JEE Main 2022 · 29 June · evening · Single correct · +4 / −1

At what temperature 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 α L = 1.4 × 10⁻⁵ K⁻¹)

  1. (a) 125.7°C
  2. (b) 91.7°C
  3. (c) 425.7°C
  4. (d) 152.7°C
Answer and solution

Answer (d) 152.7°C

Solution

Physics sir · 3 steps

The ring's diameter must stretch by just 0.011 cm. Find the temperature rise for that, then add 27.

  1. Required stretch. The ring must grow from 6.230 cm to 6.241 cm. So Δ D = 0.011 cm. A diameter is a length, so it expands with the linear coefficient α_L.

    Δ D = D α_L Δ T ⇒ Δ T = (Δ D)/(D α_L)

  2. Put in numbers. Δ T = 0.011/(6.230 × 1.4 × 10⁻⁵) = 0.011/(8.72 × 10⁻⁵) ≈ 126 K. A rise of 1 K equals a rise of 1 °C.
  3. So. The final temperature is 27 + 126 ≈ 153 °C. The closest option, within rounding, is (d), 152.7 °C.

The idea to keep. A ring's hole expands exactly like a solid disc of the same material, so use Δ D = DαΔ T on the diameter. The formula gives the rise, so add the starting temperature at the end.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Thermometry & Calorimetry

Asked about 50 times in JEE Main since 2020. This question is from 2022.