in a potentiometer arrangement, a cell of emf 1.25V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell, then the balance point shifts to 63 cm. What is the emf of the second cell?

To find the electromotive force (emf) of the second cell in a potentiometer arrangement, we can use the principle that the potential difference across the length of the wire is directly proportional to the length of the wire itself. This means that if we know the balance points for two different cells, we can set up a ratio to find the emf of the second cell.

Understanding the Potentiometer Setup

A potentiometer is a device used to measure the emf of a cell by comparing it to a known reference voltage. In this case, we have two cells with different emfs and their corresponding balance points on the potentiometer wire.

Given Information

• Emf of the first cell (E1) = 1.25 V
• Balance point for the first cell (L1) = 35 cm
• Balance point for the second cell (L2) = 63 cm

Setting Up the Proportion

According to the principle of the potentiometer, the ratio of the emfs of the two cells is equal to the ratio of their respective balance lengths. This can be expressed mathematically as:

E1 / E2 = L1 / L2

Substituting the Values

Now, we can substitute the known values into this equation:

1.25 V / E2 = 35 cm / 63 cm

Calculating the Emf of the Second Cell

To find E2, we can rearrange the equation:

E2 = E1 * (L2 / L1)

Now plug in the values:

E2 = 1.25 V * (63 cm / 35 cm)

Calculating the fraction:

63 cm / 35 cm = 1.8

Now, substituting this back into the equation:

E2 = 1.25 V * 1.8

E2 = 2.25 V

Final Result

The emf of the second cell is 2.25 V. This demonstrates how the balance point on the potentiometer wire can be used to determine the emf of different cells by maintaining a consistent ratio based on their respective lengths on the wire.