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(D) The resistance of a conductor is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional

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$2$

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(D) The resistance of a conductor is given by $R = ho \frac{l}{A}$,where $ho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.Since the conductors are made of the same material,$ho$ is constant.The cross-sectional area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,where $d$ is the diameter.Thus,$R \propto \frac{l}{d^2}$.The power delivered to a conductor connected across a constant potential difference $V$ is $P = \frac{V^2}{R}$.Therefore,$P \propto \frac{1}{R} \propto \frac{d^2}{l}$.Given for conductor $A$: $d_A = 2d_B$ and $l_A = 2l_B$.Calculating the ratio: $\frac{P_A}{P_B} = \frac{d_A^2 / l_A}{d_B^2 / l_B} = \left( \frac{d_A}{d_B} ight)^2 	imes \left( \frac{l_B}{l_A} ight) = (2)^2 	imes \left( \frac{1}{2} ight) = 4 	imes 0.5 = 2$.Hence,$P_A/P_B = 2$.

Two conductors made of the same material are connected across a common potential difference. Conductor $A$ has twice the diameter and twice the length of conductor $B$. The power delivered to the two conductors $P_A$ and $P_B$ respectively is such that $P_A/P_B$ equals to

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