A cylindrical metallic rod in thermal contact | vedclass

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Question

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

Vedclass · Accepted Answer

$\;\frac{Q}{{16}}$

Vedclass · Answer

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

$Q = \frac{{KA\left( {{T_1} - {T_2}} ight)t}}{L}$                          $,,,(i)$

Where $K$ is the thermal conductivity of the material of the rod.

Area of $cross-section$ of new rod

$A' = \pi {\left( {\frac{R}{2}} ight)^2} = \frac{{\pi {R^2}}}{4} = \frac{A}{4}$                 $...(ii)$

As the volume of the rod remains unchanged

$	herefore 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

$Q' = \frac{{KA'\left( {{T_1} - {T_2}} ight)t}}{{L'}}$                   $...(iv)$

Substituting the values of $A'$ and $L'$ from equations $(ii)$ and $(iii)$ in the above equation, we get

$Q' = \frac{{K\left( {A/4} ight)\left( {{T_1} - {T_2}} ight)t}}{4L}$

$ = \frac{1}{{16}}\frac{{KA\left( {{T_1} - {T_2}} ight)t}}{L} = \frac{1}{{16}}Q$

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