A composite metal bar of uniform section is m | vedclass

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Question

(b) If suppose ${K_{Ni}} = K \Rightarrow {K_{Al}} = 3K$ and ${K_{Cu}} = 6K.$
Since all metal bars are connected in series

So ${\left( {\frac{Q}{t}

Vedclass · Accepted Answer

${83.33^o}C$ and ${20^o}C$

Vedclass · Answer

(b) If suppose ${K_{Ni}} = K \Rightarrow {K_{Al}} = 3K$ and ${K_{Cu}} = 6K.$
Since all metal bars are connected in series

So ${\left( {\frac{Q}{t}} ight)_{Combination}} = {\left( {\frac{Q}{t}} ight)_{Cu}} = {\left( {\frac{Q}{t}} ight)_{Al}} = {\left( {\frac{Q}{t}} ight)_{Ni}}$

and $\frac{3}{{{K_{eq}}}} = \frac{1}{{{K_{Cu}}}} + \frac{1}{{{K_{Al}}}} + \frac{1}{{{K_{Ni}}}} = \frac{1}{{6K}} + \frac{1}{{3K}} + \frac{1}{K} = \frac{9}{{6K}}$

==> ${K_{eq}} = 2K$

Hence, if ${\left( {\frac{Q}{t}} ight)_{Combination}} = {\left( {\frac{Q}{t}} ight)_{Cu}}$

==> $\frac{{{K_{eq}}\,A(100 - 0)}}{{{l_{Combination}}}} = \frac{{{K_{Cu}}A(100 - {	heta _1})}}{{{l_{Cu}}}}$

==> $\frac{{2K\,A\,(100 - 0)}}{{(25 + 10 + 15)}} = \frac{{6K\,A\,(100 - {	heta _1})}}{{25}}$

==> ${	heta _1} = 83.33^\circ C$

Similar if ${\left( {\frac{Q}{t}} ight)_{Combination}} = {\left( {\frac{Q}{t}} ight)_{Al}}$

==> $\frac{{2K\,A(100 - 0)}}{{50}} = \frac{{3K\,A({	heta _2} - 0)}}{{15}}$

==> ${	heta _2} = {20^o}C$

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