Experimental Physics

JEE Main level, single correct, medium

51 of 70

Question 51 of 70

In a meter bridge, the wire of length 1m1m has a non-uniform cross-section such that, the variation dRdl\dfrac{dR}{dl} of its resistance RR with length ll is dRdl∝1l.\dfrac{dR}{dl}\propto \dfrac{1}{\sqrt{l}}. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P.P. What is the length AP?AP?

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Answer: (D) 0.25m0.25m

According to question meter bridge have non-uniform wire, where its resistance has a variation with length as dRdl∝1l\dfrac{dR}{dl}\propto \dfrac{1}{\sqrt{l}}
dRdl=kl...(1)\dfrac{dR}{dl}=\dfrac{k}{\sqrt{l}}...\left(1\right)
, where kk is proportionality constant
Let resistance of the left part APAP and right part PBPB of the wire is RAP,RPB{R}_{AP},{R}_{PB} respectively.
From balanced condition of wheat stone bridge is:
PQ=RS\dfrac{P}{Q}=\dfrac{R}{S}
So, from the above condition for given meter bridge is
R′R′=RApRpB⇒RAp=RpB...(2)\dfrac{{R}^{'}}{{R}^{'}}=\dfrac{{R}_{Ap}}{{R}_{pB}}\Rightarrow {R}_{Ap}={R}_{pB}...\left(2\right)
Now from equation (1)\left(1\right), on integrating ∫dR=k∫dll...(3)\int dR=k\int \dfrac{dl}{\sqrt{l}}...\left(3\right)
for APAP part of the wire of length extends from oo to ll from the equation (3)\left(3\right)
RAP=∫0RAP∫dR=k∫0ldll=k∫0l(l−12)dl=k[l−12+1]0l−12+1⇒RAP=2kl{R}_{AP}={\int }_{0}^{{R}_{AP}}\int dR=k{\int }_{0}^{l}\dfrac{dl}{\sqrt{l}}=k{\int }_{0}^{l}\left({l}^{-\dfrac{1}{2}}\right)dl=k\dfrac{{\left[{l}^{-\dfrac{1}{2}+1}\right]}_{0}^{l}}{-\dfrac{1}{2}+1} \Rightarrow {R}_{AP}=2k\sqrt{l}
Now, for PBPB part of the wire of length extends from ll to 1m1m, from the equation (3)\left(3\right)
RPB=∫0RPBdR=k∫l1dll=k∫l1(l−12)dl=k[l−12+1]l1−12+1⇒RPB=2k[1−l]{R}_{PB}={\int }_{0}^{{R}_{PB}}dR=k{\int }_{l}^{1}\dfrac{dl}{\sqrt{l}}=k{\int }_{l}^{1}\left({l}^{-\dfrac{1}{2}}\right)dl=k\dfrac{{\left[{l}^{-\dfrac{1}{2}+1}\right]}_{l}^{1}}{-\dfrac{1}{2}+1} \Rightarrow {R}_{PB}=2k\left[1-\sqrt{l}\right]
From equation (2)\left(2\right) RAP=RPB{R}_{AP}={R}_{PB}
∴2l=2(1−l)⇒l=1−l⇒l=12⇒l=14=0.25m∴2\sqrt{l}=2\left(1-\sqrt{l}\right) \Rightarrow \sqrt{l}=1-\sqrt{l} \Rightarrow \sqrt{l}=\dfrac{1}{2}\Rightarrow l=\dfrac{1}{4}=0.25m

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