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 of heat through the bigger rod is $$4$$ $$cals-1$$, in shorter it will be (in$$ $$cals-1)

Question

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 of heat through the bigger rod  is  $$4$$ $$cals-1$$, in shorter it will be  (in$$ $$cals-1)

Infinity Learn · Accepted Answer

Rate of flow of heat is P $$P=KA($$θ$$1-$$θ$$2)l=K$$$$π$$$$r2($$θ$$1-$$θ$$2)l$$ here K is coefficient of thermal conductivity, A is area of cross section, θ$$1$$,θ$$2$$ temperature at each end of rod l is length of rod substitute given values legibly  $$P1P2=K1K2$$.$$r1r22$$×$$l2l1=1$$ $$2122x21$$ given $$P2=4$$ calorie $$second-1$$ $$P4=14$$ rate of flow of heat through short rod is $$P=1$$ $$cals-1$$

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 of heat through the bigger rod is ${4 cals}^{-1}$ , in shorter it will be ${(in cals}^{-1})$

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

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 of heat through the bigger rod is 4 cals-1, in shorter it will be (in cals-1)

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

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 of heat through the bigger rod is ${4 cals}^{-1}$ , in shorter it will be ${(in cals}^{-1})$