Seven identical rods of material of thermal conductivity ‘K’ are connected as shown in figure. All the rods are indentical length $^{'} l^{'}$ and cross–sectional area $^{'} A^{'}$ . If the one end $^{'} A^{'}$ is kept at $100 ° C$ and the other end is kept at $0 ° C .$ The temperature of the junctions $C, D$ and E be in the steady state are $θ_{C}, θ_{E} a n d θ_{D}$ . Then,

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$θ_{C} > θ_{E} > θ_{D}$

$θ_{E} = 50 ° C & θ_{D} = 37.5 ° C$

$θ_{E} = 50 ° C, θ_{C} = 62.5 ° C & θ_{D} = 37.5 ° C$

$θ_{E} = 50 ° C, θ_{C} = 60 ° C & θ_{D} = 40 ° C$

answer is A, C.

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Detailed Solution

Use electric current methods,
$R_{1} = R_{2}, θ_{E} = 50 ° C$
$R_{5} = R_{5}, R_{3} = R_{4} + R_{5} = R_{7}$
$R_{4} + R_{6} = R_{7}$
$θ_{C} > θ_{E} > θ_{D}$
$\frac{K A (θ_{C} - 50)}{l} = \frac{K A}{l} (50 - θ_{D})$
$\frac{K A}{l} (100 - θ_{C}) = \frac{K A}{l} (θ_{C} - 50) + \frac{K A}{l} (θ_{C} - θ_{D})$
$= \frac{K A}{l} θ_{D}$
$θ_{C} + θ_{D} = 100$
$θ_{C} = 62.5 ° C$
$θ_{D} = 37.5 ° C$