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(D) Let the length of each wire be $\ell$ and the cross-sectional area be $A$. The resistance of a wire is given by $R = \rho_{res} \frac{\ell}{A}$, w

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$4$

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(D) Let the length of each wire be $\ell$ and the cross-sectional area be $A$. The resistance of a wire is given by $R = ho_{res} \frac{\ell}{A}$, where $ho_{res}$ is the resistivity.For the two wires connected in parallel, the individual resistances are $R_1 = ho_1 \frac{\ell}{A} = 6 \frac{\ell}{A}$ and $R_2 = ho_2 \frac{\ell}{A} = 3 \frac{\ell}{A}$.The equivalent resistance $R_{net}$ for two resistors in parallel is given by $\frac{1}{R_{net}} = \frac{1}{R_1} + \frac{1}{R_2}$, or $R_{net} = \frac{R_1 R_2}{R_1 + R_2}$.The combined cross-sectional area of the two wires in parallel is $2A$, and the length remains $\ell$. Thus, $R_{net} = ho \frac{\ell}{2A}$.Substituting the expressions:$ho \frac{\ell}{2A} = \frac{(6 \frac{\ell}{A}) (3 \frac{\ell}{A})}{6 \frac{\ell}{A} + 3 \frac{\ell}{A}}$$\frac{ho}{2} = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2$$ho = 4 \, \Omega \, cm$.

Two wires of same length and thickness having specific resistances $6 \, \Omega \, cm$ and $3 \, \Omega \, cm$ respectively are connected in parallel. The effective resistivity is $\rho \, \Omega \, cm$. The value of $\rho$ to the nearest integer is ..... .

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