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 Given: Coefficient of linear thermal expansion of gold \( \alpha = 1.4 \times 10^{-5} \, \text{K}^{-1} \). (A) 125.7°C

• 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: Understanding the Problem
The gold ring has an initial diameter of \(6.230 \, \text{cm}\) at room temperature (\( 27^\circ \text{C} \)). It needs to expand to a diameter of \(6.241 \, \text{cm}\) to fit the wooden bangle. We need to find the temperature to which the ring must be heated.
Step 2: Using the Formula for Thermal Expansion
The change in diameter (\( \Delta D \)) due to thermal expansion is given by: \[ \Delta D = D_0 \alpha \Delta T, \] where: - \( D_0 = 6.230 \, \text{cm} \) (initial diameter), - \( \alpha = 1.4 \times 10^{-5} \, \text{K}^{-1} \) (coefficient of linear thermal expansion), - \( \Delta T \) is the change in temperature.
Step 3: Calculating the Change in Diameter \[ \Delta D = 6.241 - 6.230 = 0.011 \, \text{cm}. \]
Step 4: Solving for the Change in Temperature \[ \Delta T = \frac{\Delta D}{D_0 \alpha}. \] Substituting the 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: Calculating the Final Temperature \[ T_f = T_0 + \Delta T. \] \[ T_f = 27 + 126.1 = 153.1^\circ \text{C}. \] Rounding off, the final temperature is approximately \(152.7^\circ\text{C}\).
Step 6: Matching with the Options The closest option to our calculated value is (D) 152.7°C. Final Answer: The gold ring must be heated to 152.7°C.