In given figure, the potentiometer wire AB has a resistance of `5 Omega` and length 10 m . The balancing length AM for the emf of 0.4 V is

In given figure, the potentiometer wire AB has a resistance of 5Omega and length 10 m. The balancing length AM for the e.m.f. of 0.4 V is

In figure,the potentiometer wire A B has a resistance of 5 Omega and length 10m. The balancing length AJ for the cell of e.m.f.of 0.4 V is

In figure AB is a potentiometer wire of length 10 m and resistance 1Omega . The balancing length for 2V potential drop is 18 m. The e.m.f. of the battery E is

A potentiometer wire has a resistance of 20 Omega and length 10 m. It is connected in series with a battery of emf 2.5 V and negligible internal resistance, and an external resistance of 5 Omega . Fall of potential in mV per metre is

In figure, AB is a potentiometer wire, length 10 m and resistance 2 Omega with open the balancing length is 5.5 m. However, on closing key K the balancing length reduces to 5 m. The initial resistance of the cell E_(1) is

A cell of e.m.f. 2 V and negligible internal resistance is connected to a potentiometer wire of resistance 10 Omega and length 4 m. The potential difference per unit length (potential gradient ) of the wire is

A cell of e.m.f. 2 V and negligible internal resistance is connected to a potentiometer wire of resistance 10 Omega and length 4 m. The potential difference per unit length (potential gradient ) of the wire is

In an experiment to find the internal resistance of a cell by a potentiometer, a balance was obtained for 50 cm length of the potentiometer wire, with a cell of e.m.f. 2V. When the cell was shunted by a resistance of 2 Omega , the balancing length of the potentiometer wire was 40 cm. What was the internal resistance of the cell ?