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(D) The resistance 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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$40\, \Omega$

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(D) The resistance 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-sectional area.Since both wires are of the same metal, $\rho$ is constant. Given that $L$ is also the same, we have $R \propto \frac{1}{A}$.Let $A_1$ be the area of the thicker wire and $A_2$ be the area of the thinner wire. Given $A_1 : A_2 = 3 : 1$, so $A_1 = 3A$ and $A_2 = A$.The resistance of the thicker wire $(R_1)$ is $10\, \Omega$. Since $R_1 = \frac{\rho L}{3A} = 10\, \Omega$, then $\frac{\rho L}{A} = 30\, \Omega$.The resistance of the thinner wire $(R_2)$ is $R_2 = \frac{\rho L}{A} = 30\, \Omega$.When joined in series, the total resistance $R_{total} = R_1 + R_2 = 10\, \Omega + 30\, \Omega = 40\, \Omega$.

Two wires of the same metal have the same length, but cross-sections are in the ratio $3 : 1$. They are joined in series. The resistance of the thicker wire is $10\, \Omega$. The total resistance of the combination will be:

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