JEE Main 2019 — Physics Mechanics

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

$$\frac{dR}{dl}=\frac{k}{\sqrt{l}}...(1)$$dldR​=l​k​...(1)

, where $k$k is proportionality constant

Let resistance of the left part $AP$AP and right part $PB$PB of the wire is ${R}_{AP},{R}_{PB}$RAP​,RPB​ respectively.

From balanced condition of wheat stone bridge is:

$$\frac{P}{Q}=\frac{R}{S}$$QP​=SR​

So, from the above condition for given meter bridge is

$$\frac{{R}^{'}}{{R}^{'}}=\frac{{R}_{Ap}}{{R}_{pB}}\Rightarrow {R}_{Ap}={R}_{pB}...(2)$$R′R′​=RpB​RAp​​⇒RAp​=RpB​...(2)

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

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

$${R}_{AP}={\int }_{0}^{{R}_{AP}}\int dR=k{\int }_{0}^{l}\frac{dl}{\sqrt{l}}=k{\int }_{0}^{l}({l}^{-\frac{1}{2}})dl=k\frac{{[{l}^{-\frac{1}{2}+1}]}_{0}^{l}}{-\frac{1}{2}+1} \Rightarrow {R}_{AP}=2k\sqrt{l}$$RAP​=∫0RAP​​∫dR=k∫0l​l​dl​=k∫0l​(l−21​)dl=k−21​+1[l−21​+1]0l​​⇒RAP​=2kl​

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

$${R}_{PB}={\int }_{0}^{{R}_{PB}}dR=k{\int }_{l}^{1}\frac{dl}{\sqrt{l}}=k{\int }_{l}^{1}({l}^{-\frac{1}{2}})dl=k\frac{{[{l}^{-\frac{1}{2}+1}]}_{l}^{1}}{-\frac{1}{2}+1} \Rightarrow {R}_{PB}=2k[1-\sqrt{l}]$$RPB​=∫0RPB​​dR=k∫l1​l​dl​=k∫l1​(l−21​)dl=k−21​+1[l−21​+1]l1​​⇒RPB​=2k[1−l​]

From equation $(2)$(2) ${R}_{AP}={R}_{PB}$RAP​=RPB​

$$\therefore 2\sqrt{l}=2(1-\sqrt{l}) \Rightarrow \sqrt{l}=1-\sqrt{l} \Rightarrow \sqrt{l}=\frac{1}{2}\Rightarrow l=\frac{1}{4}=0.25m$$∴2l​=2(1−l​)⇒l​=1−l​⇒l​=21​⇒l=41​=0.25m