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Three rods of Copper, Brass and Steel are welded together to form a $Y$- shaped structure. Area of cross -section of each rod $= 4 cm^2$. End of copper rod is maintained at $10^{\circ}C$.Lengths of the copper, brass and steel rods are $46, 13 \;and \;12\; cms$ respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivies of copper, brass and steel are $0.92, 0.26\; and\; 0.12$ CGS units respectively. Rate of heat flow through copper rod is

$\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} $

1 Answer

$\begin{align*} \text{Area of cross section A } &= 4 cm^2 \\ & = 4 \times 10^{-2}m^{-2} \end{align*}$
$t_A = 100^{\circ} c$
$t_B = 0 ^{\circ} c; \; \; \; \; t_C = 0^{\circ}c$
$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
answered Apr 7, 2014 by meena.p
edited Dec 6 by priyanka.c
 

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