Two uniform wires A and B are of the same met | vedclass

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(A) The resistance of a wire is given by $R = ho \frac{L}{A} = ho \frac{L}{\pi r^2}$. Since the mass $M = 	ext{density} 	imes 	ext{volume} = d

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$4 \, \Omega$ when the resistance of wire $A$ is $4.25 \, \Omega$

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(A) The resistance of a wire is given by $R = ho \frac{L}{A} = ho \frac{L}{\pi r^2}$. Since the mass $M = 	ext{density} 	imes 	ext{volume} = d 	imes \pi r^2 L$ is constant,we have $L \propto \frac{1}{r^2}$.Substituting this into the resistance formula: $R \propto \frac{1}{r^2 \cdot r^2} = \frac{1}{r^4}$.Therefore,$\frac{R_A}{R_B} = \left( \frac{r_B}{r_A} ight)^4$. Given $r_A = 2r_B$,we get $\frac{R_A}{R_B} = \left( \frac{1}{2} ight)^4 = \frac{1}{16}$,which implies $R_B = 16 R_A$.When $R_A$ and $R_B$ are connected in parallel,the equivalent resistance $R_{eq}$ is given by:$R_{eq} = \frac{R_A R_B}{R_A + R_B} = \frac{R_A (16 R_A)}{R_A + 16 R_A} = \frac{16 R_A^2}{17 R_A} = \frac{16}{17} R_A$.If $R_A = 4.25 \, \Omega$,then $R_{eq} = \frac{16}{17} 	imes 4.25 = \frac{16}{17} 	imes \frac{17}{4} = 4 \, \Omega$.Thus,option $(A)$ is correct.

Two uniform wires $A$ and $B$ are of the same metal and have equal masses. The radius of wire $A$ is twice that of wire $B$. The total resistance of $A$ and $B$ when connected in parallel is

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