A metallic conductor at 10⁰C connected in the left gap of meter bridge gives balancing length 40 cm. When the conductor is at 60⁰C, the balancing point shifts by ---cm, (temperature coefficient of resistance of the material of the wire is (1/220)/⁰C)

$\frac{{\mathrm{R}}_1}{\mathrm{R}}=\frac{\mathrm{\ell }}{100-\mathrm{\ell }}=\frac{40}{60}=\frac{2}{3}...\left(1\right)$
From ${\mathrm{\alpha }}_{\mathrm{R}}=\frac{{\mathrm{R}}_2-{\mathrm{R}}_1}{{\mathrm{R}}_1{\mathrm{t}}_2-{\mathrm{R}}_2{\mathrm{t}}_1}\Rightarrow \frac{1}{220}=\frac{{\mathrm{R}}_2-{\mathrm{R}}_1}{60{\mathrm{R}}_1-10{\mathrm{R}}_2}$
$\begin{array}{l}\Rightarrow 60{\mathrm{R}}_1-10{\mathrm{R}}_2=220{\mathrm{R}}_2-220{\mathrm{R}}_1\\ \Rightarrow 280{\mathrm{R}}_1=230{\mathrm{R}}_2\Rightarrow \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{23}{28}...\left(2\right)\end{array}$
From $\frac{{\mathrm{R}}_1}{\mathrm{R}0}=\frac{2}{3}\Rightarrow \mathrm{R}=\frac{3}{2}{\mathrm{R}}_1$
Substituting in (1) $\frac{{\mathrm{R}}_2}{\frac{3}{2}{\mathrm{R}}_1}=\frac{{\mathrm{\ell }}^1}{100-{\mathrm{\ell }}^1}$ 
$\begin{array}{l}\Rightarrow \frac{2}{3}\times \frac{28}{23}=\frac{{\mathrm{\ell }}^1}{100-{\mathrm{\ell }}^1}\Rightarrow \frac{2}{3}\times \frac{28}{23}=\frac{{\mathrm{\ell }}^1}{100-{\mathrm{\ell }}^1}\\ \Rightarrow 5600-556{\mathrm{\ell }}^1=69{\mathrm{\ell }}^1;125{\mathrm{\ell }}^1=5600\\ {\mathrm{\ell }}^1=\frac{5600}{125}=44.8\mathrm{cm};1^1-\mathrm{l}=44.8-40=4.8\mathrm{cm}\end{array}$