From previous question,–

Now, its given bent part has k = 780 J s^–1m^–1 °C^–1 and straight part has k =390 J s^–1m^–1°C^–1for

Also Resistance equivalent circuit was as follows-

Now,

R_{AB = ,} R_{BC = ,} R_{CD =} R_{DE =,} R_EF= and R_{BE =}

After reducing the equivalent resistance across B and E, our circuit becomes

Where

R_{1 =} R_BC+R_CD + R_DE

₌

=

Since length is in cm and conductivity in meters, so multiply with 10^-2

R_{1 =}

Again, since R₁ and R_BE are in parallel, so total heat across R1 and R_BE will be same.

ie, q₁×R₁ = q₂ ×R_BE

⇒ =

⇒ =