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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
$K _{1}: K _{3}=2: 3 ; K _{2}: K _{3}=2: 5$
$K _{1}< K _{2}< K _{3}$
$K _{1}: K _{2}=5: 2 ; K _{1}: K _{3}=3: 5$
$K _{1}> K _{2}> K _{3}$
Solution
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
$\frac{(100-70) K _{1} A }{\ell}=\frac{(70-20) K _{2} A }{\ell}=\frac{(20-0) K _{3} A }{\ell}$
$30 K _{1}=50 K _{2}=20 K _{3}$
$3 K _{1}=2 K _{3}$
$\frac{K_{1}}{K_{3}}=\frac{2}{3}=2: 3$
$5 K _{2}=2 K _{3}$
$\frac{ K _{2}}{ K _{3}}=\frac{2}{5}=2: 5$
