A cell can be balanced against `110 cm` and `100 cm` of potentiometer wire, respectively with and without being short circuited through a resistance of `10 Omega`. Its internal resistance is

A cll balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of 5 Omega . But when it is shunted by a resistance of 10 Omega , the balancing length increases by 25 cm. What is the balancing length when the cell is in an open circuit ?

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

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 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 ?

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 ?