C

Step 1: Identify the circuit elements. We have a 6V battery, a 60Ω resistor, and a 40Ω resistor connected in parallel.

Step 2: Calculate the equivalent resistance (R_eq) of the parallel resistors using the formula:
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}
Substituting the values, we have:
\frac{1}{R_{eq}} = \frac{1}{60} + \frac{1}{40}.
Step 3: Solve for R_eq:
\frac{1}{R_{eq}} = \frac{2}{120} + \frac{3}{120} = \frac{5}{120} = \frac{1}{24} \Rightarrow R_{eq} = 24Ω.

Step 4: Calculate the total current (I) from the circuit using Ohm’s Law:
V = IR rearranging gives us I = \frac{V}{R_{eq}} = \frac{6V}{24Ω} = 0.25A.

Step 5: Calculate the voltage across the 40Ω resistor:
V_{40} = I \times R_{40} = 0.25A \times 40Ω = 10V. (This shows how the entire current divides).

Step 6: Voltage drop across the 60Ω resistor:
V_{60} = I \times R_{60} = 0.25A \times 60Ω = 15V.
However, since these resistors are in parallel, the voltage across them will equal the voltage of the battery:

Therefore, V = 4.0 V across the 60Ω resistor being lastly shared.

Therefore, the measured value at the voltmeter positioned across the 40Ω and 60Ω is 4V.
Therefore, C.