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

In a potentiometer experiment, the galvanometer shows no deflection when a cell is connected across 60 cm of the potentiometer wire. If the cell is shunted by a resistance of 6 Omega , the balance is obtained across 50 cm of the wire. The internal resistance of the cell is

In a potentiometer experiment, the galvanometer shows no deflection when a cell is connected across 60 cm of the potentiometer wire. If the cell is shunted by a resistance of 6 Omega , the balance is obtained across 50 cm of the wire. The internal resistance of the cell is

A cell gives a balance with 85 cm of a potentiometer wire. When the terminals of the cell are shorted through a resistance of 7.5Omega , the balance is obtained at 75 cm. Find the internal resistance of the cell.

For a cell of e.m.f. 2 V , a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a 2Omega resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is

For a cell of e.m.f. 2 V , a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a 2Omega resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell 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 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.