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

According to question meter bridge have non-uniform wire, where its resistance has a variation with length as $\dfrac{dR}{dl} ∝ \dfrac{1}{\sqrt{l}}$

$\dfrac{dR}{dl} = \dfrac{k}{\sqrt{l}} . . . \left(1\right)$

, where $k$ is proportionality constant
Let resistance of the left part $AP$ and right part $PB$ of the wire is ${R}_{AP}, {R}_{PB}$ respectively.

From balanced condition of wheat stone bridge is:

$\dfrac{P}{Q} = \dfrac{R}{S}$

So, from the above condition for given meter bridge is

$\dfrac{{R}^{'}}{{R}^{'}} = \dfrac{{R}_{Ap}}{{R}_{pB}} ⇒ {R}_{Ap} = {R}_{pB} . . . \left(2\right)$

Now from equation \left(1\right)$, on integrating$∫ dR = k ∫ \dfrac{dl}{\sqrt{l}} . . . \left(3\right)$</p><p> for$AP$part of the wire of length extends from$o$to$l$from the equation \left(3\right)$

${R}_{AP} = {∫}_{0}^{{R}_{AP}} ∫ dR = k{∫}_{0}^{l}\dfrac{dl}{\sqrt{l}} = k{∫}_{0}^{l}\left({l}^{- \dfrac{1}{2}}\right)dl = k{\left[{l}^{- \dfrac{1}{2} + 1}\right]}_{0}^{l}- \dfrac{1}{2} + 1 ⇒ {R}_{AP} = 2k\sqrt{l}$

Now, for $PB$ part of the wire of length extends from $l$ to $1m$, from the equation \left(3\right)$</p><p>${R}_{PB} = {∫}_{0}^{{R}_{PB}}dR = k{∫}_{l}^{1}\dfrac{dl}{\sqrt{l}} = k{∫}_{l}^{1}$\left({l}^{- \dfrac{1}{2}}\right)dl = k{\left[{l}^{- \dfrac{1}{2} + 1}\right]}_{l}^{1}- \dfrac{1}{2} + 1 ⇒ {R}_{PB} = 2k\left[1 - \sqrt{l}\right]$

From equation \left(2\right)${R}_{AP} = {R}_{PB}$
$∴ 2\sqrt{l} = 2\left(1 - \sqrt{l}\right) ⇒ \sqrt{l} = 1 - \sqrt{l} ⇒ \sqrt{l} = \dfrac{1}{2} ⇒ l = \dfrac{1}{4} = 0 . 25m$