A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and the of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is :

Heat and Thermodynamics PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q64 with Solution

View written solutionFree

Correct answer: B

Understand the heat flow direction

Heat flows along the length of the cylinder, not radially outward.

So the inner solid cylinder and the outer cylindrical shell conduct heat in parallel.
Use the formula for conduction along length

For a conductor of length $L$ , cross-sectional area $A$ , and thermal conductivity $K$ ,
$\frac{Q}{t} = \frac{K A \Delta T}{L}$
Since both parts have the same length $L$ and same temperature difference $\Delta T$ , the total heat current is
$\left(\frac{Q}{t}\right)_{\text{total}} = \frac{\Delta T}{L}(K_1 A_1 + K_2 A_2)$
If the whole composite cylinder is replaced by an equivalent homogeneous cylinder of conductivity $K_{\text{eff}}$ and total area $A_1 + A_2$ , then
$\left(\frac{Q}{t}\right)_{\text{total}} = \frac{K_{\text{eff}} (A_1 + A_2) \Delta T}{L}$
Hence,
$K_{\text{eff}} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$
Find the areas
- Inner cylinder radius = $R$
$A_1 = \pi R^2$
- Outer shell occupies radii from $R$ to $2R$
$A_2 = \pi (2R)^2 - \pi R^2 = 4\pi R^2 - \pi R^2 = 3\pi R^2$
Substitute into the effective conductivity formula
$K_{\text{eff}} = \frac{K_1(\pi R^2) + K_2(3\pi R^2)}{\pi R^2 + 3\pi R^2}$ $K_{\text{eff}} = \frac{\pi R^2(K_1 + 3K_2)}{4\pi R^2}$ $K_{\text{eff}} = \frac{K_1 + 3K_2}{4}$
Match with options
$\boxed{K_{\text{eff}} = \frac{K_1 + 3K_2}{4}}$
This corresponds to Option B.
Compare with stored correct answer

Stored correct answer: B

Derived answer: B

So they agree.

View written solutionFree

Correct answer: B

Understand the heat flow direction

Heat flows along the length of the cylinder, not radially outward.

So the inner solid cylinder and the outer cylindrical shell conduct heat in parallel.
Use the formula for conduction along length

For a conductor of length $L$ , cross-sectional area $A$ , and thermal conductivity $K$ ,
$\frac{Q}{t} = \frac{K A \Delta T}{L}$
Since both parts have the same length $L$ and same temperature difference $\Delta T$ , the total heat current is
$\left(\frac{Q}{t}\right)_{\text{total}} = \frac{\Delta T}{L}(K_1 A_1 + K_2 A_2)$
If the whole composite cylinder is replaced by an equivalent homogeneous cylinder of conductivity $K_{\text{eff}}$ and total area $A_1 + A_2$ , then
$\left(\frac{Q}{t}\right)_{\text{total}} = \frac{K_{\text{eff}} (A_1 + A_2) \Delta T}{L}$
Hence,
$K_{\text{eff}} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$
Find the areas
- Inner cylinder radius = $R$
$A_1 = \pi R^2$
- Outer shell occupies radii from $R$ to $2R$
$A_2 = \pi (2R)^2 - \pi R^2 = 4\pi R^2 - \pi R^2 = 3\pi R^2$
Substitute into the effective conductivity formula
$K_{\text{eff}} = \frac{K_1(\pi R^2) + K_2(3\pi R^2)}{\pi R^2 + 3\pi R^2}$ $K_{\text{eff}} = \frac{\pi R^2(K_1 + 3K_2)}{4\pi R^2}$ $K_{\text{eff}} = \frac{K_1 + 3K_2}{4}$
Match with options
$\boxed{K_{\text{eff}} = \frac{K_1 + 3K_2}{4}}$
This corresponds to Option B.
Compare with stored correct answer

Stored correct answer: B

Derived answer: B

So they agree.