The e.m.f. E of the battery is balanced by p. d. across 75 cm of a potentiometer wire. For a standard cell of e.m.f. 1.02 V, the balancing length is 50 cm. The value of E is

The emf of a battery A is balanced by a length 80 cm on a potentio meter wire. The emf of a standard cell 1 v is balanced by 50 cm. the emf of A is

The emf of a battery A is balanced by a length of 80cm on a potentio meter wire. The emf of a standard cell 1v is balanced by 50cm. The emf of A is

The emf of a battery A is balanced by a length of 80cm on a potentio meter wire. The emf of a standard cell 1v is balanced by 50cm. The emf of A is

A 1Ω resistance in series with an ammeter is balanced by 75 cm of potentiometer wire. A standard cell of emf 1.02 V is balanced by 50 cm. The ammeter shows a reading of 1.5 A. Then the error in ammeter reading is:

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?

The e.m.f. of a cell is balanced by fall of potential along 281 cm of potentiometer wire. If a resistance of 8 Omega is connected across the cell , the balancing length is found to be 151 cm. Calculate internal resistance of a cell.

A cell of e.m.f. 1.2 V is balanced by 150 cm of potentiometer wire. When the cell is shunted by a resistance of 4 Omega , the balancing length is reduced by 30 cm. What is the internal resistance of the cell ?

A cell of e.m.f. 1 V gives a balance point at 40 cm length of a potentioeter wire. For another cell, the balance point shifts to 60 cm. Find the e.m.f. of the second cell.