Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of…

$L_{Cu} = 46 \mathrm{cm}; L_B = 13 \mathrm{cm} \& L_S=12 \mathrm{cm}$.



$\begin{array}{c}{K}_{Cu}=0.92\\ K_B=0.26\\ K_S=0.12\end{array}\}CGSunit$

We know that $R=\frac{L}{KA}$

$R_{Cu}=\frac{46}{0.92\times 4}=12.5$

$R_B=\frac{\mathrm{13}}{0.26\times 4}=12.5$

$R_S=\frac{12}{0.12\times 4}=\mathrm{25}$

Now, if $T$ is temperature of junction then

  $\frac{\mathrm{100}-T}{R_{Cu}}=\frac{T-0}{R_B}+\frac{T-0}{R_S}$

  $\frac{\mathrm{100}-T}{12.5}=\frac{T}{12.5}+\frac{T}{\mathrm{25}}$

  $\text{∴}\frac{\mathrm{100}-T}{12.5}=\frac{3T}{\mathrm{25}}$

  $200=5T$

  $\therefore T{=40}^{\circ }$

$\Rightarrow Q_{Cu}=\frac{100-T}{R_{Cu}}$

              $\text{=}\frac{100-40}{12.5}$

              $\text{=}\frac{60}{12.5}$

  $\Rightarrow Q_{Cu}=4.8 \mathrm{Cal}/\mathrm{s}$