A cylindrical metallic rod,in thermal contact | vedclass

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Question

(C) The rate of heat flow through a rod is given by $\frac{Q}{t} = \frac{kA(T_1 - T_2)}{\ell}$.Since the volume $V = A\ell$ is constant,we can write $

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$\frac{Q}{16}$

Vedclass · Answer

(C) The rate of heat flow through a rod is given by $\frac{Q}{t} = \frac{kA(T_1 - T_2)}{\ell}$.Since the volume $V = A\ell$ is constant,we can write $\ell = \frac{V}{A}$.Substituting this into the equation: $\frac{Q}{t} = \frac{kA(T_1 - T_2)}{V/A} = \frac{kA^2(T_1 - T_2)}{V}$.Since $k, T_1, T_2,$ and $V$ are constant,we have $Q \propto A^2$.Given that the radius $r_2 = \frac{r_1}{2}$,the area $A = \pi r^2$ implies $A_2 = \pi (r_1/2)^2 = \frac{A_1}{4}$.Therefore,$\frac{Q_2}{Q_1} = \left(\frac{A_2}{A_1}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.Thus,$Q_2 = \frac{Q}{16}$.

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