Three rods each of length \(l\) and cross sectional area \(A\) joined in series between two heat reservoirs…
- \(\frac{600}{7}{ }^{\circ} \mathrm{C}, \frac{400}{7}{ }^{\circ} \mathrm{C}\)
- \(\frac{600}{7}{ }^{\circ} \mathrm{C}, \frac{700}{4}{ }^{\circ} \mathrm{C}\)
- \(\frac{500}{6}{ }^{\circ} \mathrm{C}, \frac{600}{5}{ }^{\circ} \mathrm{C}\)
- \(\frac{600}{4}{ }^{\circ} \mathrm{C}, \frac{400}{7}{ }^{\circ} \mathrm{C}\)
Solution
where, \(T_1, T_2=\) temperature of junctions, \(I=\text { heat current }=\frac{d Q}{d t}\) As, \(\frac{d Q_1}{d t}=\frac{d Q_2}{d t}=\frac{d Q_3}{d t}=\frac{T_i-T_f}{k_{\text {eq }}}\) So, \(\frac{k_1 A_1\left(T_i-T_1\right)}{l_1}=\frac{k_2 A_2\left(T_1-T_2\right)}{l_2}=\frac{k_3 A_3\left(T_2-T_f\right)}{l_3}\) \(\begin{array}{rlrl} & \ddots & A_1 & =A_2=A_3 \\ & \text {and } & l_1 & =l_2=l_3 \\ \Rightarrow & 2 k\left(100-T_1\right) & =k\left(T_1-T_2\right)=0.5\left(T_2-0\right) \ldots (i) \end{array}\) Equivalent coefficient of thermal conductivity, \(\begin{aligned} & \frac{1}{k_{\mathrm{eq}}}=\frac{1}{2 k}+\frac{1}{k}+\frac{2}{k} \\ & \Rightarrow k_{\text {eq }}=\frac{2 k}{7} \\ & \because \text { Heat current, } \frac{d Q}{d t}=\frac{d Q_1}{d t} \\ & \Rightarrow \frac{k_{e q} A\left(T_i-T_f\right)}{l}=\frac{k_1 A_1\left(T_i-T_1\right)}{l_1} \end{aligned}\) here, \(l=l_1+l_2+l_3=l_1+l_1+l_1=3 l_1\) and \(A=A_1+A_2+A_3=A_1+A_1+A_1=3 A_1\) So, \(\frac{2 k}{7} \times \frac{3 A_1}{3 l_1}(100-0)=\frac{2 k A_1\left(100-T_1\right)}{l_1}\) or \(\frac{1}{7} \times 100=100-T_1\) \(\begin{array}{ll} \Rightarrow & T_1=100-\frac{100}{7} \\ \Rightarrow & T_1=\frac{600^{\circ} \mathrm{C}}{7} \end{array}\) Similarly, From Eq. (i), we get \(\begin{aligned} 2 k\left(100-T_1\right) & =0.5 k T_2 \\ \Rightarrow 200-2 \times \frac{600}{7} & =0.5 T_2 \\ \Rightarrow T_2 & =\frac{400^{\circ}}{7} \mathrm{C} \end{aligned}\) Hence, the correct option is (a).
Asked in: AP EAMCET 2019 (23 Apr Shift 1)
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