JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2020
- Hard
- MCQ
Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity \(K _{1}, K _{2},\) and \(K _{3}\), respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at \(100^{\circ} C\) and the ther at \(0^{\circ} C\) (see figure). If the joints of the rod are at \(70^{\circ} C\) and \(20^{\circ} C\) in steady state and there is no loss of energy from the surface of the rod, the correct relationship between \(K _{1}, K _{2}\) and \(K _{3}\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(K _{1}: K _{3}=2: 3 ; K _{2}: K _{3}=2: 5\)
Step-by-step Solution
Detailed explanation
Rods are identical have same length ( \(\ell\) ) and area of cross-section \((A)\)
Combination are in series, so heat current is same for all Rods
\(\left(\frac{\Delta Q }{\Delta t }\right)_{ AB }=\left(\frac{\Delta Q }{\Delta t }\right)_{ BC }=\left(\frac{\Delta Q }{\Delta t }\right)_{ CD }=\) Heat current
\(\frac{(100-70) K _{1} A }{\ell}=\frac{(70-20) K _{2} A }{\ell}=\frac{(20-0) K _{3} A }{\ell}\)
\(30 K _{1}=50 K _{2}=20 K _{3}\)
\(3 K _{1}=2 K _{3}\)
\(\frac{K_{1}}{K_{3}}=\frac{2}{3}=2: 3\)
\(5 K _{2}=2 K _{3}\)
\(\frac{ K _{2}}{ K _{3}}=\frac{2}{5}=2: 5\)
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