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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)
Heat current in $AB$ is equal to $1.5$ times of heat current in $BC$
Temperature of end $D$ is $50\,^oC$
Heat current in $AB$ is equal to heat current in $BC$
Heat current withdrawn at end $B$ is $20\ J/S$
Solution
$\mathrm{i}_{\mathrm{AB}}=\frac{\mathrm{KA}}{\mathrm{L}}(100-40), \mathrm{i}_{\mathrm{BC}}=\frac{\mathrm{KA}}{\mathrm{L}}(40-0)$
$\mathrm{i}_{\mathrm{AD}}-\mathrm{i}_{\mathrm{DC}}$ so $\mathrm{T}_{\mathrm{D}}=\frac{\mathrm{T}_{\mathrm{A}}+\mathrm{T}_{\mathrm{C}}}{2}=\frac{\mathrm{1} 00+0}{2}=50^{\circ} \mathrm{C}$
