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(C) The resistance of a conductor is given by $R = \rho \frac{L}{A_{cross}}$,where $\rho$ is the resistivity,$L$ is the length in the direction of cur

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$1/1$

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(C) The resistance of a conductor is given by $R = ho \frac{L}{A_{cross}}$,where $ho$ is the resistivity,$L$ is the length in the direction of current flow,and $A_{cross}$ is the cross-sectional area perpendicular to the current flow.For plate $A$,the length $L_A = \ell$,and the cross-sectional area $A_A = \ell 	imes t$. Thus,$R_A = ho \frac{\ell}{\ell 	imes t} = \frac{ho}{t}$.For plate $B$,the length $L_B = 2\ell$,and the cross-sectional area $A_B = 2\ell 	imes t$. Thus,$R_B = ho \frac{2\ell}{2\ell 	imes t} = \frac{ho}{t}$.Comparing the two,we get $R_A = R_B$.Therefore,the ratio $(R_A/R_B) = 1/1$.

Two square metal plates $A$ and $B$ are of the same thickness and material. The side of $B$ is twice that of $A$. These are connected as shown in series. If the resistances of $A$ and $B$ are denoted by $R_A$ and $R_B,$ then $(R_A/R_B)$ is

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