Four conducting rods are joined to make a square. All rods are identical and ends $A, B$ and $C$ are maintained at given temperatures. choose $INCORRECT$ statement for given arrangement in steady state. (value of $\frac {KA}{L}$ is $1\frac{J}{{{S^o}C}}$ , symbols , have their usual meaning)
AHeat current in $AB$ is equal to $1.5$ times of heat current in $BC$
BTemperature of end $D$ is $50\,^oC$
CHeat current in $AB$ is equal to heat current in $BC$
DHeat current withdrawn at end $B$ is $20\ J/S$
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CHeat current in $AB$ is equal to heat current in $BC$
c $\mathrm{i}_{\mathrm{AB}}=\frac{\mathrm{KA}}{\mathrm{L}}(100-40), \mathrm{i}_{\mathrm{BC}}=\frac{\mathrm{KA}}{\mathrm{L}}(40-0)$
$\mathrm{i}_{\mathrm{AD}}-\mathrm{i}_{\mathrm{DC}}$ so $\mathrm{T}_{\mathrm{D}}=\frac{\mathrm{T}_{\mathrm{A}}+\mathrm{T}_{\mathrm{C}}}{2}=\frac{\mathrm{1} 00+0}{2}=50^{\circ} \mathrm{C}$
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The length of the two rods made up of the same metal and having the same area of cross-section are $0.6 m$ and $0.8 m$ respectively. The temperature between the ends of first rod is ${90^o}C$ and ${60^o}C$ and that for the other rod is $150^oC$ and ${110^o}C$. For which rod the rate of conduction will be greater
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