The emf of a battery A is balanced by a length of 80cm on a potentio meter wire. The emf of a standard cell 1v is balanced by 50cm. The emf of A is

Options

A. 2v
B. 1.4v
C. 1.5v
D. 1.6v

Answer

To solve the problem, we will use the relationship between the electromotive force (emf) of a battery and the length of the wire on a potentiometer. The emf is directly proportional to the length of the wire used to balance it.

### Step-by-Step Solution:

1. **Understanding the Proportionality**:
The emf (E) of a battery is directly proportional to the length (L) of the potentiometer wire used to balance it. This can be expressed mathematically as:
\[ \frac{E_A}{E_B} = \frac{L_1}{L_2} \]
where:
- \(E_A\) is the emf of battery A,
- \(E_B\) is the emf of the standard cell (1V),
- \(L_1\) is the length of the wire for battery A (80 cm),
- \(L_2\) is the length of the wire for the standard cell (50 cm).

2. **Substituting Known Values**:
We know that \(E_B = 1 \text{ V}\), \(L_1 = 80 \text{ cm}\), and \(L_2 = 50 \text{ cm}\). Substituting these values into the equation gives:
\[ \frac{E_A}{1} = \frac{80}{50} \]

3. **Calculating the emf of Battery A**:
Rearranging the equation to solve for \(E_A\):
\[ E_A = \frac{80}{50} \text{ V} \]
Simplifying this fraction:
\[ E_A = \frac{8}{5} \text{ V} = 1.6 \text{ V} \]

4. **Final Answer**:
Therefore, the emf of battery A is:
\[ E_A = 1.6 \text{ V} \]