JEE Main 27-Aug-2021 Shift 1 Solved Paper

Given, thermal resistance of rod CD,
RCD=10.0kW−1 Temperature of end A,TA=200°C
Temperature of end B,TB=100°C
Temperature of end D,TD=125°C
As rods are given as identical, then
RCD=RCB=10kW−1
As, C is mid-point of AB, and as we know
R ∝ length
RAC=RBC=5kW−1
Now, we know that heat current is given as

Q

t

=

ΔT

R

Here, ∆T is temperature difference and R is thermal resistance.
Now, for point C, consider temperature of point C as T.
Consider at point C heat is coming in from A and going out at point B and D.
Therefore, we can write,

TA−T

RAC

=

T−TD

RCD

=

T−TB

RBC

⇒

200−T

RAC

=

T−125

RCD

+

T−100

RBC

⇒

200−T

5

=

T−125

10

,

T−100

5

⇒400−2T=T−125+2T−200
5T=725°C
T=145°C
The heat current in CD is
ICD=

T−TD

RCD

=

145−125

10

=

20

10

=2W
Thus, the value of P is 2.