The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is `0.5 Omega`. If the balance point is obtained at I = 30 cm from the positive end, the e.m.f. of the battery is .
where i is the current in the potentiometer wire.

Options

A. `(30E)/(100.5)`
B. `(30E)/(100-0.5)`
C. `(30(E-0.5i))/100`
D. `(30E)/(100)`

Explanation

To solve the problem, we need to find the e.m.f. of the battery using the information provided about the potentiometer and the balance point. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a potentiometer wire of length \( L = 100 \, \text{cm} \) and a standard cell with e.m.f. \( E \, \text{volts} \). The battery we want to measure has an internal resistance of \( r = 0.5 \, \Omega \), and the balance point is at \( l = 30 \, \text{cm} \).

### Step 2: Determine the Potential Drop Across the Potentiometer Wire
The potential drop across the entire potentiometer wire is proportional to its length. The potential drop across the 30 cm length of the potentiometer wire can be calculated as follows:
\[ V = \frac{l}{L} \times E = \frac{30 \, \text{cm}}{100 \, \text{cm}} \times E = 0.3E \]

### Step 3: Relate the Potential Drop to the Battery's E.M.F.
At the balance point, the potential drop across the potentiometer wire (which is \( 0.3E \)) is equal to the e.m.f. of the battery plus the potential drop across its internal resistance. The current \( i \) flowing through the potentiometer wire can be expressed as:
\[ V = E_{battery} - i \cdot r \]
Where \( E_{battery} \) is the e.m.f. of the battery we want to find.

### Step 4: Calculate the Current \( i \)
The current \( i \) can be determined using the relationship of the potential drop across the potentiometer wire. Since \( V = 0.3E \) and the internal resistance \( r = 0.5 \, \Omega \), we can express the current as:
\[ i = \frac{V}{r} = \frac{0.3E}{0.5} = 0.6E \]

### Step 5: Substitute \( i \) Back into the Equation
Now we can substitute \( i \) back into the equation relating the e.m.f. of the battery:
\[ 0.3E = E_{battery} - (0.6E \cdot 0.5) \]
\[ 0.3E = E_{battery} - 0.3E \]
\[ E_{battery} = 0.3E + 0.3E = 0.6E \]

### Final Answer
The e.m.f. of the battery is:
\[ E_{battery} = 0.6E \, \text{volts} \]