Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1 = 300 K and T2 100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K1/K2 =.

Question

Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1 = 300 K and T2 100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K1/K2 =. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/10da8cf934fd46066934b6504e8e9f33-.png" /> We have in steady state, (((200-300/L)/k1 π r2))+(((200-100/L)/k2 π(2 r)2)) ⇒ (k1 π r2 × 100/L)=(100 k2 π × 4 r2/L) ⇒ (k1/k2)=4

We have in steady state, (k1​πr2L200−300​​)+(k2​π(2r)2L200−100​​) ⇒Lk1​πr2×100​=L100k2​π×4r2​ ⇒k2​k1​​=4

We have in steady state,
$(\frac{\frac{200 - 300}{L}}{k _{1} π r ^{2}}) + (\frac{\frac{200 - 100}{L}}{k _{2} π ( 2 r ) ^{2}})$
$\Rightarrow \frac{k _{1} π r ^{2} \times 100}{L} = \frac{100 k _{2} π \times 4 r ^{2}}{L}$
$\Rightarrow \frac{k _{1}}{k _{2}} = 4$