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(B) Given: Resistivity $\rho$ is the same,length $l$ is the same.Let the thicker wire have area $A_1$ and radius $r_1$,and the thinner wire have area

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$\frac{8}{5}\, \Omega$

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(B) Given: Resistivity $ho$ is the same,length $l$ is the same.Let the thicker wire have area $A_1$ and radius $r_1$,and the thinner wire have area $A_2$ and radius $r_2$.Since the thinner wire has half the thickness,$r_2 = \frac{r_1}{2}$.Area $A = \pi r^2$,so $A_2 = \pi (\frac{r_1}{2})^2 = \frac{1}{4} A_1$.Using the resistance formula $R = ho \frac{l}{A}$,we have $R \propto \frac{1}{A}$.Therefore,$\frac{R_1}{R_2} = \frac{A_2}{A_1} = \frac{1}{4}$.Given the thinner wire resistance $R_2 = 8\, \Omega$,we get $\frac{R_1}{8} = \frac{1}{4}$,which implies $R_1 = 2\, \Omega$.For a parallel combination,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.$R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{2 	imes 8}{2 + 8} = \frac{16}{10} = \frac{8}{5}\, \Omega$.

Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of $8\, \Omega$,the resistance of the combination is equal to

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