In a potentiometer arrangement a cell of emf `1.25` V givesn a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shift to 63 cm, what is the emf of the second cell ?

In a potentiometer a cell of emf 1.5 V balances at a lengths of 270 c. if another cellbalances at 360 cm for the same current its emf will be

Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Omega 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

show 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.

In the determination of internal resistance of a cell of e.in.f. 1.5 V with potentiometer, the balancing length is 75 cm in the open circuit and when a resistance of 10 Omega is used the balancing length is found to be 65 cm. Calculate the internal resistance of the cell.

A potentiometer wire has a length 400 cm and a resistance of 20 Omega . It is connected in series with an external resistance R and a cell of e.m.f. 1.5 V and internal resistance 1 Omega . A source of e.m.f.6 mV is balanced against a length of 30 cm of the potentiometer wire. Find R. According to the principle of potentiometer Vpropl . i.e., V = ilp where l'is the balancing length & p the resistance per cm of the potentiometer wire.