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QJEEM19Ph123d: JEE Main 2019 Physics - Class 12 JEEM19Ph123d Solution: In a meter bridge, the wire of length 1 m has a non-uniform…

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

JEE Main 2019 (Online) 12th January Morning Slot Physics - Current Electricity Question 253 English

Multiple Choice Options

  • A 0.3 m
  • B 0.25 m
  • C 0.35 m
  • D 0.2 m

Step-by-Step Verified Solution

For the given wire :

dR = C dℓℓ{{d\ell } \over {\sqrt \ell }} ,

where C = constant.

Let resistance of part

AP is R1 and PB is R2

∴\therefore    R′R′=R1R2{{R'} \over {R'}} = {{{R_1}} \over {{R_2}}}   or    R1 = R2     

By balanced WSB concept.

Now   ∫dR=c∫dℓℓ\int {dR = c\int {{{d\ell } \over {\sqrt \ell }}} }

∴\therefore   R1 = C ∫0ℓℓ−1/2dℓ\int\limits_0^\ell {{\ell ^{ - 1/2}}d\ell } = C.2. ℓ{\sqrt \ell }

R2 = C ∫ℓ1ℓ−1/2dℓ\int\limits_\ell ^1 {{\ell ^{ - 1/2}}d\ell } = C.(2 −- 2 ℓ{\sqrt \ell } )

Putting R1 = R2

C2 ℓ{\sqrt \ell } = C(2 −- 2 ℓ{\sqrt \ell } )

∴\therefore   2 ℓ{\sqrt \ell } = 1

ℓ{\sqrt \ell } = 12{1 \over 2}

i.e. ℓ\ell = 14{1 \over 4} m      ⇒\Rightarrow   0.25 m

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