In a Wheatstone's metre bridge, resistances R and S are connected...

For a metre bridge, the balancing condition with resistances $R$ and $S$ in the left and right gaps respectively is given by:

$\dfrac{R}{S} = \dfrac{l}{100 - l}$

When the resistances are interchanged, the new balancing length becomes $100 - l$. It is given that the null point shifts to the left by $20\ \text{cm}$, so the new balancing length is $l - 20$.

Equating the two expressions for the new balancing length:

$100 - l = l - 20$

$2l = 120 \Rightarrow l = 60\ \text{cm}$

Substituting $l = 60\ \text{cm}$ into the initial balancing condition:

$\dfrac{R}{S} = \dfrac{60}{100 - 60} = \dfrac{60}{40} = \dfrac{3}{2}$

$\Rightarrow S = \dfrac{2}{3}R$

It is given that the series combination of $R$ and $S$ is $2\ \text{k}\Omega = 2000\ \Omega$.

$R + S = 2000$

$R + \dfrac{2}{3}R = 2000$

$\dfrac{5R}{3} = 2000$

$R = 2000 \times \dfrac{3}{5} = 1200\ \Omega$

Answer: $1200\ \Omega$