n identical resistance are taken in which $\frac{\mathrm{n}}{2}$ resistors are joined in series in the left gap and the remaining $\frac{\mathrm{n}}{2}$ resistances are joined in parallel in the right gap of a metre bridge. Balancing length in cm is

Meter bridge is shown in the figure below,

$\text{R\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{R}}_{\text{1}}}{\text{2}}\text{+}\frac{{\text{R}}_{\text{1}}}{\text{2}}\text{\hspace{0.17em}+\hspace{0.17em}}\frac{{\text{R}}_{\text{1}}}{\text{2}}\text{\hspace{0.17em}+\hspace{0.17em}}\mathrm{......}\frac{\text{n}}{\text{2}}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}times\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{R}}_{\text{1}}\text{n}}{\text{2}}$

and $\frac{\text{1}}{\text{S}}\text{=}\frac{\text{1}}{{\text{R}}_{\text{1}}}\text{+}\frac{\text{1}}{{\text{R}}_{\text{1}}}\text{\hspace{0.17em}+\hspace{0.17em}}\frac{\text{1}}{{\text{R}}_{\text{1}}}\text{\hspace{0.17em}+\hspace{0.17em}}\mathrm{.......}\frac{\text{n}}{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}times\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}}\frac{\text{n}}{{\text{2R}}_{\text{1}}}$

$\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}s\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{2R}}_{\text{1}}}{\text{n}}$

If be the balancing length in the meter bridge wire, then

$\frac{R}{S}=\frac{1}{100-l}\Rightarrow \frac{\frac{R_{1}n}{2}}{\frac{2R_1}{n}}=\frac{l}{100-l}$

$l=\frac{{\text{100n}}^{\text{2}}}{{\text{n}}^{\text{2}}\text{+4}}$