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 ?

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.

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?

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 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

In an experiment to find the internal resistance of a cell by a potentiometer, a balance was obtained for 50 cm length of the potentiometer wire, with a cell of e.m.f. 2V. When the cell was shunted by a resistance of 2 Omega , the balancing length of the potentiometer wire was 40 cm. What was the internal resistance of the cell ?