A rod CD of thermal resistance 10.0 KW ⁻¹ is joined at the middle of an identical rod AB as shown in figure. The end A, B and D are maintained at 200 ° C, 100 ° C and 125 ° C respectively. The heat current in CD is P watt. The value of P is.................. Numerical answer type. Work it out, then open the answer.

Answer: 2. C is the middle of AB, so RAC = RCB = 5 K/W and RCD = 10 K/W. Heat balance at C: (200-T)/(5) + (100-T)/(5) + (125-T)/(10) = 0 400 - 2T + 200 - 2T + 125 - T = 0 ⇒ T = 145° C P = (145-125)/(10) = 2 W

A rod CD of thermal resistance 10.0 KW ⁻¹ is joined at the middle of… | JEE Main 2021 (27 Aug, Shift 1)

Q40JEE Main 2021 · 27 AUG · SHIFT 1Medium

Heat Conduction

A rod CD of thermal resistance 10.0 KW ${}^{-1}$ is joined at the middle of an identical rod AB as shown in figure. The end A, B and D are maintained at 200 $^\circ$ C, 100 $^\circ$ C and 125 $^\circ$ C respectively. The heat current in CD is P watt. The value of P is ..................

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

C is the middle of AB, so $R_{AC} = R_{CB} = 5$ K/W and $R_{CD} = 10$ K/W.
Heat balance at C: $\dfrac{200-T}{5} + \dfrac{100-T}{5} + \dfrac{125-T}{10} = 0$
$400 - 2T + 200 - 2T + 125 - T = 0 \Rightarrow T = 145^\circ$ C
$P = \dfrac{145-125}{10} = 2$ W

Report a mistake

Heat Conduction question from JEE Main 2021 (27 Aug, Shift 1)

More questions on Heat Conduction