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(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length of the conducto

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maximum when the battery is connected across $1 	ext{ cm} 	imes 0.5 	ext{ cm}$ faces

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(A) The resistance $R$ of a conductor is given by the formula $R = ho \frac{L}{A}$,where $ho$ is the resistivity,$L$ is the length of the conductor in the direction of current flow,and $A$ is the cross-sectional area perpendicular to the current flow.From the relation $R \propto \frac{L}{A}$,we see that for a fixed volume of the material,the resistance is proportional to the square of the length $(L^2)$ or inversely proportional to the square of the area $(A^2)$.Alternatively,for a fixed rod,the resistance is maximum when the current flows through the smallest cross-sectional area $A$,which results in the longest effective length $L$.The three possible cross-sectional areas are:$1$. $A_1 = 10 	ext{ cm} 	imes 1 	ext{ cm} = 10 	ext{ cm}^2$$2$. $A_2 = 10 	ext{ cm} 	imes 0.5 	ext{ cm} = 5 	ext{ cm}^2$$3$. $A_3 = 1 	ext{ cm} 	imes 0.5 	ext{ cm} = 0.5 	ext{ cm}^2$The resistance is maximum when the cross-sectional area $A$ is minimum. The minimum area is $0.5 	ext{ cm}^2$,which corresponds to the faces of dimensions $1 	ext{ cm} 	imes 0.5 	ext{ cm}$.

$A$ metal rod of length $10 \text{ cm}$ and a rectangular cross-section of $1 \text{ cm} \times 0.5 \text{ cm}$ is connected to a battery across opposite faces. The resistance will be

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