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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 $ \large\frac{l_A}{l_b}$ of their respective lengths must be

$\begin {array} {1 1} (A)\;2 & \quad (B)\;1 \\ (C)\;\large\frac{1}{2} & \quad (D)\;\large\frac{1}{4} \end {array}$


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$ R_1 = \large\frac{\rho_Al_A}{\pi R_A^2}$
$ \large\frac{\rho_Bl_B}{\pi R_B^2}$
$ \large\frac{l_A}{l_B}$ $ = \large\frac{\rho_BR_A^2}{\rho_AR_B^2}$ $\large\frac{2\rho_AR_A^2}{\rho_A4R_A^2}$
$ \Rightarrow \large\frac{l_A}{l_B}$ $ = 2$
Ans : (A)
answered Feb 15, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
How in the world 2/4=2? The answer should be 1/2 i.e. (C)

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