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(C) Let the side length of plate $A$ be $L$. Then the side length of plate $B$ is $2L$.Since both are square plates of thickness $t$, the cross-sectio

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(C) Let the side length of plate $A$ be $L$. Then the side length of plate $B$ is $2L$.Since both are square plates of thickness $t$, the cross-sectional area $A_{cs}$ through which current flows is given by $A_{cs} = 	ext{side} 	imes t$.For plate $A$: $L_A = L$ (length in direction of current), $A_{cs,A} = L 	imes t$.For plate $B$: $L_B = 2L$ (length in direction of current), $A_{cs,B} = 2L 	imes t$.Using the resistance formula $R = \frac{ho L}{A_{cs}}$:$R_A = \frac{ho L}{Lt} = \frac{ho}{t}$$R_B = \frac{ho (2L)}{(2L)t} = \frac{ho}{t}$Therefore, $\frac{R_A}{R_B} = \frac{ho/t}{ho/t} = 1$.

Two square-shaped metal plates $A$ and $B$ of the same thickness $(t)$ and of the same material are connected as shown in the figure. The side of $B$ is twice that of $A$. If the resistances of $A$ and $B$ are $R_A$ and $R_B$ respectively, then $\frac{R_A}{R_B}$ is:

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