Two wires A and B of same material and same mass have radius 2r and r. If resistance of wire A is $34\Omega$ , then resistance of B will be
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$R = \rho \frac{1}{A}$ and mass m = volume (V) $\times$ density (d) = (Al)d Since wires have same material so $\rho$ and d is same for both. Also they have same mass $\Rightarrow$ Al = constant $\Rightarrow l \propto \frac{1}{A}$ $\Rightarrow \frac{R_1}{R_2} = \frac{l_1}{l_2} \times \frac{A_2}{A_1} = \left(\frac{A_2}{A_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^4$ $\Rightarrow \frac{34}{R_2} = \left(\frac{r}{2r}\right)^4 \Rightarrow R_2 = 544\Omega$
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