In which of the follwing arrangements of resistors does the meter `M`, which has a resistance of `2 Omega`, give the largest reading when the same potential difference is appliced between points `P` and `Q`?

Options

A.
B.
C.
D.

Explanation

To determine in which arrangement of resistors the meter `M` (with a resistance of `2 Ω`) gives the largest reading when the same potential difference is applied between points `P` and `Q`, we will analyze each arrangement step by step.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We need to find the arrangement of resistors that allows the maximum current to flow through the meter `M` when a potential difference (voltage) is applied across points `P` and `Q`.

2. **Analyzing Arrangement A:**
- In Arrangement A, assume we have resistors of `1 Ω`, `1 Ω`, and the meter `M` of `2 Ω` in series.
- The total resistance \( R_A = 1 + 1 + 2 = 4 \, \Omega \).
- The current \( I_A = \frac{V_{PQ}}{R_A} = \frac{V_{PQ}}{4} \).

3. **Analyzing Arrangement B:**
- In Arrangement B, we have two resistors of `1 Ω` and `2 Ω` in parallel, and the meter `M` in series.
- The equivalent resistance of the parallel resistors \( R_{parallel} = \frac{1 \cdot 2}{1 + 2} = \frac{2}{3} \, \Omega \).
- The total resistance \( R_B = R_{parallel} + R_M = \frac{2}{3} + 2 = \frac{8}{3} \, \Omega \).
- The current \( I_B = \frac{V_{PQ}}{R_B} = \frac{V_{PQ}}{\frac{8}{3}} = \frac{3V_{PQ}}{8} \).

4. **Analyzing Arrangement C:**
- In Arrangement C, the meter `M` is in parallel with two resistors of `1 Ω` each.
- The equivalent resistance of the two `1 Ω` resistors in parallel \( R_{parallel} = \frac{1 \cdot 1}{1 + 1} = \frac{1}{2} \, \Omega \).
- The total resistance \( R_C = \frac{1}{2} \, \Omega \) (since the meter is in parallel).
- The current \( I_C = \frac{V_{PQ}}{R_C} = \frac{V_{PQ}}{\frac{1}{2}} = 2V_{PQ} \).

5. **Analyzing Arrangement D:**
- In Arrangement D, we have resistors of `1 Ω` and `2 Ω` in series with the meter `M`.
- The total resistance \( R_D = 1 + 2 + 2 = 5 \, \Omega \).
- The current \( I_D = \frac{V_{PQ}}{R_D} = \frac{V_{PQ}}{5} \).

6. **Comparing the Currents:**
- \( I_A = \frac{V_{PQ}}{4} \)
- \( I_B = \frac{3V_{PQ}}{8} \)
- \( I_C = 2V_{PQ} \)
- \( I_D = \frac{V_{PQ}}{5} \)

7. **Finding the Maximum Current:**
- Among the calculated currents, \( I_C = 2V_{PQ} \) is the largest current.
- Therefore, the arrangement that gives the largest reading on the meter `M` is Arrangement C.

### Conclusion:
The arrangement of resistors that allows the meter `M` to give the largest reading is **Arrangement C**.