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(D) $1$. The total resistance of the wire is $16\,\Omega$. When bent into a square,each side has a resistance of $R = 16/4 = 4\,\Omega$.$2$. Let the s

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(D) $1$. The total resistance of the wire is $16\,\Omega$. When bent into a square,each side has a resistance of $R = 16/4 = 4\,\Omega$.$2$. Let the square be $ABCD$. The sides $AB, BC, CD,$ and $DA$ each have a resistance of $4\,\Omega$.$3$. $A$ wire of $16\,\Omega$ is connected between opposite corners $B$ and $D$.$4$. The circuit now consists of two parallel branches connected between $A$ and $C$. The upper branch has two $4\,\Omega$ resistors in series ($AB$ and $BC$),giving $4 + 4 = 8\,\Omega$. The lower branch has two $4\,\Omega$ resistors in series ($AD$ and $DC$),giving $4 + 4 = 8\,\Omega$.$5$. The $16\,\Omega$ resistor is connected between $B$ and $D$. Since the potential at $B$ and $D$ is the same due to the symmetry of the Wheatstone bridge $(4/4 = 4/4)$,no current flows through the $16\,\Omega$ resistor.$6$. Thus,the effective resistance between $A$ and $C$ is the parallel combination of the two $8\,\Omega$ branches: $R_{eq} = (8 	imes 8) / (8 + 8) = 64 / 16 = 4\,\Omega$.

$A$ uniform wire of $16\,\Omega$ is bent into the form of a square. Two opposite corners of the square are connected by a wire of resistance $16\,\Omega$. The effective resistance between the other two opposite corners is ............... $\Omega$.

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