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 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

A Daniel cell is balanced on 125 cm length of a potentiometer wire. Now the cells is short-circuited by a resistance 2 ohm and the balance is obtained at 100 cm . The internal resistance of the Dainel cell is

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 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

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

In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across 0.52 m of the potentiometer wire. If the cell is shunted by a resistance of 5Omega balance is obtained when the cell connected across 0.4m of the wire. Find the internal resistance of the cell.

In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2Omega , the balancing length becomes 120 cm.The internal resistance of the cell 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