Question

At what temperature should a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm? Both the diameters have been measured at room temperature (27 °C).
Given: Coefficient of linear thermal expansion of gold \( \alpha = 1.4 \times 10^{-5} \, \text{K}^{-1} \).

• A. 125.7°C
• B. 91.7°C
• C. 425.7°C
• D. 152.7°C

Hint:

The linear thermal expansion formula \( \Delta L = L_0 \alpha \Delta T \) is used to calculate the change in length (or diameter) of an object due to a temperature change. Ensure that the units of \( \Delta T \) are consistent (Kelvin or Celsius).

Answer 1

The Correct Option is D

Step 1: Problem Identification
A gold ring with an initial diameter of \(6.230 \, \text{cm}\) at \( 27^\circ \text{C} \) requires expansion to \(6.241 \, \text{cm}\) to fit a wooden bangle. The objective is to determine the necessary final temperature.
Step 2: Thermal Expansion Formula Application
The formula for the change in diameter (\( \Delta D \)) due to thermal expansion is provided as:\[\Delta D = D_0 \alpha \Delta T,\]where:- \( D_0 = 6.230 \, \text{cm} \) denotes the initial diameter,- \( \alpha = 1.4 \times 10^{-5} \, \text{K}^{-1} \) is the coefficient of linear thermal expansion,- \( \Delta T \) represents the temperature increment.
Step 3: Diameter Change Calculation \[\Delta D = 6.241 - 6.230 = 0.011 \, \text{cm}.\]
Step 4: Temperature Change Determination \[\Delta T = \frac{\Delta D}{D_0 \alpha}.\]Substituting known values:\[\Delta T = \frac{0.011}{6.230 \times 1.4 \times 10^{-5}}.\]\[\Delta T = \frac{0.011}{8.722 \times 10^{-5}} = 126.1 \, \text{K}.\]
Step 5: Final Temperature Computation \[T_f = T_0 + \Delta T.\]\[T_f = 27 + 126.1 = 153.1^\circ \text{C}.\]The final temperature, rounded, is approximately \(152.7^\circ\text{C}\).
Step 6: Option Correlation The computed value most closely aligns with option (D) 152.7°C.Final Answer: The gold ring must achieve a temperature of 152.7°C.