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Question

If the wire has resistivity ρ and cross sectional area A, the equivalent resistance between P and Q is?
695755_426a3253334046ac84ac553c9b1a9984.png

A
ρl2A
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B
2ρlA
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C
2ρlA
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D
ρlA
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Solution

The correct option is A ρl2A

We need to find the total length of the wire through which we can calculate the resultant resistance.
In the figure, VA=VF {Some potential }
that there be no current through arm AF and hence we can calculate that, similarly there will be no current through arm CD because of the same potential drop. Hence, we can also calculate that part.
Now, we will calculate the resistance of each arm,
RPA=ρl2A=RPF
Also, resistance ABC is in parallel connection with resistance AC.
1R1=1RABC==1RAC=Aρl+2Aρl
=A(2+1)ρl
R1=ρlA(2+1)
As we can deduce from the figure, that resistance due to PA,(ABC,AC),Cθ are in series hence, their total ressitance is given by,
R2=RPA+R1+RCθ=ρl2A+ρlA(2+1)+ρl2A
=ρlA(1+12+1)
=ρlA(2+22+1)
Similarly, we will calculate the resistance for part PFEDθ which will be same as R2
Now, R2 and R3 are resistance in parallel
1Req=1R2+1R3=2R2
Req=R22=ρl2A(2+22+1)
Req=ρl2AΩ

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