b
\(\mathrm{R}_{1}+\mathrm{R}_{2}=1000\)

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

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 \('l'{\rm{in}}\,{\rm{eqn}}(i)\)

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

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

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

\(\Rightarrow 20 \mathrm{R}_{1}=11000\)

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