Question

A block of metal A is connected in series with another block of metal B such that the two metal blocks have the same area of cross sections. The thermal conductivity of metal A is \(K\) and the free end of metal A is at \(80^\circ C\). The temperature of the interface is \(60^\circ C\) and the free end of metal B is at \(20^\circ C\). Assuming the two metals have the same thickness, the conductivity of metal B is:

• A. \(2K\)
• B. \(4K\)
• C. \(\frac{K}{2}\)
• D. \(\frac{K}{4}\)

Hint:

In steady heat flow, heat passing per second is same through all sections.

Answer 1

The Correct Option is C

Step 1: Heat flow condition.
In steady state, heat flow through both rods is same.

Step 2: Heat conduction formula.
\[ Q = \frac{KA\Delta T}{L} \]

Step 3: Apply for both metals.
\[ \frac{K(80-60)}{L} = \frac{K_B(60-20)}{L} \]

Step 4: Simplify.
\[ K \times 20 = K_B \times 40 \]

Step 5: Solve.
\[ K_B = \frac{K}{2} \]

Step 6: Final conclusion.
\[ \boxed{\frac{K}{2}} \] Hence, correct answer is option (C).