3K, 2K અને K ઉષ્મા વાહકતા ધરાવતા સળિયાને આકૃત | vedclass

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Question

$\frac{{dQ}}{{dt}} = KA\frac{{\Delta T}}{L}$

$For\,the\,first\,rod,\left( {\frac{{dQ}}{{dt}}} ight) = \frac{{3KA}}{L}\left( {100 - 	heta } igh

Vedclass · Accepted Answer

$70$

Vedclass · Answer

$\frac{{dQ}}{{dt}} = KA\frac{{\Delta T}}{L}$

$For\,the\,first\,rod,\left( {\frac{{dQ}}{{dt}}} ight) = \frac{{3KA}}{L}\left( {100 - 	heta } ight)$

$Similarly,{\left( {\frac{{dQ}}{{dt}}} ight)_2} = 2K\frac{A}{L}\left( {	heta  - 50} ight)$

${\left( {\frac{{dQ}}{{dt}}} ight)_3} = K\frac{A}{L}\left( {	heta  - 20} ight)$

$Now,{\left( {\frac{{dQ}}{{dt}}} ight)_1} = {\left( {\frac{{dQ}}{{dt}}} ight)_2} + {\left( {\frac{{dQ}}{{dt}}} ight)_3}$

$ \Rightarrow \,\,3\left( {100 - 	heta } ight) = 2\left( {	heta  - 50} ight) + \left( {	heta  - 20} ight)$

$ \Rightarrow 	heta  = {70^ \circ }$

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