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JEE Main 2022 (29 Jun, Shift 2) · Thermometry & Thermal Expansion

Thermometry & Thermal Expansion question from JEE Main 2022 (29 Jun, Shift 2)

Q43JEE Main 2022 · 29 JUN · SHIFT 2Medium

Thermometry & Thermal Expansion

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∘^\circC).

(Given: coefficient of linear thermal expansion of gold α\alphaL = 1.4 ×\times 10−-5 K−1{}^{-1})

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Answer: (d)

  1. Ring diameter must grow by ΔD=6.241−6.230=0.011\Delta D = 6.241 - 6.230 = 0.011 cm.
  2. ΔT=ΔDDα=0.0116.230×1.4×10−5≈126\Delta T = \dfrac{\Delta D}{D\alpha} = \dfrac{0.011}{6.230 \times 1.4\times10^{-5}} \approx 126 K
  3. T=27+126≈153∘CT = 27 + 126 \approx 153^\circ\text{C}
  4. So the answer is (d) 152.7∘C152.7^\circ\text{C}.
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