Unknown Resistance Determination in Metre Bridge with Balance Shift

To find the unknown resistance, we apply the balancing condition of a metre bridge.

Let the resistance in the left gap initially be $R_1 = 2\ \Omega$ and the unknown resistance in the right gap be $R_2 = X$, where $X > 2\ \Omega$.

Let $l_1$ be the distance of the balance point from the left end of the $100\text{ cm}$ wire. The condition for bridge balance is given by:
$\frac{R_1}{R_2} = \frac{l_1}{100 - l_1}$
Substituting the values:
$\frac{2}{X} = \frac{l_1}{100 - l_1} \quad \text{--- (1)}$

Since $X > 2\ \Omega$, we have $\frac{2}{X} < 1$, which implies:
$l_1 < 100 - l_1 \implies l_1 < 50\text{ cm}$

When the two resistances are interchanged, the unknown resistance $X$ is placed in the left gap and $2\ \Omega$ in the right gap. Let $l_2$ be the new balance length from the left end:
$\frac{X}{2} = \frac{l_2}{100 - l_2} \quad \text{--- (2)}$

Taking the reciprocal of equation (1):
$\frac{X}{2} = \frac{100 - l_1}{l_1}$

Equating this to equation (2):
$\frac{l_2}{100 - l_2} = \frac{100 - l_1}{l_1}$
By symmetry, this gives:
$l_2 = 100 - l_1$

We are given that the balance point shifts by $20\text{ cm}$. Since $X > 2\ \Omega$, the balance point moves to the right ($l_2 > l_1$), so the shift is:
$l_2 - l_1 = 20\text{ cm}$

Substituting $l_2 = 100 - l_1$:
$(100 - l_1) - l_1 = 20$
$100 - 2l_1 = 20$
$2l_1 = 80 \implies l_1 = 40\text{ cm}$

Now, substituting $l_1 = 40\text{ cm}$ back into equation (1):
$\frac{2}{X} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3}$
$X = 3\ \Omega$

Therefore, the unknown resistance is $3\ \Omega$.

Correct Option: (A)