A potentiometer is used for the compaisem of e.m.f. of two cells `E_(1)` and `E_(2)`. For cell `E_(1)` the no deflection point os obtained at `20 cm` and for `E_(2)` the no deflection point is obtained at `30 cm`. The ratio of their e.m.f.'s will be

Options

A. `2//3`
B. `3//2`
C. 1
D. 2

Answer

To find the ratio of the electromotive forces (e.m.f.) of the two cells \( E_1 \) and \( E_2 \) using a potentiometer, we can follow these steps:

### Step 1: Understand the principle of the potentiometer
A potentiometer measures the e.m.f. of a cell by balancing it against a known potential difference. The no deflection point indicates that the potential difference across the potentiometer wire is equal to the e.m.f. of the cell.

### Step 2: Set up the equations for the two cells
Let \( V \) be the potential drop per unit length of the potentiometer wire. For cell \( E_1 \), the no deflection point is obtained at \( 20 \, \text{cm} \):
\[ E_1 = 20 \, \text{cm} \times V \]
For cell \( E_2 \), the no deflection point is obtained at \( 30 \, \text{cm} \):
\[ E_2 = 30 \, \text{cm} \times V \]

### Step 3: Write the ratio of the e.m.f.s
To find the ratio of the e.m.f.s of the two cells, we can divide the two equations:
\[ \frac{E_1}{E_2} = \frac{20 \, \text{cm} \times V}{30 \, \text{cm} \times V} \]

### Step 4: Simplify the ratio
The potential drop per unit length \( V \) cancels out:
\[ \frac{E_1}{E_2} = \frac{20}{30} = \frac{2}{3} \]

### Step 5: Conclude the result
Thus, the ratio of the e.m.f.s of the two cells is:
\[ \frac{E_1}{E_2} = \frac{2}{3} \]

### Final Answer
The ratio of their e.m.f.s \( E_1 : E_2 = 2 : 3 \).

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