JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2022
- Medium
- Numerical
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 ........... \(^{\circ} C\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(45\)
Step-by-step Solution
Detailed explanation
\(\frac{d \theta}{d t}=\frac{K_{1} A_{1}}{l_{1}}\left(T_{1}-T\right)=\frac{K_{2} A_{2}}{l_{2}}\left( T - T _{2}\right)\)
\(\frac{450- T }{ T -0}=\frac{K_{2} A _{2} 1_{1}}{K_{1} A _{1} 1_{2}}=9 \times \frac{1}{2} \times 2\)
\(450- T =9 T \Rightarrow T =45^{\circ}\,C\)
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