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A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat $Q$ in time $t$. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time $t$ ?
$\frac{Q}{4}\;$
$\;\frac{Q}{{16}}$
$\;2Q$
$\;\frac{Q}{2}$
Solution
The amount of heat flows in time $t$ through a cylindrical metallic rod of length $L$ and uniform area of $cross-section\,A(=\pi\,R^2)$ with its ends maintained at temperatures $T_1$ and $T_2$ $(T_1>T_2)$ is given by
$Q = \frac{{KA\left( {{T_1} – {T_2}} \right)t}}{L}$ $,,,(i)$
Where $K$ is the thermal conductivity of the material of the rod.
Area of $cross-section$ of new rod
$A' = \pi {\left( {\frac{R}{2}} \right)^2} = \frac{{\pi {R^2}}}{4} = \frac{A}{4}$ $…(ii)$
As the volume of the rod remains unchanged
$\therefore AL = A'L'$
Where $L'$ is the length the new rod
$or\,\,\,\,L' = L\frac{A}{{A'}}$ $,,,(iii)$
$ = 4L$ $(Using (ii))$
Now, the amount of heat flows in same time $t$ in the new rod with its ends maintained at the same temperatures $T_1$ and $T_2$ is given by
$Q' = \frac{{KA'\left( {{T_1} – {T_2}} \right)t}}{{L'}}$ $…(iv)$
Substituting the values of $A'$ and $L'$ from equations $(ii)$ and $(iii)$ in the above equation, we get
$Q' = \frac{{K\left( {A/4} \right)\left( {{T_1} – {T_2}} \right)t}}{4L}$
$ = \frac{1}{{16}}\frac{{KA\left( {{T_1} – {T_2}} \right)t}}{L} = \frac{1}{{16}}Q$
