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(D) Given: Ratio of lengths,$\frac{\ell_1}{\ell_2} = \frac{2}{3}$ and ratio of radii,$\frac{r_1}{r_2} = \frac{8}{9}$.Since the material is the same,re

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$32: 27$

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(D) Given: Ratio of lengths,$\frac{\ell_1}{\ell_2} = \frac{2}{3}$ and ratio of radii,$\frac{r_1}{r_2} = \frac{8}{9}$.Since the material is the same,resistivity $ho$ is constant.The resistance of a wire is given by $R = ho \frac{\ell}{A} = ho \frac{\ell}{\pi r^2}$.Therefore,the ratio of resistances is $\frac{R_1}{R_2} = \frac{\ell_1}{\ell_2} 	imes \left( \frac{r_2}{r_1} ight)^2$.Substituting the values: $\frac{R_1}{R_2} = \frac{2}{3} 	imes \left( \frac{9}{8} ight)^2 = \frac{2}{3} 	imes \frac{81}{64} = \frac{27}{32}$.According to Ohm's law,$I = \frac{V}{R}$. Since $V$ is constant,$I \propto \frac{1}{R}$.Thus,$\frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{32}{27}$.The ratio of electric currents is $32: 27$.

Two wires made of the same material have lengths in the ratio $2: 3$ and radii in the ratio $8: 9$. If the same potential difference is applied across the ends of the wires,the ratio of the electric currents flowing through them is

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