InterviewSolution
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InterviewSolution
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A circular ring (centre O) of radius a, and of uniform cross section is made up of three different metallic rods AB, BC and CA (joined together at the points A, B and C in pairs) of thermal conductivityies `alpha_1`,`alpha_2` and `alpha_3` respectively (see diagram). The junction A, B and C are maintained at the temperatures `100^@C`,`50^@C` and `0^@C`, respectively. All the rods are of equal lengths and cross sections. Under steady state conditions, assume that no heat is lost from the sides of the rods. Let `Q_1`,`Q_2` and `Q_3` be the rates of transmission of heat along the three rods AB, BC and CA. ThenA. `Q_1=Q_2=Q_3` and all are transmitted in the clockwise senseB. `Q_1` and `Q_2` flow in clockwise sense and `Q_3` in the anticlockwise sense.C. `Q_1:Q_2:Q_3::alpha_1:alpha_2:2alpha_3`D. `(Q_1)/(alpha_1)+(Q_2)/(alpha_2)=(Q_3)/(alpha_3)` |
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Answer» Correct Answer - B::C::D The flow of heat will always be in the direction of the temperature gradient from higher to lower temperature Hence `Q_1` in rod AB, `Q_2` in rod BC will both be in clockwise sense while `Q_3` in CA will be in anti-clockwise sense Also we have if L is the length of each rod and A its area of cross-section `Q_1=(alpha_1A(100-50))/(L)=(50alpha_1)(A)/(L)` `Q_2=(alpha_2A(50-0))/(L)=(50alpha_2)(A)/(L)` `Q_3=(alpha_3A(100-0))/(L)=(100alpha_3)(A)/(L)` Hence `Q_1:Q_2:Q_3::alpha_1:alpha_2:2alpha_3` Also `(Q_1)/(alpha_1)+(Q_2)/(alpha_2)=(50(A)/(L))+(50(A)/(L))=(100A)/(L)=(Q_3)/(alpha_3)` |
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