Question

in a potentiometer arrangement a cell of emf 1.20 V gives a balance point at 30 cm length of the wire .this cell is now replaced by another cell of unknown emf .if the ratio of emf of the two cell is 1.5.calculate the difference in the balancing length of the Potentiometer wire in the two cases .​

Answer 1

E1/E2=1.5

E1=1.2V

L1=30cm

L2=?

E1/E2=L1/L2

L2/L1=E2/E1=1/1.5

L2=1/1.5×30cm

=20 cm

difference in the balancing length of potentiometer wire

=30 cm-20cm

=10cm

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Answer 2

HERE IS YOUR ANSWER

E1/E2=1.5

E1=1.2V

L1=30cm

L2=? ???

E1/E2=  L1/L2

L2/L1=   E2/E1=1/1.5

L2  = 1 /1.5×30cm

=   20 cm

NOW

Difference in the balancing length of potentiometer wire  

=    30 -20 = 10

=    10cm

hence Difference in the balancing length of potentiometer wire is 10 cm