Question

For a standard cell of 2 volt, the null point is obtained If the length of potentiometer wire is a 40cm What is the EMF of a known battery of the null point is obtained at 60cm​

Answer 1

Given:

V = 2V

L1 = 40 cm

L2 = 60 cm

To find:

E =?

Explanation:

Let V  be the potential across the balance point and one end of the wire.

Hence according to the principle of the potentiometer,

V ∝ L                                    ...(1)

Let E be emf of a cell employed in the circuit between the ends of the potentiometer wire of length L.

E ∝ L                                    ...(2)

Divide eq (1) by (2)

$\displaystyle \frac{V}{E} = \frac{L_1}{L_2}$

$\displaystyle \frac{2}{E} = \frac{40}{60}$

E = 30 V

EMF of a battery when the null point is obtained at 60cm is 30 V.​