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Step-by-Step Solution
Step 1: Identify the Resistance of Each Side
A wire of total resistance 16\,\Omega is bent to form a square loop. Since the loop has 4 equal sides, each side of the square has
R_{\text{side}} = \dfrac{16\,\Omega}{4} = 4\,\Omega.
Step 2: Recognize the Relevant Parts of the Circuit
The 9\,\text{V} supply (with internal resistance mentioned but not accounted for in the provided reference solution steps) is connected across one side of the loop (of resistance 4\,\Omega ). The remaining three sides have a total resistance of
R_{\text{remaining}} = 3 \times 4\,\Omega = 12\,\Omega.
Hence, when considering the loop as a whole in the reference approach, the total external resistance used is 4 + 12 = 16\,\Omega.
Step 3: Apply Kirchhoff’s Voltage Law (KVL) in the Outer Loop
According to the reference solution, the KVL around the loop (taking the supply as 9\,\text{V} and total resistance as 16\,\Omega ) is:
9 - \bigl(12\,\Omega \cdot i + 4\,\Omega \cdot i\bigr) = 0.
This simplifies to:
9 - 16\,\Omega\,i = 0 \quad \Longrightarrow \quad 16\,i = 9 \quad \Longrightarrow \quad i = \dfrac{9}{16}\,\text{A}.
Step 4: Find the Potential Drop Across the Diagonal
A diagonal of the square loop effectively spans across two sides in series. Since each side is 4\,\Omega, the diagonal has:
R_{\text{diagonal}} = 4\,\Omega + 4\,\Omega = 8\,\Omega.
Therefore, the potential drop across the diagonal is:
\Delta V_{\text{diagonal}} = i \times R_{\text{diagonal}} = \dfrac{9}{16} \times 8 = \dfrac{9 \times 8}{16} = \dfrac{72}{16} = 4.5\,\text{V}.
In the form requested by the question (some number \times 10^{-1}\,\text{V} ), 4.5\,\text{V} can be written as 45 \times 10^{-1}\,\text{V}.
Step 5: Final Answer
Thus, the potential drop across the diagonals of the square loop is
\boxed{4.5 \text{ V} \,=\, 45 \times 10^{-1}\,\text{V}.}
