Question

Consider two rods of the same length and different specific heats (S1, S2), conductivities (K1, K2), and area of cross-sections (A1, A2) and both having temperature T1 and T2 at their ends. If the rate of loss of heat due to conduction is equal, then

• A. K1A1 = K2A2
• B. $\frac{K_1A_1}{S_1}$ = $\frac{K_2A_2}{S_2}$
• C. K2A1 = K1A2
• D. $\frac{K_2A_1}{S_2}$ = $\frac{K_1A_2}{S_1}$

Answer 1

Answer: (A)

K1A1 = K2A2

Answer 2

The correct option is (A) : K1A1 = K2A2

Use $\frac{dQ}{dt}=\frac{KA}{L}(T_1-T_2)$