In a potentiometer arrangment, a cell of emf 1.25 V 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 63.0 cm`, what is the emf of the second cell ?

In a potentiometer experiment , a particular cell produces balance in the wire at 560 cm . If a 10Omega resistance is connected with the cell in parallel ,then this balanced lendth change to 412 cm . Find out the internal resistance of the cell .

The length of a potentimeter wire is 10cm . A cell of emf E is balanced at a lenth (10)/(3) cm from the positive end of the wire .If the length of the wire is increased by 5 cm , at what distance (in cm ) from positive end will the same cell give a balance point ?

In a potentimeter arrangment for determining the emf of a cell , the balance point of the cel in open circuit is 350 cm . When a resistance of 9Omega is used in the external circuit of h the balance point shift to 300cm . Determine the internal reistance of the cell .

Fig. shows a 2.0V potentiometer used for the determination of internal resistance of 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 experiment of calibration of voltmeter, a standard cell of e.m.f. 1.1 Volt is balanced against 440 cm of potentiometer wire. The potential difference across the ends of a resistance is found to balance against 220 cm of the wire. The corresponding reading of voltmeter is 0.5 volt. The error in the reading of voltmeter will be

Fig. shows a potentiometer with a cell of 2.0V and internal resistance 0.40Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02V (for very moderate currents upto a few pA) gives a balance point at 67.3 cm length of the wire. Io ensure very low currents drawn from the standard cell, a very high resistance of 600kOmega is put in series with it which is shorted close to the balance point. The standard cell is then, replaced by a cell of unknown emf e and the balance point found similarly turns out to be at 82.3 cm length of the wire, Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

Fig. shows a potentiometer with a cell of 2.0V and internal resistance 0.40Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02V (for very moderate currents upto a few pA) gives a balance point at 67.3 cm length of the wire. to ensure very low currents drawn from the standard cell, a very high resistance of 600kOmega is put in series with it which is shorted close to the balance point. The standard cell is then, replaced by a cell of unknown emf e and the balance point found similarly turns out to be at 82.3 cm length of the wire, Is the balance point affected by this high resistance?

Fig. shows a potentiometer with a cell of 2.0V and internal resistance 0.40Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02V (for very moderate currents upto a few pA) gives a balance point at 67.3 cm length of the wire. Io ensure very low currents drawn from the standard cell, a very high resistance of 600kOmega is put in series with it which is shorted close to the balance point. The standard cell is then, replaced by a cell of unknown emf e and the balance point found similarly turns out to be at 82.3 cm length of the wire, What purpose does the high resistance of 600kOmega have?