Resistance of potentiometer wire is `0.1Omega//cm`. A cell of emf 1.5V is balanced at 300cm on this potentiometer wire. Calculate the current and balancing length for another cell of emf 1.4V on the same potentiometer wire.

The resistance per unit length of a wire is 1Omega//"m" . A Leclanche cell of e.m.f. 1.45 V is balanced against 2.9 m length of potentiometer wire. The current through the wire is

A potentiometer wire has a resistance per unit length of 0.1 Omega//m . A cell of emf 1.5V balance against 300 cm length of the wire. Find the current through potentiometer wire.

In a potentiometer experiment, e.m.f. of a cell is balanced by a length of 150 cm on the potentiometer wire. When the cell is shunted by a 10 Omega resistance, the balancing length reduces to 90 cm. What is the internal resistance of the cell?

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

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

In figure,the potentiometer wire A B has a resistance of 5 Omega and length 10m. The balancing length AJ for the cell of e.m.f.of 0.4 V is

When a balance point is obtained in a potentiometer for finding the internal resistance of a cell, the current through the potentiometer wire is due to

When a resister of 10Omega is connected across a cell, its terminal potential difference is balanced by 150cm of potentiometer wire and when 20Omega resistance is connected across the cell, terminal potential difference is balanced by 175cm of the same potentiometer wire. Find the balancing length when the cell is in open circuit and also find the internal resistance of the cell.

When a resistor of 5 Omega resistance is connected across a cell,its terminal potential difference is balanced by 150 cm of the wire and when a resistor of 10 Omega resistance is connected across cell, the terminal p.d. is balanced by 175 cm of same potentiometer wire. Find the balancing length when the cell is in open circuit and the internal resistance of acell.

A potentiometer wire of length 10 m and resistance 9 ohm is connected to a battery of e.m.f. 2.1 volt having internal resistance 1.5 Omega . Find the potential gradient along the wire and the balancing length for a cell of e.m.f. 1.08 volt .