Question

Three rods of same length and area of cross-section but of thermal conductivities $K,2K$ and $3K$ are connected in series in the same order. The free end of the first rod is at $0^{\circ }C$ and the free end of the third rod is at ${55}^{\circ }C$. Instead state, the temperature difference across the middle rod is:-

• A. ${15}^{\circ }C$
• B. ${30}^{\circ }C$
• C. ${10}^{\circ }C$
• D. ${25}^{\circ }C$

Answer 1

The correct option is A ${15}^{\circ }C$

$\frac{1}{keq}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}$ [ $\because$ rods are in series ]

$\frac{1}{keq}=\frac{1}{k}+\frac{1}{2k}+\frac{1}{3k}$

$\frac{1}{keq}=\frac{6+3+2}{6k}$

$keq=\frac{6k}{11}$

So, $Q=\frac{A\left(\frac{6k}{11}\right)(55-0)}{2}$

$=3\times \frac{2}{11}\times 55\frac{Ak}{l}=O\frac{Ak}{l}$

Let the temp difference across $2k$ rod be $△T$, then

$Q=\frac{A\left(2K\right)}{l}△T$

$\Rightarrow 30\frac{Ak}{l}=2\frac{Ak}{l}△5$

$\Rightarrow △T={15}^{o}C$

Option $-A$ is correct.