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Two rods A and B of equal cross-sectional area and equal length L are joined in series between a hot source at T1 = 100 deg C and a cold sink at T2 = 0 deg C. The lateral surfaces are insulated, and rod A has three times the thermal conductivity of rod B (K_A = 3*K_B). Find the ratio of their thermal resistances, R_A / R_B.
- R_A / R_B = 1/3
- R_A / R_B = 3
- R_A / R_B = 3/4
- R_A / R_B = 4/3
Correct answer: R_A / R_B = 1/3
Solution
Thermal resistance is R = L/(K*A). Since rods A and B have the same length L and the same area A, the ratio of resistances depends only on conductivities: R_A/R_B = (L/(K_A*A)) / (L/(K_B*A)) = K_B/K_A. With K_A = 3*K_B this gives R_A/R_B = K_B/(3*K_B) = 1/3. The better conductor (A) has the smaller thermal resistance.
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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- A continuous-flow calorimeter is used in two separate experiments to find the specific heat of a liquid. In the first trial, supplying 60 W raises the liquid temperature by 10 K. In the second trial, the power is doubled to 120 W, but the same 10 K rise is maintained by tripling the flow rate. Assuming heat loss to surroundings is the same in both trials, what is the power lost to the surroundings?
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