Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity $\mathrm{K}_{1}, \mathrm{~K}_{2}$, and $\mathrm{K}_{3}$, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at $100^{\circ} \mathrm{C}$ and the other at $0^{\circ} \mathrm{C}$ (see figure). If the joints of the rod are at $70^{\circ} \mathrm{C}$ and $20^{\circ} \mathrm{C}$ in steady state and there is no loss of energy from the surface of the rod, the correct relationship between $\mathrm{K}_{1}, \mathrm{~K}_{2}$ and $\mathrm{K}_{3}$ is :
Correct Option: 1
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
$\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{AB}}=\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{BC}}=\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{CD}}=$ Heat current
$\frac{(100-70) \mathrm{K}_{1} \mathrm{~A}}{\ell}=\frac{(70-20) \mathrm{K}_{2} \mathrm{~A}}{\ell}=\frac{(20-0) \mathrm{K}_{3} \mathrm{~A}}{\ell}$
$30 \mathrm{~K}_{1}=50 \mathrm{~K}_{2}=20 \mathrm{~K}_{3}$
$3 \mathrm{~K}_{1}=2 \mathrm{~K}_{3}$
$\frac{\mathrm{K}_{1}}{\mathrm{~K}_{3}}=\frac{2}{3}=2: 3$
$5 \mathrm{~K}_{2}=2 \mathrm{~K}_{3}$
$\frac{\mathrm{K}_{2}}{\mathrm{~K}_{3}}=\frac{2}{5}=2: 5$
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