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12 rods of same thermal resistance $(R_{T})$ are joined as shown in figure. The ends P and Q are maintained at constant temperature $T_{1} and T_{2} (T_{1} > T_{2})$ . The steady state heat energy flow rate is

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$\frac{{({T_1} - {T_2})}}{{{R_T}}}$

$\frac{{4({T_1} - {T_2})}}{{5{R_T}}}$

$\frac{4}{{3{R_T}}}({T_1} - {T_2})$

$\frac{1}{8}\frac{{({T_1} - {T_2})}}{{{R_T}}}$

answer is C.

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Detailed Solution

R_eq between P and Q

$\Rightarrow \,\,\,\,\,{R_{eq}} = \frac{{3{R_T}}}{4}$

$\therefore \,\,\,\,\,({T_1} - {T_2}) = (H)\,{R_{eq}}$

$\therefore \,\,\,\,\,H = \frac{{4({T_1} - {T_2})}}{{3{R_T}}}$

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