In a meter bridge experiment, initially the jockey is at null point. Now resistance $R_1\text{\hspace{0.17em}and\hspace{0.17em}}{R}_2$  is interchanged. Shift in the position of  jockey is ____cm

Meter bridge works on the principle of wheat stone bridge
From diagram  $R_1=2\Omega$, $R_2=3\Omega$
$\text{Initially\hspace{0.17em}}\frac{R_1}{R_2}=\frac{2}{3}=\frac{l}{100-l}\Rightarrow l=40\text{\hspace{0.17em}cm}$

$\text{Now}\frac{R_2}{R_1}=\frac{3}{2}=\frac{l^{\text{'}}}{100-l^{\text{'}}}$

$\Rightarrow l^{\text{'}}=60\text{\hspace{0.17em}cm}$

Shift, $l^{\text{'}}-l=60-40=20\text{\hspace{0.17em}cm}$

Therefore, the correct answer is 20.00