Two rods A and B of same cross-sectional are A and length l connected in series between a source (T?

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10-2.Transmission of Heat

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Two rods $A$ and $B$ of same cross-sectional are $A$ and length $l$ connected in series between a source $(T_1 = 100^o C)$ and a sink $(T_2 = 0^o C)$ as shown in figure. The rod is laterally insulated If $T_A$ and $T_B$ are the temperature drops across the rod $A$ and $B$, then

A

$\frac{{{T_A}}}{{{T_B}}} = \frac{3}{1}$

B

$\frac{{{T_A}}}{{{T_B}}} = \frac{1}{3}$

C

$\frac{{{T_A}}}{{{T_B}}} = \frac{3}{4}$

D

$\frac{{{T_A}}}{{{T_B}}} = \frac{4}{3}$

Solution

The heat flowing through a rod is given by:

$\dot{ Q }= KA \frac{\Delta T }{1}$

Here the rods are in series, i.e. the rods have the same heat flowing through them.

$3 KA \frac{ T _{ A }}{1}= KA \frac{ T _{ B }}{1}$

i.e.$\frac{ T _{ A }}{ T _{ B }}=\frac{1}{3}$

Standard 11

Physics

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