JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 24)
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 ................. .
_27th_August_Morning_Shift_en_24_1.png)
Answer
2
Explanation
Rods are identical so
RAB = RCD = 10 Kw$$-$$1
C is mid-point of AB, so
RAC = RCB = 5 Kw$$-$$1
at point C
$${{200 - T} \over 5} = {{T - 125} \over {10}} + {{T - 100} \over 5}$$
2(200 $$-$$ T) = T $$-$$ 125 + 2(T $$-$$ 100)
400 $$-$$ 2T = T $$-$$ 125 + 2T $$-$$ 200
$$T = {{725} \over 5}$$ = 145$$^\circ$$C
$${I_h} = {{145 - 125} \over {10}}w = {{20} \over {10}}w$$
Ih = 2w
_27th_August_Morning_Shift_en_24_2.png)
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