$\Rightarrow \mathrm{R}_{2}=1000-\mathrm{R}_{1}$

On balancing condition

$\mathrm{R}_{1}(100-I)=\left(1000-\mathrm{R}_{1}\right) l$      ...$(i)$

On Interchaanging resistance balance point shifts left by $10 \,cm$

On balancing condition $\left(1000-R_{1}\right)(110-l)=R_{1}(l-10)$

or, $\mathrm{R}_{1}(l-10)=\left(1000-\mathrm{R}_{1}\right)(110-l) \ldots(\mathrm{ii})$

Dividing eqn $(i)$ by $(ii)$

$\frac{100-l}{l-10}=\frac{l}{110-l}$

$\Rightarrow \quad(100-l)(110-l)=l(l-10)$

$ \Rightarrow 11000 - 100l - 110l + {l^2} = {l^2} - 10l$

$\Rightarrow 11000=200 l$

or, $l=55$

Putting the value of $'l'{\rm{in}}\,{\rm{eqn}}(i)$

$\mathrm{R}_{1}(100-55)=\left(1000-\mathrm{R}_{1}\right) 55$

$\Rightarrow \mathrm{R}_{1}(45)=\left(1000-\mathrm{R}_{1}\right) 55$

$\Rightarrow \mathrm{R}_{1}(9)=\left(1000-\mathrm{R}_{1}\right) 11$

$\Rightarrow 20 \mathrm{R}_{1}=11000$

$\therefore \quad \mathrm{R}_{1}=550\, \mathrm{K\Omega}$