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

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

In the figure shown wire AB has a length of 100 cm and resistance 8Omega . Find the balance point length l .

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 ?

Calculate the value of unknown potential V for the given potentiometer circuit .The total length 400 cm of potentiometer wire has a resistance of 10 Omega and the balance point in obtained at a length of 240 cm

A cell of e.m.f. 1.2 V is balanced by 150 cm of potentiometer wire. When the cell is shunted by a resistance of 4 Omega , the balancing length is reduced by 30 cm. What is the internal resistance of the cell ?

A potentiometer wire has resistance of per unit length of 0.1Omega//m . A cell of e.m.f. 1.5V balances against 300cm length of the wire. Find the current in the potentiometer wire.