The potentiometer wire is of length 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of `20Omega`, the null point of it is found to be at 1000 cm.The resistance of potentiometer wire is

The length of a potentiometer wire is 600 cm and it carries a current of 40 mA . For a cell of emf 2V and internal resistance 10Omega , the null point is found to be at 500 cm . On connecting a voltmeter acros the cell, the balancing length is decreased by 10 cm The resistance of the voltmeter is

The length of a potentiometer wire is 600 cm and it carries a current of 40 mA . For a cell of emf 2V and internal resistance 10Omega , the null point is found to be ast 500 cm . On connecting a voltmeter acros the cell, the balancing length is decreased by 10 cm The voltmeter reading will be

The length of a potentiometer wire AB is 600 cm , and it carries a constant current of 40 mA from A to B . For a cell of emf 2 V and internal resistance 10 Omega , the null point is found at 500 cm from A . When a voltmeter is connected across the cell, the balancing length of the wire is decreased by 10 cm . Reading of the voltmeter is

The length of a potentiometer wire AB is 600 cm , and it carries a constant current of 40 mA from A to B . For a cell of emf 2 V and internal resistance 10 Omega , the null point is found at 500 cm from A . When a voltmeter is connected across the cell, the balancing length of the wire is decreased by 10 cm . Potential gradient along AB is

There is a potentiometer wire of length 1200 cm and 60 mA current is flowing in it. A battery of emf 5V and internal resistance of 20 ohm is balanced on potentiometer wire with balancing length 1000 cm. The resistance of potentiometer wire is

The length of the potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of emf 2 V and internal resistance 10Omega is balanced on this potentiometer the balance length is found to be 500 cm. The resistance of potentiometer wire will be

In a potentiometer circuit, the emf of driver cell is 2V and internal resistance is 0.5 Omega . The potentiometer wire is 1m long . It is found that a cell of emf 1V and internal resistance 0.5 Omega is balanced against 60 cm length of the wire. Calculate the resistance of potentiometer wire.

A potentiometer wire of length 10 m and resistance 30 Omega is connected in series with a battery of emf 2.5 V , internal resistance 5 Omega and an external resistance R . If the fall of potential along the potentiometer wire is 50 mu V mm^(-1) , then the value of R is found to be 23 n Omega . What is n ?

AB is a potentiometer wire of length 100 cm and its resistance is 10 Omega . It is connected in series with a battery of emf 2V and resistance 40 Omega . If a source of unknown emf E is balanced by 40 cm length of the potentiometer wire, the value of E is :-

The wire of potentiometer has resistance 4 Omega and length 1m. It is connected to a cell of e.m.f. 2V and internal resistance 1 Omega . The current flowing through the potentiometer wire is