Question

A composite metal bar of uniform section is made up of length 25 cm of copper, 10 cm of nickel and 15 cm of aluminium. Each part being in perfect thermal contact with the adjoining part. The copper end of the composite rod is maintained at ${100}^{\circ }$ and the aluminium end at $0^{\circ }$. The whole rod is covered with belt so that there is no heat loss occurs at the sides. If $K_{Cu}=2K_{Al}$ and $K_{Al}=3K_{Ni}$, then what will be the temperatures of $Cu-Ni$ and $Ni-Al$ junctions respectively.

• A. Temperature of $Cu-Ni$ is ${78.26}^{\circ }$
• B. Temperature of $Ni-Al$ is ${26.08}^{\circ }$
• C. Temperature of $Cu-Ni$ is ${50.3}^{\circ }$
• D. Temperature of $Ni-Al$ is ${53.40}^{\circ }$

Answer 1

Answer: (A), (B)

The correct options are
A Temperature of $Cu-Ni$ is ${78.26}^{\circ }$
B Temperature of $Ni-Al$ is ${26.08}^{\circ }$


If suppose $K_{Ni}=K\Rightarrow K_{Al}=3K$ and $K_{Cu}=6K$

Since all metal bars are connected in series

So ${\left(\frac{Q}{t}\right)}_{combination}={\left(\frac{Q}{t}\right)}_{Cu}={\left(\frac{Q}{t}\right)}_{Al}={\left(\frac{Q}{t}\right)}_{Ni}$

and $\frac{l_1+l_2+l_3}{K_{eq}}=\frac{l_1}{K_{Cu}}+\frac{l_2}{K_{Al}}+\frac{l_3}{K_{Ni}}=\frac{25}{6K}+\frac{15}{3K}+\frac{10}{K}$

$\Rightarrow K_{eq}=\frac{60}{23}K$

Hence, if ${\left(\frac{Q}{t}\right)}_{Combination}={\left(\frac{Q}{t}\right)}_{Cu}$

$\Rightarrow \frac{K_{eq}A(100-0)}{l_{Combination}}=\frac{K_{Cu}A(100-{\theta }_1)}{l_{Cu}}$

$\Rightarrow \frac{60KA(100-0)}{23(25+10+15)}=\frac{6KA(100-{\theta }_1)}{25}\Rightarrow {\theta }_1={78.26}^{\circ }C$

Similar if ${\left(\frac{Q}{t}\right)}_{Combination}={\left(\frac{Q}{t}\right)}_{Al}$

$\Rightarrow \frac{60KA(100-0)}{23\times 50}=\frac{3KA({\theta }_2-0)}{15}\Rightarrow {\theta }_2={26.08}^{\circ }C$