A cylindrical metallic rod in thermal contact | vedclass

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Question

(C) The rate of heat conduction through a rod is given by $\frac{Q}{t} = \frac{kA(T_1 - T_2)}{l}$,where $A = \pi r^2$ is the cross-sectional area and

Vedclass · Accepted Answer

$\frac{Q}{16}$

Vedclass · Answer

(C) The rate of heat conduction through a rod is given by $\frac{Q}{t} = \frac{kA(T_1 - T_2)}{l}$,where $A = \pi r^2$ is the cross-sectional area and $l$ is the length.When the rod is melted and reshaped,the volume $V = A \cdot l = \pi r^2 l$ remains constant.Let the new radius be $r' = \frac{r}{2}$.Since $V = \pi (r')^2 l' = \pi (\frac{r}{2})^2 l' = \pi \frac{r^2}{4} l'$,we have $\pi r^2 l = \pi \frac{r^2}{4} l'$,which gives the new length $l' = 4l$.The new area is $A' = \pi (r')^2 = \pi (\frac{r}{2})^2 = \frac{A}{4}$.The heat conducted by the new rod in time $t$ is $Q' = \frac{kA' (T_1 - T_2) t}{l'}$.Substituting $A' = \frac{A}{4}$ and $l' = 4l$:$Q' = \frac{k (A/4) (T_1 - T_2) t}{4l} = \frac{1}{16} \frac{kA(T_1 - T_2) t}{l}$.Since $Q = \frac{kA(T_1 - T_2) t}{l}$,we get $Q' = \frac{Q}{16}$.

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