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(B) Since the wires are connected in series,the current $I$ flowing through both wires is the same.Using Ohm's law,$V = IR$,so $V \propto R$.Given $\f

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$\frac{1}{2}$

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(B) Since the wires are connected in series,the current $I$ flowing through both wires is the same.Using Ohm's law,$V = IR$,so $V \propto R$.Given $\frac{V_A}{V_B} = \frac{3}{2}$ and $\frac{L_A}{L_B} = \frac{6}{1}$.The resistance $R$ is given by $R = ho \frac{L}{A} = ho \frac{L}{\pi r^2}$.Since the material is the same,$ho$ is constant,so $R \propto \frac{L}{r^2}$.Therefore,$\frac{V_A}{V_B} = \frac{R_A}{R_B} = \frac{L_A}{r_A^2} 	imes \frac{r_B^2}{L_B}$.Substituting the values: $\frac{3}{2} = \frac{6}{1} 	imes \left( \frac{r_B}{r_A} ight)^2$.$\left( \frac{r_B}{r_A} ight)^2 = \frac{3}{2} 	imes \frac{1}{6} = \frac{3}{12} = \frac{1}{4}$.Taking the square root,$\frac{r_B}{r_A} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

Two wires $A$ and $B$ made of the same material and having their lengths in the ratio $6 : 1$ are connected in series. The potential differences across the wires are $3\,V$ and $2\,V$ respectively. If $r_A$ and $r_B$ are the radii of $A$ and $B$ respectively,then $\frac{r_B}{r_A}$ is

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