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Q.

Two rods A and B of same length and radius are joined together end to end. The thermal conductivity of A and B are 2K and K. Under steady state conditions, if temperature difference between the open ends of A and B is 36⁰C, the temperature difference across 'A' is

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a

12⁰C

b

18⁰C

c

24⁰C

d

9⁰C

answer is A.

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Detailed Solution

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$\large \frac Qt=\frac {\Delta Q}{R_t}$ where Δθ = Temp. difference; R_t = Thermal resistance = l\kA
For the rod A, $\large \frac Qt=\frac {\theta_A}{(R_t)_A}=\frac {\theta_A}{l / 2kA}$
For the rod B, $\large \frac Qt=\frac {\theta_B}{(R_t)_B}=\frac {\theta_B}{l / kA}$
$\large \therefore\frac {\theta_A}{l / 2kA}=\frac {\theta_B}{l / kA}$
⇒ Δθ_B = 2.Δθ_A
NowΔθ_A + Δθ_B = 36⁰C
Δθ_A + 2.Δθ_A = 36⁰C ⇒ Δθ_A = 12⁰C

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