A 16Ω wire is bent to form a square loop. A cell of 9 V is connected across one of its sides. The potential difference between the diagonals of the square loop is:

6 V

The correct answer is Option (1) → 6 V

Let the square sides be AB, BC, CD, DA. Total resistance = 16 Ω ⇒ each side R_side = 4 Ω.

Battery 9 V is connected across side AB, so node potentials: take A = 0, B = 9 V (battery across AB).

The branch AB is one branch (R=4 Ω). The other branch A→D→C→B is three sides in series (R=12 Ω). Both branches are in parallel across the 9 V source.

Currents:

$I_{AB}=\frac{9}{4}=2.25\ \text{A}$,

$I_{ADC B}=\frac{9}{12}=0.75\ \text{A}$.

Potential drops along B→C→D→A (following the 0.75 A branch): each side drops

$\Delta V_{\text{side}}=I\cdot R=0.75\times4=3\ \text{V}$.

Thus potentials: $V_B=9\ \text{V},\ V_C=9-3=6\ \text{V},\ V_D=6-3=3\ \text{V},\ V_A=0\ \text{V}$.

The potential difference between the diagonally opposite points A and C is

$V_C-V_A=6-0=6\ \text{V}$.

Answer: 6 V