On interchanging the resistances, the balance point of a meter bridge shifts to the left by \(10\ cm\). The resistance of their series combination is \(1\ k\Omega\). How much was the resistance on the left slot before interchanging the resistances?  ..................  \(\Omega\)

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

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 \('l'{\rm{in}}\,{\rm{eqn}}(i)\)

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

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

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

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

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