In a potentiometer arrangement, a cell of emf `1.25V` gives a balance point at `35.0` cm length of the wire. If the cell is replaced by another cell and the balance point shifts to `63.0cm,` what is the emf of the second cell ?

In a potentiometer experiment of a cell of emf 1.25 V gives balancing length of 30 cm. If the cell is replaced by another cell, balancing length is found to be 40 cm. What is the emf of second cell ?

The balance point of a potentiometer with two cells in series is at l= 250 cm. And if one cell is reversed the balance point is at l_1 = 150 cm. The ratio of emfs of the two cells is

The figure shows a 2.0V potentiometer used for the determination of internal resistance of a 1.5V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Omega is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

The null point in a potentiometer with a cell of e.m.f. xi is obtained at a distance l on the wire. Then :

With a resistance P Omega in the left gap and another resistance Q Omega in the right gap of a meter bridge, the balance point is obtained at 40 cm. When a resistance of 6 Omega is connected across Q, the balance point shifts to 50 cm. The value of resistance P is

In a potentiometer experiment, the balancing point with a cell is at a length 240 cm. on shunting the cell with a resistance of 2 Omega the balancing length becomes 120 cm. The internal resistance of the cell is

In the potentiometer for measuring the emf of the cell, at null point no current flows through :