NEET Questions

The emf of a cell is E V, and its internal resistance is $1\Omega$. A resistance of $4\Omega$ is joined to battery in parallel. This is connected in secondary circuit of potentiometer. The balancing length is 160 cm. If 1V cell balances for 100 cm of potentiometer wire, the emf of cell $E$ is

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf’s is:

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf’s is:

A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F, because the method involves: