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NEET UGPhysicsModerateNEET 2025

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is T1T_1 and that at the right junction is T2T_2. The ratio T1/T2T_1/T_2 is

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Sol. In series, Req.=R1+R2+R3R_{eq.} = R_1 + R_2 + R_3

=l2KA+lKA+l2KA= \frac{l}{2KA} + \frac{l}{KA} + \frac{l}{2KA}

=4l2KA= \frac{4l}{2KA}

Req.=2lKAR_{eq.} = \frac{2l}{KA}

In series rate of heat flow is same

∴3T−T1R1=3T−TReq.\therefore \frac{3T-T_1}{R_1} = \frac{3T-T}{R_{eq.}}

(3T−T1)2KAl=(2T)KA2l\frac{(3T-T_1)2KA}{l} = \frac{(2T)KA}{2l}

⇒6T−2T1=T\Rightarrow 6T-2T_1 = T

⇒2T1=5T\Rightarrow 2T_1 = 5T

⇒T1=5T2\Rightarrow T_1 = \frac{5T}{2} ...(1)

Now, equate heat flow rate in 3rd3^{rd} section & total section

T2−TR3=3T−TReq.\frac{T_2-T}{R_3} = \frac{3T-T}{R_{eq.}}

⇒(T2−T)(2KA)l=2T(KA)2l\Rightarrow \frac{(T_2-T)(2KA)}{l} = \frac{2T(KA)}{2l}

⇒2T2−2T=T\Rightarrow 2T_2-2T = T

⇒T2=3T2\Rightarrow T_2 = \frac{3T}{2} ...(2)

By equation (1) and equation (2)

T1T2=5T×22×3T=53\frac{T_1}{T_2} = \frac{5T \times 2}{2 \times 3T} = \frac{5}{3}

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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