Step-by-step Solution:

1. Principle of a Metre Bridge: A metre bridge works on the principle of a balanced Wheatstone bridge. If $R_L$ is the resistance in the left gap, $R_R$ is the resistance in the right gap, and the balancing length from the left end is l, then for a wire of total length 100 cm, the condition for a balanced bridge is: $\frac{R_L}{R_R} = \frac{l}{100 - l}$

2. Initial Setup (Case 1): Let the known resistance $R_1 = 2 \Omega$ be connected in the left gap and the unknown resistance R be connected in the right gap. We are given that $R > 2 \Omega$. Let the balancing length from the left end be $l_1$ cm. The equation for the balanced bridge is: $\frac{2}{R} = \frac{l_1}{100 - l_1} \quad \cdots (1)$ Since R > 2, we have 2/R < 1, which implies $l_1 / (100 - l_1) < 1$. This means $l_1 < 100 - l_1$, or $2l_1 < 100$, so $l_1 < 50$ cm.

3. After Interchanging Resistances (Case 2): The resistances are interchanged. Now, the unknown resistance R is in the left gap and the known resistance $R_1 = 2 \Omega$ is in the right gap. The problem states that the balance point shifts by 20 cm. Since $l_1 < 50$ cm, and now the resistance in the left gap (R) is greater than the resistance in the right gap ($2 \Omega$), the new balancing length $l_2$ must be greater than 50 cm. Therefore, the new balancing length is $l_2 = l_1 + 20$ cm. The equation for the balanced bridge in this case is: $\frac{R}{2} = \frac{l_2}{100 - l_2}$ Substituting $l_2 = l_1 + 20$: $\frac{R}{2} = \frac{l_1 + 20}{100 - (l_1 + 20)} = \frac{l_1 + 20}{80 - l_1} \quad \cdots (2)$

4. Solving for $l_1$ and R: We now have a system of two equations with two unknowns (R and $l_1$). From equation (1), we can express R in terms of $l_1$: $R = 2 \left( \frac{100 - l_1}{l_1} \right)$ Substitute this expression for R into equation (2): $\frac{1}{2} \left[ 2 \left( \frac{100 - l_1}{l_1} \right) \right] = \frac{l_1 + 20}{80 - l_1}$ $\frac{100 - l_1}{l_1} = \frac{l_1 + 20}{80 - l_1}$ Now, cross-multiply to solve for $l_1$: $(100 - l_1)(80 - l_1) = l_1(l_1 + 20)$ $8000 - 100l_1 - 80l_1 + l_1^2 = l_1^2 + 20l_1$ $8000 - 180l_1 = 20l_1$ $8000 = 200l_1$ $l_1 = \frac{8000}{200} = 40 \text{ cm}$

5. Calculating the Unknown Resistance R: Now that we have $l_1 = 40$ cm, we can substitute this value back into the expression for R from equation (1): $R = 2 \left( \frac{100 - l_1}{l_1} \right) = 2 \left( \frac{100 - 40}{40} \right)$ $R = 2 \left( \frac{60}{40} \right) = 2 \left( \frac{3}{2} \right)$ $R = 3 \Omega$

6. Conclusion: The unknown resistance is $3 \Omega$. This matches option A. We can verify our result. If $R=3 \Omega$, then in Case 1: $2/3 = l_1/(100-l_1) \implies 200-2l_1 = 3l_1 \implies 5l_1=200 \implies l_1 = 40$ cm. In Case 2: $3/2 = l_2/(100-l_2) \implies 300-3l_2 = 2l_2 \implies 5l_2=300 \implies l_2=60$ cm. The shift is $l_2 - l_1 = 60 - 40 = 20$ cm, which confirms our calculation.