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Thermal Properties of Matter

JEE Main level, numerical answer, medium

98 of 126

Question 98 of 126

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 $200^{\circ}C,100^{\circ}C$ and $125^{\circ}C$ respectively. The heat current in $CD$ is $PW$ . The value of $P$ is

Show answer and solution

Answer: 2

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

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

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

725=5t

t=145ºC

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

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JEE Main 2025 Apr 2 Shift 1 question paper

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