A Cubical frame is made by connecting 12 identical uniform conducting rods as shown in the figure. In the steady state the temperature of junction A is $100^{0} C$ while that the G is $0^{0} C$ . Then,

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Temperature of F is $40^{0} C$

H will be Hotter than B

Temperature of E is $50^{0} C$

Temperature of D is ${66.67}^{0} C$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} A B : 100 - T_{1} = R \frac{Δ Q}{Δ t} - - - - (1) \\ (o r) 100 - T_{1} = R i \\ B C : T_{1} - T_{2} = R \frac{Δ Q}{2 Δ t} - - - - (2) \\ C A : T_{2} = R \frac{Δ Q}{Δ t} - - - - (3) \end{array}$

$\begin{array}{l} (2) \div (3) \\ \frac{T_{1} - T_{2}}{T_{2}} = \frac{\frac{R Δ Q}{2 Δ t}}{\frac{R Δ Q}{Δ t}} \\ \Rightarrow \frac{T_{1} - T_{2}}{T_{2}} = \frac{1}{2} \\ \Rightarrow 2 T_{1} - 2 T_{2} = T_{2} \\ \Rightarrow 2 T_{1} = 3 T_{2} \\ \Rightarrow T_{1} = \frac{3 T_{2}}{2} - - - - (4) \end{array}$

$\begin{array}{l} F r o m (1) a n d (3) \\ \Rightarrow 100 - T_{1} = T_{2} - - - (5) \\ f r o m (4) a n d (5) \\ \Rightarrow 100 - \frac{3 T_{2}}{2} = T_{2} \\ \Rightarrow 100 = \frac{5 T_{2}}{2} \\ \Rightarrow T_{2} = 40 ° C \\ s u b s t i t u t e i n (4) \\ \Rightarrow T_{1} = \frac{3 (40)}{2} = 60 ° C \end{array}$