JEE Main 2019 — Experimental Physics Question with Solution

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

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

, 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:

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

So, from the above condition for given meter bridge is

$\frac{R^{\text{'}}}{R^{\text{'}}}=\frac{R_{Ap}}{R_{pB}} \Rightarrow R_{Ap} = R_{pB} ...\left(2\right)$

Now from equation $\left(1\right)$, on integrating  $\int dR=k\int \frac{dl}{\sqrt{l}} ...\left(3\right)$

 for $AP$ part of the wire of length extends from $o$ to $l$ from the equation $\left(3\right)$

$$\begin{gathered} R_{AP} ={\int }_0^{R_{AP}}\int dR=k{\int }_0^l\frac{dl}{\sqrt{l}} =k{\int }_0^l \left(l^{-\frac{1}{2}}\right) dl =k \frac{{\left[l^{-{\displaystyle \frac{1}{2}}+1}\right]}_0^l}{-\frac{1}{2}+1} \\ \Rightarrow R_{AP} =2k\sqrt{l} \end{gathered}$$

Now, for $PB$ part of the wire of length extends from $l$ to $1 m$, from the equation $\left(3\right)$

$$\begin{gathered} R_{PB} ={\int }_0^{R_{PB}}dR=k{\int }_l^1\frac{dl}{\sqrt{l}} =k{\int }_l^1 \left(l^{-\frac{1}{2}}\right) dl =k \frac{{\left[l^{-{\displaystyle \frac{1}{2}}+1}\right]}_l^1}{-\frac{1}{2}+1} \\ \Rightarrow R_{PB} =2k\left[1-\sqrt{l}\right] \end{gathered}$$ 

From equation $\left(2\right)$ $R_{AP}=R_{PB}$
$$\begin{gathered} \therefore 2\sqrt{l}=2\left(1-\sqrt{l}\right) \\ \Rightarrow \sqrt{l}=1-\sqrt{l} \\ \Rightarrow \sqrt{l}=\frac{1}{2} \Rightarrow l=\frac{1}{4}=0.25 m \end{gathered}$$