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(B) 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 $1 \ cm 	imes 0.5 \ cm$ faces.

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(B) 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.To maximize the resistance $R$,we need to maximize the ratio $\frac{l}{A}$. This means we need the largest possible length $l$ and the smallest possible cross-sectional area $A$.The dimensions of the rod are $10 \ cm$,$1 \ cm$,and $0.5 \ cm$.If the battery is connected across the $1 \ cm 	imes 0.5 \ cm$ faces,the current flows through the length $l = 10 \ cm$. The area of cross-section is $A = 1 \ cm 	imes 0.5 \ cm = 0.5 \ cm^2$.In this case,the ratio $\frac{l}{A} = \frac{10}{0.5} = 20 \ cm^{-1}$.Comparing this with other configurations,this configuration yields the maximum value for the ratio $\frac{l}{A}$,hence the resistance is maximum.

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

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