In a meter bridge, the wire of length 1m has a non-uniform

(3) We have given

$\frac{\mathrm{dR}}{\mathrm{d} \ell} \propto \frac{1}{\sqrt{\ell}} \Rightarrow \frac{\mathrm{dR}}{\mathrm{d} \ell}=\mathrm{k} \times \frac{1}{\sqrt{\ell}}$ (where $\mathrm{k}$ is constant)

$\mathrm{dR}=\mathrm{k} \frac{\mathrm{d} \ell}{\sqrt{\ell}}$

Let $R_{1}$ and $R_{2}$ be the resistance of AP and PB respectively.

Using wheatstone bridge principle

$\therefore \frac{\mathrm{R}^{\prime}}{\mathrm{R}^{\prime}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ or $\mathrm{R}_{1}=\mathrm{R}_{2}$

Now, $\int \mathrm{d} \mathrm{R}=\mathrm{k} \int \frac{\mathrm{d} \ell}{\sqrt{\ell}}$

$\therefore \mathrm{R}_{1}=\mathrm{k} \int_{0}^{\ell} \ell^{-1 / 2} \mathrm{~d} \ell=\mathrm{k} \cdot 2 \cdot \sqrt{\ell}$

$\mathrm{R}_{2}=\mathrm{k} \int_{\ell}^{1} \ell^{-1 / 2} \mathrm{~d} \ell=\mathrm{k} \cdot(2-2 \sqrt{\ell})$

Putting $\mathrm{R}_{1}=\mathrm{R}_{2}$

$\mathrm{k} 2 \sqrt{\ell}=\mathrm{k}(2-2 \sqrt{\ell})$

$\therefore 2 \sqrt{\ell}=1$

$\sqrt{\ell}=\frac{1}{2}$

i.e., $\ell=\frac{1}{4} \mathrm{~m} \Rightarrow 0.25 \mathrm{~m}$