A cell of itnernal resitance `1.5 Omega` and of e.m.f. `1.5` volt balances `500 cm` on a potentiomter wire. If a wirr of 15 `Omega` is connected between the balance point and the cell, then tha balance point will shift

Options

A. To zero
B. By 500 cm
C. By 750 cm
D. None of the above

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Initial Conditions
We have a cell with:
- Internal resistance (R_internal) = 1.5 Ω
- Electromotive force (e.m.f) = 1.5 V
- Initial balance point on the potentiometer wire = 500 cm

### Step 2: Analyze the Effect of Adding a Resistor
When a wire of resistance 15 Ω is connected between the balance point and the cell, we need to determine how this affects the balance point.

### Step 3: Calculate the Total Resistance
The total resistance in the circuit when the 15 Ω wire is connected becomes:
- Total Resistance (R_total) = R_internal + R_additional
- R_total = 1.5 Ω + 15 Ω = 16.5 Ω

### Step 4: Calculate the Current in the Circuit
Using Ohm's Law (V = IR), we can find the current (I) flowing through the circuit:
- I = e.m.f / R_total
- I = 1.5 V / 16.5 Ω
- I ≈ 0.0909 A

### Step 5: Determine the New Balance Point
The potential drop across the 15 Ω wire can be calculated using Ohm's Law:
- V_drop = I * R_additional
- V_drop = 0.0909 A * 15 Ω
- V_drop ≈ 1.3635 V

### Step 6: Compare the Potential Drops
The initial potential drop across the potentiometer wire was:
- V_initial = e.m.f = 1.5 V

With the additional resistance, the potential drop across the potentiometer wire will now be:
- V_new = V_initial - V_drop
- V_new = 1.5 V - 1.3635 V ≈ 0.1365 V

### Step 7: Calculate the New Balance Point
Using the ratio of the potential drops, we can find the new balance point (L_new):
- L_new / 500 cm = V_new / V_initial
- L_new = 500 cm * (0.1365 V / 1.5 V)
- L_new ≈ 45.5 cm

### Conclusion
The new balance point shifts to approximately 45.5 cm from the original balance point.