1.

Two rods A and B of same cross-sectional area A and length L connected in series between a source(T1 = 100° C) and a sink(T2 = 0° C) as shown in figure. The rod is laterally insulated. The ratio of the thermal resistance of the rod is?​

Answer»

\blue{\bold{\underline{\underline{Answer:}}}}

\green{\therefore{\frac{R_{A}}{R_{B}}=\frac{1}{3}}}\\

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

\green{ \underline \bold{<klux>GIVEN</klux> : }} \\ : \implies T_{1} = 100^{ \circ} \: C \\ \\ : \implies T_{2} = 0^{ \circ} \: C \\ \\ : \implies L_{1} = L_{2} = L \\ \\ : \implies A_{1} = A_{2} = A \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \frac{ R_{A}}{ R_{B} } = ?

• ACCORDING to given QUESTION :

\bold{As \: we \: know \: that} \\ \circ \: \text{Thermal \: resistance (R) }= \frac{ \text{Length(l)}}{ \text{Thermal \: conductivity(k)} \times \text{Cross \: section \: area(A)}} \\ \\ : \implies R_{A} = \frac{ l_{1} }{ k_{1} \times A_{1} } \\ \\ : \implies R_{A} = \frac{l}{3k \times A} - - - - - (1) \\ \\ \bold{for \: R_{B} : } \\ : \implies R_{B} = \frac{ l_{2} }{ k_{2} \times A_{2} } \\ \\ : \implies R_{B} = \frac{l}{k \times A} - - - - - (2) \\ \\ \bold{Dividing \:(2) \: upon \: (1)} \\ : \implies \frac{ R_{A}}{R_{B} } = \frac{ \frac{l}{3k \times A} }{ \frac{l}{k \times A} } \\ \\ : \implies \frac{ R_{A}}{R_{B} } = \frac{l \times k \times A}{l \times 3k \times A} \\ \\ \green{ : \implies \frac{ R_{A}}{R_{B} } = \frac{1}{3} }


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