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.

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 of length 10 m is connected in series with a battery the emf of a cell balances against 250 cm length of wire if length of potentiometer wire is increased by 1 m then new balancing length of wire will be

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.

A potentiometer wire of length 100 cm has a resistance of 10Omega . It is connected in series with a resistance and an accumulator of e.m.f. 2 V and negligible internal resistance. A source of e.m.f. 10 mV is balanced against a 40 cm length of the potentiometer wire. The value of the external resistance is

In given figure, the potentiometer wire AB has a resistance of 5Omega and length 10 m. The balancing length AM for the e.m.f. of 0.4 V is

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

An accumulator of 5 volt is connected through a resistance of 40Omega to a potentiometer wire 10 m long and of resistance 10Omega . For a cell, the null point is found at a length of 8 m from the common terminal. The current through the wire is

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?

A potential drop per unit length along a wire is 5 xx 10^-3 V//m . If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.