A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2 R . The thermal conductivity of the material of the inner cylinder is $\mathrm{K}_{1}$ and that of the outer cylinder is $\mathrm{K}_{2}$. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:

(D) $\frac{\mathrm{K}_{1}+3 \mathrm{~K}_{2}}{4}$

(A) $\frac{\mathrm{K}_{1}+\mathrm{K}_{2}}{2}$

(B) $\mathrm{K}_{1}+\mathrm{K}_{2}$

(C) $\frac{2 \mathrm{~K}_{1}+3 \mathrm{~K}_{2}}{5}$

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Heat & Thermodynamics

A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2 R . The thermal conductivity of the material of the inner cylinder is

K_{1}

and that of the outer cylinder is

K_{2}

. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:

(A) $\frac{K _{1} + K _{2}}{2}$
(B) $K_{1} + K_{2}$
(C) $\frac{2 K _{1} + 3 K _{2}}{5}$
(D) $\frac{K _{1} + 3 K _{2}}{4}$ ✓ Correct

Answer: (D)

Solution

Heat flows along the length of the cylinder, so the inner core and the outer shell behave like two conductors carrying heat side by side, a parallel arrangement in which the conductances add. The effective conductance of the whole cross-section equals the sum of the two separate conductances,

k_{eff} A = k_{1} A_{1} + k_{2} A_{2}

. The core has area

π R^{2}

and the shell has area

π ((2 R)^{2} - R^{2}) = 3 π R^{2}

, while the total area is

π (2 R)^{2} = 4 π R^{2}

. Substituting,

k_{eff} (4 π R^{2}) = k_{1} (π R^{2}) + k_{2} (3 π R^{2})

, and cancelling

π R^{2}

gives

4 k_{eff} = k_{1} + 3 k_{2}

, so

k_{eff} = \frac{k _{1} + 3 k _{2}}{4}

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Heat & Thermodynamics

Solution

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