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The ends of the two rods of conductivities , radii and lengths in the ratio of 1:2 are maintained at the same temperature difference. If the rate of flow through the bigger rod is 12" cal s"^(-1), in shorter it will be (in "cal s"^(-1)) : |
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Answer» Solution :`H=(Q)/(t)=(KA(T_(1)-T_(2)))/(d)` `therefore H_(1)=(k_(1)A_(1)(T_(1)-T_(2)))/(d_(1))` `H_(2)=(k_(2)A_(2)(T_(1)-T_(2)))/(d_(2))` `:.(H_(1))/(H_(2))=(k_(1))/(k_(2))*(A_(1))/(A_(2))xx(d_(2))/(d_(1))=(1)/(2)xx(1/2)^(2)xx2/1=(1)/(4)` `:.H_(1)=H_(2)xx1/4=12xx1/4=3" CAL s"^(-1)`. Thus correct choice is (c ). |
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