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A copper rod and a steel rod, each of equal length and equal cross-sectional area, are joined end to end. The free end of the copper rod is maintained at 0 degree C and the free end of the steel rod is maintained at 100 degree C. Find the temperature at the junction. (Thermal conductivity of copper = 390 W/(m*deg C) and of steel = 46 W/(m*deg C).)
- 5.3 deg C
- 10.6 deg C
- 20.1 deg C
- 15 deg C
Correct answer: 10.6 deg C
Solution
At steady state, the heat current is the same in both rods. Let T be the junction temperature. For copper (0 deg C end): Q/t = K_Cu * A * (T - 0) / L = 390 * A * T / L. For steel (100 deg C end): Q/t = K_St * A * (100 - T) / L = 46 * A * (100 - T) / L. Setting equal: 390T = 46(100 - T) = 4600 - 46T. 390T + 46T = 4600. 436T = 4600. T = 4600/436 = 10.55 deg C, approximately 10.6 deg C.
Related JEE Advanced Physics questions
- Match the temperature of a blackbody listed in Group-I to the corresponding statement in Group-II, and select the correct answer.
[Given: Wien’s constant = 2.9 × 10⁻³ m-K and hc/e = 1.24 × 10⁻⁶ V-m]
Group-I:
(P) 2000 K
(Q) 3000 K
(R) 5000 K
(S) 10000 K
Group-II:
(1) The peak wavelength of emitted radiation can cause photoelectron ejection from a metal with a work function of 4 eV.
(2) The peak wavelength of emitted radiation falls within the visible spectrum.
(3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup.
(4) The energy radiated per unit area is one-sixteenth of that emitted by a blackbody at 6000 K.
(5) The peak wavelength of emitted radiation is suitable for imaging human bones.
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