A material 'B' has twice the specific resista | vedclass

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(A) Given: $\rho_B = 2\rho_A$ and $d_B = 2d_A$.Since the wires are circular,the area of cross-section is $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4

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$2$

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(A) Given: $\rho_B = 2\rho_A$ and $d_B = 2d_A$.Since the wires are circular,the area of cross-section is $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.For the resistances to be equal,$R_B = R_A$.Using the formula $R = \frac{\rho l}{A}$,we have $\frac{\rho_B l_B}{A_B} = \frac{\rho_A l_A}{A_A}$.Substituting the values: $\frac{2\rho_A l_B}{\frac{\pi (2d_A)^2}{4}} = \frac{\rho_A l_A}{\frac{\pi d_A^2}{4}}$.Simplifying the equation: $\frac{2\rho_A l_B}{4 A_A} = \frac{\rho_A l_A}{A_A} \Rightarrow \frac{2 l_B}{4} = l_A$.Therefore,$\frac{l_B}{l_A} = \frac{4}{2} = 2$.

$A$ material '$B$' has twice the specific resistance of '$A$'. $A$ circular wire made of '$B$' has twice the diameter of a wire made of '$A$'. Then,for the two wires to have the same resistance,the ratio $\frac{l_B}{l_A}$ of their respective lengths must be

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