In a meter bridge,the wire of length 1, m has | vedclass

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(C) Given $\frac{dR}{d\ell} = \frac{k}{\sqrt{\ell}}$,where $k$ is a constant.Integrating from $\ell = 0$ to $\ell = 1\, m$,the total resistance $R_{AB

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$0.25$

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(C) Given $\frac{dR}{d\ell} = \frac{k}{\sqrt{\ell}}$,where $k$ is a constant.Integrating from $\ell = 0$ to $\ell = 1\, m$,the total resistance $R_{AB}$ of the wire is:$R_{AB} = \int_{0}^{1} \frac{k}{\sqrt{\ell}} d\ell = k [2\sqrt{\ell}]_{0}^{1} = 2k$.Let the length $AP = L$. The resistance of segment $AP$ is $R_{AP} = \int_{0}^{L} \frac{k}{\sqrt{\ell}} d\ell = k [2\sqrt{\ell}]_{0}^{L} = 2k\sqrt{L}$.The resistance of segment $PB$ is $R_{PB} = \int_{L}^{1} \frac{k}{\sqrt{\ell}} d\ell = k [2\sqrt{\ell}]_{L}^{1} = 2k(1 - \sqrt{L})$.For a balanced meter bridge with equal resistances $R'$ in the gaps,the condition is $\frac{R'}{R_{AP}} = \frac{R'}{R_{PB}}$,which implies $R_{AP} = R_{PB}$.Therefore,$2k\sqrt{L} = 2k(1 - \sqrt{L})$.$\sqrt{L} = 1 - \sqrt{L} \implies 2\sqrt{L} = 1 \implies \sqrt{L} = 0.5$.$L = (0.5)^2 = 0.25\, m$.

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