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Physics-

General

Easy

Question

Five rods of same dimensions are arranged as shown in the figure. They have thermal conductivities K₁, K₂, K₃, K₄ and K₅. When points A and B are maintained at different temperatures, no heat flows through the central rod if

$K_{1} = K_{4} and K_{2} = K_{3}$
$K_{1} K_{4} = K_{2} K_{3}$
$K_{1} K_{2} = K_{3} K_{4}$
$\frac{K_{1}}{K_{4}} = \frac{K_{2}}{K_{3}}$

The correct answer is: $K subscript 1 end subscript K subscript 4 end subscript equals K subscript 2 end subscript K subscript 3 end subscript$

For no current flow between C and D
$open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript A C end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript C B end subscript$ Þ $fraction numerator K subscript 1 end subscript A left parenthesis theta subscript A end subscript minus theta subscript C end subscript right parenthesis over denominator l end fraction equals fraction numerator K subscript 2 end subscript A left parenthesis theta subscript C end subscript minus theta subscript B end subscript right parenthesis over denominator l end fraction$
Þ $fraction numerator theta subscript A end subscript minus theta subscript C end subscript over denominator theta subscript C end subscript minus theta subscript B end subscript end fraction equals fraction numerator K subscript 2 end subscript over denominator K subscript 1 end subscript end fraction$ ...(i)
Also $open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript A D end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript D B end subscript$ Þ $fraction numerator K subscript 3 end subscript A left parenthesis theta subscript A end subscript minus theta subscript D end subscript right parenthesis over denominator l end fraction equals fraction numerator K subscript 4 end subscript A left parenthesis theta subscript D end subscript minus theta subscript B end subscript right parenthesis over denominator l end fraction$
Þ $fraction numerator theta subscript A end subscript minus theta subscript D end subscript over denominator theta subscript D end subscript minus theta subscript B end subscript end fraction equals fraction numerator K subscript 4 end subscript over denominator K subscript 3 end subscript end fraction$ ...(ii)
It is given that $theta subscript C end subscript equals theta subscript D end subscript comma$ hence from equation (i) and (ii) we get $fraction numerator K subscript 2 end subscript over denominator K subscript 1 end subscript end fraction equals fraction numerator K subscript 4 end subscript over denominator K subscript 3 end subscript end fraction$ Þ $K subscript 1 end subscript K subscript 4 end subscript equals K subscript 2 end subscript K subscript 3 end subscript$

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