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Heat and Thermodynamics question

2025 · 7 Apr · Shift 2 · Q71
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Heat and Thermodynamics question

2025 · 7 Apr · Shift 2 · Q71

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivites of these rods are: LALB=12,rArB=2\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}=\frac{1}{2}, \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=2LB​LA​​=21​,rB​rA​​=2 and KAKB=12\frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}=\frac{1}{2}KB​KA​​=21​. The free ends of rods A and B are maintained at 400 K , 200 K , respectively. The temperature of rods interface is ‾K\underline{\hspace{2cm}}K​K, when equilibrium is established.
Numerical answer
View written solutionFree

Correct answer: 360

  1. Use steady-state heat conduction

When two rods are joined in series and steady state is established, the heat current through both rods is the same:

KAAA(400−T)LA=KBAB(T−200)LB\frac{K_A A_A (400 - T)}{L_A} = \frac{K_B A_B (T - 200)}{L_B}LA​KA​AA​(400−T)​=LB​KB​AB​(T−200)​

where TTT is the interface temperature.

  1. Write area ratio using radii

For cylindrical rods,

A=πr2A = \pi r^2A=πr2

Given:

rArB=2  ⟹  AAAB=(rArB)2=4\frac{r_A}{r_B} = 2 \implies \frac{A_A}{A_B} = \left(\frac{r_A}{r_B}\right)^2 = 4rB​rA​​=2⟹AB​AA​​=(rB​rA​​)2=4
  1. Use given ratios

Given:

LALB=12,KAKB=12\frac{L_A}{L_B} = \frac{1}{2}, \qquad \frac{K_A}{K_B} = \frac{1}{2}LB​LA​​=21​,KB​KA​​=21​

Now,

KAAA/LAKBAB/LB=KAKB⋅AAAB⋅LBLA\frac{K_A A_A / L_A}{K_B A_B / L_B} = \frac{K_A}{K_B} \cdot \frac{A_A}{A_B} \cdot \frac{L_B}{L_A}KB​AB​/LB​KA​AA​/LA​​=KB​KA​​⋅AB​AA​​⋅LA​LB​​

Substitute values:

=12⋅4⋅2=4= \frac{1}{2} \cdot 4 \cdot 2 = 4=21​⋅4⋅2=4

So,

KAAALA=4⋅KBABLB\frac{K_A A_A}{L_A} = 4 \cdot \frac{K_B A_B}{L_B}LA​KA​AA​​=4⋅LB​KB​AB​​
  1. Apply heat current equality

Thus,

4(400−T)=T−2004(400 - T) = T - 2004(400−T)=T−200

Solve:

1600−4T=T−2001600 - 4T = T - 2001600−4T=T−200 1800=5T1800 = 5T1800=5T T=360 KT = 360\,\text{K}T=360K
  1. Final answer

The interface temperature is:

360\boxed{360}360​
  1. Comparison with stored answer

Stored correct answer = 360360360

Our derived answer matches the stored answer.

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