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Standard XII

Chemistry

Applications of Colloids

Two rods of s...

Question

Two rods of same length have their ends maintained at same temperature. If $s_{1}$ and $s_{2}$ are the cross-sectional area, $K_{1}$ and $K_{2}$ their thermal conductivities, $C_{1}$ and $C_{2}$ their heat capacities, $ρ_{1}$ and $ρ_{2}$ their densities the condition for the rate of flow of heat to be the same in the rods will be

K_{1} s_{1} = K_{2} s_{2}

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\frac{K_{1}}{s_{1}} = \frac{K_{2}}{s_{2}}

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\frac{K_{1}}{K_{2}} = \frac{s_{1} C_{1} ρ_{1}}{s_{2} C_{2} ρ_{2}}

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\frac{s_{1} K_{1}}{s_{2} K_{2}} = \frac{C_{2} ρ_{2}}{C_{1} ρ_{1}}

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Solution

The correct option is A $K_{1} s_{1} = K_{2} s_{2}$
Rate of flow of heat is $K A \frac{d θ}{d x} = K s \frac{d θ}{d x}$ as $A = s$ given
As both the rods are of same length and one end of the rods is at same temp, $\frac{d θ}{d x}$ will be same for both.
Rate of flow of heat is same for both the rods as given.
Hence, $K_{1} s_{1} \frac{d θ}{d x} = K_{2} s_{2} \frac{d θ}{d x}$
$\Rightarrow K_{1} s_{1} = K_{2} s_{2}$

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