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A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two end conduct an amount of heat ' $Q_1$ ' in time ' $t$ ' the metallic rod is melted and the material is formed into a rod of length four times the length of original rod. The amount of heat conducted by the new rod when placed in thermal contact with the same two reservoirs in time $t$ is ' $Q_2$ '. Then $\frac{Q_1}{Q_2}$ is
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$16$
The heat conduction rate is given by:
$\frac{Q}{t}=\frac{k A\left(T_1-T_2\right)}{l}$
Considering volume conservation:
$\begin{aligned} & A_1 l_1=A_2 l_2 \\ & A_1 l_1=A_2\left(4 l_1\right) \\ & \Rightarrow A_2=\frac{A_1}{4} \\ & \frac{Q_2}{t}=\frac{k\left(\frac{A_1}{4}\right)\left(T_1-T_2\right)}{4 l_1}=\frac{1}{16} \frac{k A_1\left(T_1-T_2\right)}{l_1}=\frac{1}{16} \frac{Q_1}{t} \\ & \Rightarrow Q_2=\frac{1}{16} Q_1\end{aligned}$
$\frac{Q}{t}=\frac{k A\left(T_1-T_2\right)}{l}$
Considering volume conservation:
$\begin{aligned} & A_1 l_1=A_2 l_2 \\ & A_1 l_1=A_2\left(4 l_1\right) \\ & \Rightarrow A_2=\frac{A_1}{4} \\ & \frac{Q_2}{t}=\frac{k\left(\frac{A_1}{4}\right)\left(T_1-T_2\right)}{4 l_1}=\frac{1}{16} \frac{k A_1\left(T_1-T_2\right)}{l_1}=\frac{1}{16} \frac{Q_1}{t} \\ & \Rightarrow Q_2=\frac{1}{16} Q_1\end{aligned}$
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