Two cells when connected in series are balanced on 8 m on a potentiometer. If the cells are connected with polarities of one of the cells reversed, then they balance on 2 m. The ratio of emf’s of two cells isA. 3:4B. 4:3C. 3:5D. 5:3

To find the ratio of the electromotive forces (EMFs) of the two cells, we can use the principle of a potentiometer. When two cells are connected in series and balanced on a potentiometer, the potential difference across the length of the wire is directly proportional to their EMFs.

Let E1 be the EMF of the first cell and E2 be the EMF of the second cell. When the cells are connected in series and balanced at 8 m on the potentiometer, we have:

E1 + E2 = 8 ...(1)

Now, when the cells are connected with the polarities of one of the cells reversed and balanced at 2 m on the potentiometer, the potential difference is still proportional to their EMFs. However, the total potential difference is in the opposite direction due to the reversed polarity. So, we have:

E1 - E2 = 2 ...(2)

Now, we can solve this system of equations (equation 1 and equation 2) to find the values of E1 and E2:

Adding equation 1 and equation 2:

(E1 + E2) + (E1 - E2) = 8 + 2

2E1 = 10

E1 = 5

Substitute the value of E1 back into equation 1 to find E2:

5 + E2 = 8

E2 = 8 - 5

E2 = 3

So, the EMF of the first cell (E1) is 5 units, and the EMF of the second cell (E2) is 3 units.

The ratio of the EMFs of the two cells is E1:E2 = 5:3.

Therefore, the correct answer is option B: 4:3.