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(A) The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sec

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$\frac{2}{9} R$

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(A) The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.Since the cross-section is circular,$A = \pi r^2$.For the first wire: $R = \rho \frac{L}{\pi r^2} \quad (1)$For the second wire: $L_2 = 2L$ and $r_2 = 3r$.The new resistance $R_2 = \rho \frac{2L}{\pi (3r)^2} = \rho \frac{2L}{9 \pi r^2} \quad (2)$Dividing equation $(2)$ by equation $(1)$:$\frac{R_2}{R} = \frac{\rho \frac{2L}{9 \pi r^2}}{\rho \frac{L}{\pi r^2}} = \frac{2}{9}$Therefore,$R_2 = \frac{2}{9} R$.

$A$ metal wire of length $L$ and radius $r$ has a resistance $R$. If a wire of the same metal of length $2L$ and radius $3r$ is taken,then what will be its resistance?

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