Rate of flow of heat through a cylindrical rod is ' H₁ '. The temperature at the ends of the rod are T₁ and T₂ . If all the dimensions of the rod become double and the temperature difference remains the same and if the rate of flow of heat becomes ' H₂ ' then H₂ =

Step-by-step solution

The rate of flow of heat through a rod is given by H = KA(T₁ - T₂) L For a cylindrical rod of radius r and length L , the area of cross-section is A = r^2 . Initial rate of heat flow is H₁ = K r^2 (T₁ - T₂) L When all dimensions are doubled, the new radius is r' = 2r and the new length is L' = 2L . The new rate of heat flow is H₂ = K (2r)^2 (T₁ - T₂) 2L H₂ = 4 K r^2 (T₁ - T₂) 2L H₂ = 2 ( K r^2 (T₁ - T₂) L ) H₂ = 2H₁ Answer: 2 H ₁