The wire AB of a meter bridge changes linearly form radius r to 2r from left end to right end. Where should the free end of the galvanometer be connected on AB so that the deflection in the galvanometer is zero?

Let the galvanometer be connected at a point $\mathrm{x}={\mathrm{x}}_1$ from end A where $x=0.$ Let ${\mathrm{R}}_1,{\mathrm{R}}_2$ are the resistances of left and right parts of that point. Consider an element of thickness $dx$ at a distance $x$ from end A of radius $r_x$

From the figure,

$\frac{r_x-r}{x}=\frac{2r-r}{l}\Rightarrow r_x=\frac{r}{l}x+r$

$\text{ as }\mathrm{l}=1\mathrm{m},{\mathrm{r}}_{\mathrm{x}}=\frac{\mathrm{r}}{1}\mathrm{x}+\mathrm{r}=\mathrm{r}(1+\mathrm{x})$

Resistance of this element will be ${\mathrm{dR}}_{\mathrm{x}}=\frac{\rho \mathrm{dx}}{\pi {\mathrm{r}}_{\mathrm{x}}^2}$

$\therefore {\mathrm{R}}_1={\int }_0^{{\mathrm{x}}_1} \frac{\rho \mathrm{dx}}{\pi {\mathrm{r}}_{\mathrm{x}}^2}=\frac{\rho }{\pi }{\int }_0^{{\mathrm{x}}_1} \frac{\mathrm{dx}}{\left[\mathrm{r}\right(1+\mathrm{x}){]}^2}$

$=\frac{\rho }{\pi r^2}{\int }_0^{x_1} (1+x{)}^{-2}dx=\frac{\rho }{\pi r^2}\left[1-\frac{1}{1+x_1}\right]$

${\mathrm{R}}_2=\frac{\rho }{\pi }{\int }_{{\mathrm{x}}_1}^1 \frac{\mathrm{dx}}{{\mathrm{r}}_{\mathrm{x}}^2}=\frac{\rho }{\pi {\mathrm{r}}^2}{\int }_{{\mathrm{x}}_1}^1 (1+\mathrm{x}{)}^{-2}\mathrm{dx}$

$=\frac{\rho }{\pi r^2}\left[\frac{1}{1+{\mathrm{x}}_1}-\frac{1}{1+1}\right]=\frac{\rho }{\pi r^2}\left[\frac{1}{1+{\mathrm{x}}_1}-\frac{1}{2}\right]$

For zero deflection of the galvanometer,

$\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{4}{4}\Rightarrow {\mathrm{R}}_1={\mathrm{R}}_2$

$\therefore \frac{\rho }{\pi r^2}\left[1-\frac{1}{1+{\mathrm{x}}_1}\right]=\frac{\rho }{\pi {\mathrm{r}}^2}\left[\frac{1}{1+{\mathrm{x}}_1}-\frac{1}{2}\right]$

$\Rightarrow {\mathrm{x}}_1=\frac{1}{3}\mathrm{m}=33.33\mathrm{cm}$