(A) In steady state,the rate of heat flow through both rods is equal.Let $T$ be the temperature at the junction.$\left(\frac{kA(T_1 - T)}{l}\right)_{\

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(A) In steady state,the rate of heat flow through both rods is equal.Let $T$ be the temperature at the junction.$\left(\frac{kA(T_1 - T)}{l}\right)_{\text{steel}} = \left(\frac{kA(T - T_2)}{l}\right)_{\text{copper}}$Since the area $A$ is the same for both rods,it cancels out:$\frac{50(100 - T)}{0.1} = \frac{400(T - 0)}{0.2}$$500(100 - T) = 2000(T)$$100 - T = 4T$$5T = 100$$T = 20^{\circ}C$

Class 11 Physics 10-2.Heat Transfer Heat Conduction and Thermal Conductivity

Medium

$A$ cylindrical steel rod of length $0.10 \,m$ and thermal conductivity $50 \,Wm^{-1}K^{-1}$ is welded end to end to a copper rod of thermal conductivity $400 \,Wm^{-1}K^{-1}$ and of the same area of cross-section but $0.20 \,m$ long. The free end of the steel rod is maintained at $100^{\circ}C$ and that of the copper rod at $0^{\circ}C$. Assuming that the rods are perfectly insulated from the surroundings,the temperature at the junction of the two rods is ................... $^{\circ}C$.

KVPY 2012 All KVPY

A
$20$
B
$30$
C
$40$
D
$50$

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10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

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Three rods made of the same material and having the same cross-sectional area but different lengths $10\, cm, 20\, cm$,and $30\, cm$ are joined as shown. The temperature of the junction is ......... $^oC$.

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