First slide

Conduction

Question

Two rods of same length and areas of cross-section A₁ and A₂ have their ends maintained at same temperature. If K₁ and K₂ are their thermal conductivities, C₁ and C₂are the specific heats of their material and $ρ_{1}, ρ_{2}$ their densities, then for the rate of flow by conduction through them to be equal:

Moderate

A

$\frac{A_{1}}{A_{2}} = \frac{K_{1}}{c_{1} ρ_{1}} \frac{c_{2} ρ_{2}}{K_{2}}$
B

$\frac{A_{1}}{A_{2}} = \frac{K_{1}}{K_{2}}$
C

$\frac{A_{1}}{A_{2}} = \frac{K_{1} c_{1} ρ_{1}}{K_{2} c_{2} ρ_{2}}$
D

$\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$

Solution

$\frac{∆ Q}{∆ t} = \frac{K_{1} A_{1} (T_{1} - T_{2})}{l} = \frac{K_{2} A_{2} (T_{1} - T_{2})}{l} \Rightarrow \frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$

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