Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

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

Answers (1)

best_answer


Let 'T' be the temperature at point C

H_{3}= P= \frac{\left ( T-125 \right )}{R_{3}}
P= \frac{\left ( T-125 \right )}{10}\rightarrow \left ( 1 \right )
From Kirchoff's current law (analogy) At junction C, H_{1}+H_{2}+H_{3}= 0
\because R_{AB}= R_{CD}\: \left ( Identical \right )
R_{2}= R_{1}= \frac{R_{3}}{2}
\frac{\left ( T-100 \right )}{R_{1}}+\frac{\left ( T-200 \right )}{R_{2}}+\frac{\left ( T-125 \right )}{R_{3}}= 0
\frac{\left ( T-100 \right )}{\frac{R_{3}}{2}}+\left ( \frac{ T-200 }{\frac{R_{3}}{2}} \right )+\left ( \frac{ T-125 }{R_{3}} \right )= 0
2T-200+2T-400+T-125= 0
5T= 725
T= 145^{\circ}C
From eqn (1)
P= \left ( \frac{145-125}{10} \right )= 2

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question