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.

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 emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Omega , the balancing length is reduced by 20 cm. Find the internal resistance of the cell.

A 2.0 V potentiometer is used to determine the internal resistance of 1.5 V cell. The balance point of the cell in the circuit is 75 cm . When a resistor of 10Omega is connected across cel, the balance point sifts to 60 cm . The internal resistance of the cell is

In potentiometer experiment, a cell is balanced by length 120 cm. When a cell is shunted by resistance of 5Omega , the balancing length is 80 cm. The internal resistance of cell is

A potentiometer wire of length 4 m and resistance 8 Omega is connected in, series with a battery of e.m.f. 2 V and negligible internal resistance. If the e.m.f. of cell balances against length of 217 cm of the wire, find the e.m.f. of cell. When the cell is shunted by assistance of 15 Omega ,the balancing length is reduced to 200 cm. Find the internal resistance of the 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

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

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

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Omega the balance point is found at 40 cm of the wire from the same end. The internal resistance of the cell is