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NEET 2025
Exam
NEET
Year
2025
Subject
Physics
Chapter
Thermal Properties of Matter

NEET 2025 · Physics · Thermal Properties of Matter

Three identical heat conducting rods are connected in series as...

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K2 K while that in the middle has thermal conductivity KK. The left end of the combination is maintained at temperature 3T3 T and the right end at TT. 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 Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal

Options

  1. A32\frac{3}{2}
  2. B43\frac{4}{3}
  3. C53\frac{5}{3}
  4. D54\frac{5}{4}

Correct answer

C. 53\frac{5}{3}

Step-by-step solution

 In  steady state ⇒(dQdt)1=(dQdt)2=(dQdt)32KA(3T−T1)L=KA(T1−T2)L=2KA(T2−T)L6T−2T1=T1−T2=2T2−2T6T−2T1=T1−T26T+T2=3T1....(i)T1−T2=2T2−2T2T−3T2=−T1.....(ii)\begin{aligned} & \text{\text{ In } steady state } \Rightarrow\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_2=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_3 \\ & \frac{2 \mathrm{KA}\left(3 \mathrm{ T}-\mathrm{T}_1\right)}{\mathrm{L}}=\frac{\mathrm{KA}\left(\mathrm{ T}_1-\mathrm{T}_2\right)}{\mathrm{L}}=\frac{2 \mathrm{KA}\left(\mathrm{ T}_2-\mathrm{T}\right)}{\mathrm{L}} \\ & 6 \mathrm{ T}-2 \mathrm{ T}_1=\mathrm{T}_1-\mathrm{T}_2=2 \mathrm{ T}_2-2 \mathrm{ T} \\ & 6 \mathrm{ T}-2 \mathrm{ T}_1=\mathrm{T}_1-\mathrm{T}_2 \\ & 6 \mathrm{ T}+\mathrm{T}_2=3 \mathrm{ T}_1 ....(i)\\ & \mathrm{ T}_1-\mathrm{T}_2=2 \mathrm{ T}_2-2 \mathrm{ T} \\ & 2 \mathrm{ T}-3 \mathrm{ T}_2=-\mathrm{T}_1.....(ii) \end{aligned} Multiply equation (2) By ' 3 ' 6T−9T2=−3T1....(iii)Equation(1)−(3)6 \mathrm{ T}-9 \mathrm{ T}_2=-3 \mathrm{ T}_1 ....(iii) Equation (1) - (3)10T2=6T1T1T2=53\begin{aligned} & 10 \mathrm{ T}_2=6 \mathrm{ T}_1 \\ & \frac{\mathrm{ T}_1}{\mathrm{ T}_2}=\frac{5}{3} \end{aligned}$$
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