1.

In a potentiometer whose wire resistance is `10Omega`. The potential fall per `cm` is `V "volts"` is reduced to `(V)/(4)"volt"//cm`. The resistance that must be connected in series with the potentiometer wire isA. `40Omega`B. `30Omega`C. `20Omega`D. `10Omega`

Answer» Correct Answer - A
`(V)/((V)/(4))=(R^(1))/(10)implies R^(1)=40Omega`


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