$\begin{array}{1 1}(A)\;4.8\;cal/s \\ (B)\;6.0\;cal/s \\(C)\;1.2 \;cal/s \\(D)\;2.4\;cal/s \end{array} $

$\begin{align*} \text{Area of cross section A } &= 4 cm^2 \\ & = 4 \times 10^{-2}m^{-2} \end{align*}$

$t_A = 100^{\circ} c$

$t_B = 0 ^{\circ} c; \; \; \; \; t_C = 0^{\circ}c$

$l_{cu} = 46 cm;\;\;\; l_B = 13 cm; \;\;\;l_S = 12 cm; $

$K_{cu}= 0.92, \;\;\; K_B = 0.26; \; \; \; K_S = 0.12$

the temperature at the junction point be 't'

$H{cu} = H_B + H_S$

$\begin{align*} \implies \frac{K_{cu} A (t_A - t)}{l_{cu}} = \frac{K_B (t - t_B)}{l_B} + \frac{K_S (t - t_S)}{l_S}\end{align*}$

$\begin{align*} \implies \frac{0.92 (100 - t)}{46} = \frac{0.26 (t- 0)}{13} + \frac{0..12 (t-0)}{12} \end{align*}$

On solving we get

$t = 40^{\circ} c$

$\begin{align*} H_{cu} &= \frac{K_{cu} A(100 - 40) }{l_{cu}} \\ & = \frac{0.92 \times 4 \times 60}{46} \\ & =4.80\; cal/s \end{align*}$

$\therefore$ Rate of heat flow through copper rod is 4.80 cal/s

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