Two rods A and B of different materials but same cross section are joined as shown in figure- The free end of A is maintained at 100- C and the free end of B is maintained at 0- C- If l2 - 2l1- K1 - 2K2 and rods are thermally insulated from sides to prevent heat losses- then the temperature - of the junction of the two rods is

Consider the two sections like two resistances R_A & R_B. R_A=l_1k_1A nbsp; nbsp; nbsp;R_B=2l_1k_12A=4R_A nbsp; So θ= [ R_BR_A+R_B ][100 0] θ=80^∘ ...

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Byju's Answer

Standard XIII

Physics

Conduction Law

Two rods A an...

Question

Two rods $A$ and $B$ of different materials but same cross section are joined as shown in figure. The free end of $A$ is maintained at $100^{\circ} C$ and the free end of $B$ is maintained at $0^{\circ} C$ . If $l_{2} = 2 l_{1}, K_{1} = 2 K_{2}$ and rods are thermally insulated from sides to prevent heat losses, then the temperature $θ$ of the junction of the two rods is

80^{\circ} C

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60^{\circ} C

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40^{\circ} C

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20^{\circ} C

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Solution

The correct option is A $80^{\circ} C$
Consider the two sections like two resistances $R_{A} & R_{B}$ .
$R_{A} = \frac{l_{1}}{k_{1} A}$
$R_{B} = \frac{2 l_{1}}{\frac{k_{1}}{2} A} = 4 R_{A}$
So $θ = [\frac{R_{B}}{R_{A} + R_{B}}] [100 - 0]$
$θ = 80^{\circ} C$

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The correct option is A 80∘CConsider the two sections like two resistances RA & RB. RA=l1k1A RB=2l1k12A=4RA So θ=[RBRA+RB][100−0] θ=80∘C

The correct option is A $80^{\circ} C$
Consider the two sections like two resistances $R_{A} & R_{B}$ .
$R_{A} = \frac{l_{1}}{k_{1} A}$
$R_{B} = \frac{2 l_{1}}{\frac{k_{1}}{2} A} = 4 R_{A}$
So $θ = [\frac{R_{B}}{R_{A} + R_{B}}] [100 - 0]$
$θ = 80^{\circ} C$