Question

Consider two rods 1 and 2 of same length. They have different specific heats \((C_1,C_2)\), thermal conductivities \((K_1,K_2)\) and area of cross-section \((A_1,A_2)\) respectively. Both the rods have temperatures \((T_1,T_2)\) at their ends. If their rate of loss of heat due to conduction is equal, then

• A. \(A_1K_2 = A_2K_1\)
• B. \(A_1K_1 = A_2K_2\)
• C. \(\frac{A_1K_1}{C_1} = \frac{A_2K_2}{C_2}\)
• D. \(\frac{A_1K_2}{C_1} = \frac{A_2K_1}{C_2}\)

Hint:

Rate of conduction is \(\dfrac{KA\,\Delta T}{L}\); specific heat plays no role.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The rate of heat flow through a rod is \(\dfrac{Q}{t}=\dfrac{KA(T_1-T_2)}{L}\).

Step 2: Key Formula or Approach
Both rods have the same length and the same end temperatures, so \(L\) and \(\Delta T\) cancel.

Step 3: Detailed Explanation
Equal rates give \(K_1A_1=K_2A_2\).
The specific heat does not appear in the steady conduction rate, so it does not enter the relation.

Final Answer:
The relation is \(A_1K_1=A_2K_2\), option (B). \[ \boxed{A_1K_1=A_2K_2\ \text{(B)}} \]