Question

A potentiometer has 400 cm long wire which is connected to an auxiliary of steady voltage 4 V. A Leclanche cell gives null point at 140 cm and Daniel cell at 100 cm. i)Compare the emfs of the two cells. ii) If the length of wire is increased by 100 cm, find the position of the null point with the first cell.

Answer 1

i) EMF of Leclanche cell > Emf of Daniel cell

ii) The position of the null point with the first cell is 112 cm.

Explanation:

The emf of the cell is given by the formula:

EMF of cell = Potential gradient × Null point on the wire

Where,

Potential gradient = 400 cm

i) The emf of Leclanche cell:

EMF of Leclanche cell = 400 × 140

∴ EMF of Leclanche cell = 56000 V

The emf of Daniel cell:

Emf of Daniel cell = 400 × 100

∴ Emf of Daniel cell = 40000 V

EMF of Leclanche cell > Emf of Daniel cell

ii) The first cell given in the question is Leclanche cell.

The emf of the Leclanche cell when length is increased by 100 cm is,

EMF of Leclanche cell = (400 + 100) × Null point

56000 V = 500 × Null point

Null point = 56000/500

∴ Null point = 112 cm

Answer 2

Answer:

A. 7:5

B. 175 cm

Explanation:

Attached below