If K_{1} and K_{2} are thermal conductivities L_{1} and L _{2} are lengths and A _{1} and A _{2} are

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10-2.Transmission of Heat

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If $K_{1}$ and $K_{2}$ are the thermal conductivities $L_{1}$ and $L _{2}$ are the lengths and $A _{1}$ and $A _{2}$ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_{2}}{K_{1}}=9, \frac{A_{1}}{A_{2}}=2, \frac{L_{1}}{L_{2}}=2$.

Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steel - copper junction in the steady state will be ........... $^{\circ} C$

A

$18$

B

$14$

C

$45$

D

$150$

(JEE MAIN-2022)

Solution

$\frac{d \theta}{d t}=\frac{K_{1} A_{1}}{l_{1}}\left(T_{1}-T\right)=\frac{K_{2} A_{2}}{l_{2}}\left( T – T _{2}\right)$

$\frac{450- T }{ T -0}=\frac{K_{2} A _{2} 1_{1}}{K_{1} A _{1} 1_{2}}=9 \times \frac{1}{2} \times 2$

$450- T =9 T \Rightarrow T =45^{\circ}\,C$

Standard 11

Physics

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