Question

Six identical conducting rods are joined as shown in figure. Points A and D are maintained at temperatures 200⁰C and 20⁰C respectively. The temperature of junction B will be –

• A. 120⁰C
• B. 100⁰C
• C. 140⁰ C
• D. 80⁰C

Answer 1

Answer: (C)

140⁰ C

Answer 2

$\frac{200–T_{B}}{R}$ = $\frac{T_{C}–20}{R}$

200 – T_B = T_C – 20

220 = T_B + T_C

T_B + T_C = 220 ….(i)

$\frac{\tau}{2}$ = $\frac{T_{B}–T_{C}}{2R}$

$\frac{200–T_{B}}{R}$ = $\frac{T_{B}–T_{C}}{R}$

200 – T_B = T_B – T_C

200 – T_B = T_B – (220 – T_B)

420 = 3 T_B

T_B = 140 ⁰ C