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(D) The resistance of a wire is given by $R = \frac{ho \ell}{A}$.Since the mass $m = 	ext{density} 	imes 	ext{volume} = ho_d 	imes A 	imes \e

Vedclass · Accepted Answer

$32$

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(D) The resistance of a wire is given by $R = \frac{ho \ell}{A}$.Since the mass $m = 	ext{density} 	imes 	ext{volume} = ho_d 	imes A 	imes \ell$ is constant,and the material is the same (same density $ho_d$),the volume $V = A \ell$ is constant.Thus,$\ell = \frac{V}{A}$.Substituting this into the resistance formula: $R = \frac{ho V}{A^2}$.Since $ho$ and $V$ are constant,$R \propto \frac{1}{A^2}$.Since $A = \pi r^2$,we have $R \propto \frac{1}{r^4}$.Therefore,$\frac{R_A}{R_B} = \left( \frac{r_B}{r_A} ight)^4$.Given $r_A = 2.0 \ mm$,$r_B = 4.0 \ mm$,and $R_B = 2 \ \Omega$:$\frac{R_A}{2} = \left( \frac{4.0}{2.0} ight)^4 = (2)^4 = 16$.$R_A = 16 	imes 2 = 32 \ \Omega$.

Two wires $A$ and $B$ are made up of the same material and have the same mass. Wire $A$ has a radius of $2.0 \ mm$ and wire $B$ has a radius of $4.0 \ mm$. The resistance of wire $B$ is $2 \ \Omega$. The resistance of wire $A$ is . . . . . . $\Omega$.

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