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Two conducting rods A and B of same length and cross-sectional area are connected (i) In series (ii) In parallel as shown. In both combination a temperature difference of 100°C is maintained. If thermal conductivity of A is 3K and that of B is K then the ratio of heat current flowing in parallel combination to that flowing in series combination is

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$\frac{1}{1}$

$\frac{3}{{16}}$

$\frac{{16}}{3}$

$\frac{1}{3}$

answer is A.

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Detailed Solution

$Heat\,\,current\,\,H = \frac{{\Delta \theta }}{R} \Rightarrow \frac{{{H_P}}}{{{H_S}}} = \frac{{{R_S}}}{{{R_P}}}$

$In\,\,first\,\,case\,\,{R_S} = {R_1} + {R_2} = \frac{l}{{(3K)A}} + \frac{l}{{KA}} = \frac{4}{3}\frac{l}{{KA}}$

$In\,\,second\,\,case\,\,:{R_P} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{\frac{l}{{(3K)A}} \times \frac{l}{{KA}}}}{{\left( {\frac{l}{{(3K)A}} + \frac{l}{{KA}}} \right)}} = \frac{l}{{4KA}}$

$\therefore \frac{{{H_P}}}{{{H_S}}} = \frac{{\frac{{4l}}{{3KA}}}}{{\frac{l}{{4KA}}}} = \frac{{16}}{3}$

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