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A poterntimeter wire is 100 cm long and ...

A poterntimeter wire is 100 cm long and a constant potential difference is maintained acrooss it . Two cells are connected in seires first to support one another and then in opposite direction . The balance points are obtained at 50 cm and 10 cm from the positive end the wire in the two cases . The ratio of emf's is

A

`5:4`

B

`3:4`

C

`3:2`

D

`5:1`

Text Solution

Verified by Experts

The correct Answer is:
C
if the emf of the two cells are `E_(1)and E_(2)` then ,
`E_(1)+E_(2)=50k`
and `e_(1)-E_(2)=10k` [ where k = condtant ]
`therefore(E_(1)+E_(2))/(E_(1)-E_(2))=(5)/(1)`
`"or, " (E_(1))/(E_(2))=(5+1)/(5-1)=(3)/(2)`
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CHHAYA PUBLICATION-KIRCHHOFF'S LAWS AND ELECTRICAL MEASUREMENT -EXAMINATION ARCHIVE (NEET)
  1. A poterntimeter wire is 100 cm long and a constant potential differenc...

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