Residence of 100 cm long potentiometer wire is `10 Omega`, it is connected to a battery (2 volt) and a resistance R in series. A source of 10 mV gives null point at 40 cm length, then external resistance R is

Resistance of 100 cm long potentiometer wire is 10 Omega . It is connected with external resistance R and a cell with emf 2V and negligible internal resistance. While balancing 10 mV emf in the secondary circuit , null point is obtained at 40 cm . Find this external resistance.

AB is a potentiometer wire of length 100 cm and its resistance is 10 Omega . It is connected in series with a battery of emf 2V and resistance 40 Omega . If a source of unknown emf E is balanced by 40 cm length of the potentiometer wire, the value of E is :-

To the potentiometer wire of L and 10Omega resistance, a battery of emf 2.5 volts and a resistance R are connected in series. If a potential difference of 1 volt is balanced across L/2 length, the value of R is Omega will be

To the potentiometer wire of L and 10Omega resistance, a battery of emf 2.5 volts and a resistance R are connected in series. If a potential difference of 1 volt is balanced across L/2 length, the value of R is Omega will be

To the potentiometer wire of L and 10Omega resistance, a battery of emf 2.5 volts and a resistance R are connected in series. If a potential difference of 1 volt is balanced across L/2 length, the value of R is Omega will be

The resistance of a 10 m long potentiometer wire is 20 Omega . It is connected in series with a 3 V battery and 10 Omega resistor. The potential difference between two points separated by distance 30 cm is equal to .......

In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:-