Question

) In a potentiometer experiment, balancing length for first cell and second cell are 2.5 m and 2 m respectively, when cells are used separately. Hence E1/E2 is​

Answer 1

Given:

In a potentiometer experiment, balancing length for first cell and second cell are 2.5 m and 2 m respectively, when cells are used separately.

To find:

Ratio of E1 and E2.

Calculation:

Let the voltage drop per unit length on the potentiometer wire be denoted as V:

When E1 cell is used in the potentiometer , the balancing length comes as 2.5 cm.

$\therefore \: E1 = V \times (balancing \: length)$

$= > \: E1 = V \times2.5$

$= > \: E1 = \dfrac{5V}{2} \: volt$

When E2 is used in the potentiometer wire the balancing length comes as 5 cm.

$\therefore \: E2 = V \times (balancing \: length)$

$= > \: E2 = V \times2$

$= > \: E2 = 2V \: volt$

So, the required ratio:

$\therefore \: E1 : E2 = \dfrac{5}{2} : 2$

$= > \: E1 : E2 = 5 : 4$

So, the ratio is E1 : E2 = 5 : 4.