha \) and indicates the amount of expansion per unit length per degree of temperature change. This value varies between materials, reflecting their atomic bonds and structural properties.

• Steel has a coefficient of linear expansion (\( \alpha_{\text{steel}} \)) of \( 11 \times 10^{-6} \) per °C, meaning it expands only slightly with temperature changes.
• Copper has a higher coefficient (\( \alpha_{\text{copper}} \)) of \( 17 \times 10^{-6} \) per °C, indicating it expands more than steel under similar conditions.

Understanding these coefficients helps predict how measurements will change with temperature, which is vital in precise applications, such as instrument calibration or construction.

Linear Expansion Formula

The Linear Expansion Formula helps calculate how much an object will expand due to temperature changes. It is expressed as:\[ \Delta L = L_0 \alpha \Delta T \]Where:

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

This formula is used to compute the extent of expansion for both the steel scale and copper wire in our exercise. By plugging in the respective values for copper and steel, we can determine how much each material expands, helping us understand their new dimensions at the increased temperature of 40°C.

For instance, using the Linear Expansion Formula for the copper wire:

• Original length \( L_0 = 80 \text{ cm} \)
• Temperature change \( \Delta T = 40 - 20 = 20 \text{ °C} \)
• \( \alpha_{\text{copper}} = 17 \times 10^{-6} \) per °C
• Calculate the expansion: \( \Delta L_{\text{copper}} = 80 \times 17 \times 10^{-6} \times 20 \)
• This results in a change of \( 0.0272 \text{ cm} \)

This formula allows predictions about size changes due to thermal effects, which are crucial for accuracy in engineering and design tasks.

Temperature Change Calculations

Temperature Change Calculations are fundamental in predicting how objects behave when heated or cooled. They help us determine the extent of expansion or contraction, impacting how accurate an initial measurement remains at different temperatures.

To calculate the impact of temperature changes on both the steel scale and copper wire:

• Identify \( \Delta T \), the change in temperature, which in this example is \( 20 \)°C (from 20°C to 40°C).
• Apply the Linear Expansion Formula to measure the change in length due to thermal expansion.

Specifically, for our exercise:

• The steel scale expanded by 0.0176 cm, calculated using its coefficient of linear expansion \( \alpha_{\text{steel}} \).
• The copper wire expanded by 0.0272 cm, as determined by its higher \( \alpha_{\text{copper}} \).

Ultimately, understanding these changes leads to accurate measurement adjustment, critical in environments that undergo temperature fluctuations. Such calculations ensure that precision equipment functions correctly, despite ambient changes, enhancing the reliability of measurements in science and industry.