JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2021 · 27 August · Morning Shift · Physics · Thermal Properties of Matter

A rod CD of thermal resistance 10 . 0 KW ^ - 1 is joined at the...

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 ends $A, B$ and $D$ are maintained at $200 ° C, 100 ° C$ and $125 ° C$ respectively. The heat current in $CD$ is $PW$ . The value of $P$ is

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

Correct answer

Step-by-step solution

${Q}_{AC} + {Q}_{DC} = {Q}_{CB}$

$\dfrac{KA ( 200 - t )}{\dfrac{ℓ}{2}} + \dfrac{KA ( 125 - t )}{ℓ} = \dfrac{( t - 100 ) KA}{\dfrac{ℓ}{2}}$

$400 - 2t + 125 - t = 2t - 200$

$725 = 5t$

$t = 145 º C$

${Q}_{CD} = \dfrac{KA ( 145 - 125 )}{ℓ} = \dfrac{20}{\dfrac{ℓ}{KA}} = 2W$

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