Wires A and B have identical lengths and have | vedclass

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(D) The rate of heat conduction is given by the formula: $\frac{Q}{t} = \frac{KA\Delta 	heta}{l}$.Since the rate of heat flow $\frac{Q}{t}$,the lengt

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${K_A} = {K_B}/4$

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(D) The rate of heat conduction is given by the formula: $\frac{Q}{t} = \frac{KA\Delta 	heta}{l}$.Since the rate of heat flow $\frac{Q}{t}$,the length $l$,and the temperature difference $\Delta 	heta$ are the same for both wires,we have: $K_A A_A = K_B A_B$.Given that the cross-sections are circular,the area $A = \pi r^2$. Therefore,$A_A = \pi r_A^2$ and $A_B = \pi r_B^2$.Substituting the given condition $r_A = 2r_B$,we get $A_A = \pi (2r_B)^2 = 4\pi r_B^2 = 4A_B$.Now,substituting this into the equality: $K_A (4A_B) = K_B A_B$.Dividing both sides by $4A_B$,we get: $K_A = \frac{K_B}{4}$.

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