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(A) The resistance $R$ 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

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Double the resistance of the second wire

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(A) The resistance $R$ of a wire is given by the formula $R = ho \frac{L}{A}$,where $ho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area. Since the area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we can write $R = ho \frac{4L}{\pi d^2}$. Let the first wire have length $L_1 = L$ and diameter $d_1 = d$. Its resistance is $R_1 = ho \frac{4L}{\pi d^2}$. For the second wire,$L_2 = 2L$ and $d_2 = 2d$. Its resistance is $R_2 = ho \frac{4(2L)}{\pi (2d)^2} = ho \frac{8L}{4\pi d^2} = ho \frac{2L}{\pi d^2}$. Comparing $R_1$ and $R_2$: $R_1 = 2 	imes (ho \frac{2L}{\pi d^2}) = 2 R_2$. Thus,the resistance of the first wire is double the resistance of the second wire.

Two wires are made of the same material. The length and diameter of the second wire are both double those of the first wire. What will be the resistance of the first wire compared to the second wire?

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