JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 23)
Three conductors of same length having thermal conductivity $k_1, k_2$ and $k_3$ are connected as shown in figure.
_22nd_January_Morning_Shift_en_23_1.png)
Area of cross sections of $1^{\text {st }}$ and $2^{\text {nd }}$ conductor are same and for $3^{\text {rd }}$ conductor it is double of the $1^{\text {st }}$ conductor. The temperatures are given in the figure. In steady state condition, the value of $\theta$ is _________ ${ }^{\circ} \mathrm{C}$. (Given : $\mathrm{k}_1=60 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}, \mathrm{k}_2=120 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}, \mathrm{k}_3=135 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ )
Explanation
_22nd_January_Morning_Shift_en_23_2.png)
$${R_1} = {{2L} \over {{K_1}A}}$$
(As $${l_1} = {l_2} = {l_3} = l$$
$${A_1} = {A_2} = A/2$$
$${A_3} = A$$)
$${R_2} = {{2L} \over {{K_2}A}}$$
$${R_3} = {L \over {{K_3}A}}$$
$${I_1} + {I_2} = {I_3}$$
$$ \Rightarrow {{100 - \theta } \over {{{2L} \over {{K_1}A}}}} + {{100 - \theta } \over {{{2L} \over {{K_2}A}}}} = {{\theta - 0} \over {{L \over {{K_3}A}}}}$$
$$ \Rightarrow {{100 - \theta } \over 2}({K_1} + {K_2}) = {K_3}\theta $$
$$ \Rightarrow 50({K_1} + {K_2}) = \theta \left[ {{K_3} + {{{K_1} + {K_2}} \over 2}} \right]$$
$$ \Rightarrow 50(60 + 120) = \theta \left[ {135 + {{60 + 120} \over 2}} \right]$$
$$ \Rightarrow 50 \times 80 = \theta (135 + 90)$$
$$ \Rightarrow \theta = {{50 \times 180} \over {225}}$$
$$ \Rightarrow \theta = 40^\circ C$$
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