JEE PHYSICS MCQ: THERMAL PROPERTIES OF MATTER

Option A (2.43) is incorrect. Option B (3.43) is incorrect. Option C (4.45) is incorrect. Option D (5.43) is incorrect. There appears to be an error in the provided solution. The correct answer should be 3.43. The provided solution calculates the equivalent resistance incorrectly, leading to an incorrect thermal conductivity. The correct approach involves using the formula for equivalent thermal conductivity for rods in series: $\frac{L}{K_{eq}A} = \frac{L_1}{K_1A} + \frac{L_2}{K_2A}$. Since $L_1 = L_2 = L/2$, we have $\frac{L}{K_{eq}A} = \frac{L/2}{3A} + \frac{L/2}{4A}$, which simplifies to $\frac{1}{K_{eq}} = \frac{1}{6} + \frac{1}{8} = \frac{4+3}{24} = \frac{7}{24}$. Thus, $K_{eq} = \frac{24}{7} \approx 3.43$.