The balance point of a potentiometer with two cells in series is at `l= 250` cm. And if one cell is reversed the balance point is at `l_1 = 150` cm. The ratio of emfs of the two cells is

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

The null point in a potentiometer with a cell of e.m.f. xi is obtained at a distance l on the wire. Then :

In a potentiometer arrangement, a cell of emf 1.25V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0cm, what is the emf of the second cell ?

In a potentiometer experiment, two cells of emf E_1 and E_2 are used in series in the secondary circuit. The balancing length is found to be 58 cm. If the polarity of E_2 is reversed, the balancing length becomes 29 cm. The ratio E_1//E_2 of the emfs of two cells is :

With a resistance P Omega in the left gap and another resistance Q Omega in the right gap of a meter bridge, the balance point is obtained at 40 cm. When a resistance of 6 Omega is connected across Q, the balance point shifts to 50 cm. The value of resistance P is

In a potentiometer experiment of a cell of emf 1.25 V gives balancing length of 30 cm. If the cell is replaced by another cell, balancing length is found to be 40 cm. What is the emf of second cell ?

Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 ohm is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance, in ohm is

Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 ohms is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in ohms is