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Question

A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ε and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in fig. shows no deflection is :

Shift 1 Jan 10 JEE Main 2019 Solved Paper For Physics


A

(13/24)L

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B

(5/12)L

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C

(11/24)L

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D

(11/12)L

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Solution

The correct option is A

(13/24)L


Step 1: Given data

Resistance =12r

Internal resistance of cell =3r

Step 2: Find potential drop

According to ohm law, Potential, V is directly proportional to the current, I

V=IRI=VR [R is the resistance.]

Current in potentiometer wire, i

i=(ε/12r+r)i=(ε/13r) [ ε is the EMF]

Voltage across the wire, VAJ

VAJ=i×RAJVAJ=i×(12rL)x [RAJ resistance across the wire, x is the length AJ, L is the length of the potentiometer.]

As null point is at J, so potential drop across length is equal to EMF of the cell

VAJ=ε2[(ε12r+r)×(12rL)x]=ε2(ε13r)×(12rL)x=ε212ε13xL=ε2x=13L24

Hence, option A is correct.


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