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(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the mass $m = 	ext{Volume} 	imes 	ext{Density} = (A \cdot l) \cdot d$,and bot

Vedclass · Accepted Answer

$544$

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(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the mass $m = 	ext{Volume} 	imes 	ext{Density} = (A \cdot l) \cdot d$,and both wires have the same material and mass,the product $A \cdot l$ is constant.Thus,$l = \frac{m}{A \cdot d} \propto \frac{1}{A}$.Substituting this into the resistance formula: $R = ho \frac{l}{A} \propto \frac{1}{A^2}$.Since $A = \pi r^2$,we have $R \propto \frac{1}{(\pi r^2)^2} \propto \frac{1}{r^4}$.Therefore,$\frac{R_A}{R_B} = \left( \frac{r_B}{r_A} ight)^4$.Given $R_A = 34\,\Omega$,$r_A = 2r$,and $r_B = r$:$\frac{34}{R_B} = \left( \frac{r}{2r} ight)^4 = \left( \frac{1}{2} ight)^4 = \frac{1}{16}$.$R_B = 34 	imes 16 = 544\,\Omega$.

Two wires $A$ and $B$ of the same material and same mass have radii $2r$ and $r$,respectively. If the resistance of wire $A$ is $34\,\Omega$,then the resistance of wire $B$ will be ............ $\Omega$.

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