A potentiometer wire has resistance `40Omega` and its length is 10m. It is connected by a resistance of `760Omega` in series if emf of battery is 2V then potential gradient is:-

A potentiometer wire has a length 2 m and resistance of 10 Omega . It is coonnected in series with a resistance of 990 Omega and a cell of e.m.f. 2 V. The potential gradient along the wire is

A potentiometer wire of length 10m and resistance 20Omega is connected in series with another resistance of 80Omega and a battery of emf 4V . The potential gradient on the wire will be (in mV//cm )

A potentiometer wire 10m long has a resistance of 40Omega . It is connected in series with a resistances box and a 2 v storage cell. If the potential gradient along the wire is 0.01(V)/(m) the resistance unplugged in the box is

The wire of potentiometer has resistance 4 Omega and length 1m. It is connected to a cell of e.m.f. 2V and internal resistance 1 Omega . The current flowing through the potentiometer wire is

A potentiometer wire 10 m long has a resistance 40ohms. It is put in series to a resistance 760ohms and connected to a 2volt battery. The potential gradient in the wire is –

A potentiometer wire has length 10m and resistance 10Omega . It is connected to battery of EMF 11 volt and internal resistance 1Omega , then the potential gradient in the wire is

The wire of potentiometer has resistance 4Omega and length Im. It is connected to a cell of EMF 2V and internal resistance 1Omega . Ifa cell ofEMF 1.2V is balanced byit, the balancing length will be:

The wire of potentiometer has resistance 4Omega and length 1m . It is connected to a cell of emf 2 volt and internal resistance 1Omega . If a cell of emf 1.2 volt is balanced by it, the balancing length will be

A potentiometer wire is 10 m long and has a resistance of 18Omega . It is connected to a battery of emf 5 V and internal resistance 2Omega . Calculate the potential gradient along the wire.