To solve the problem, we will analyze the circuit step by step and derive the necessary equations to find the balancing length of the wire that produces zero deflection in the galvanometer.
### Step 1: Understanding the Circuit
We have:
- An accumulator (battery) with an EMF of \(2 \, V\) and negligible internal resistance connected across a uniform wire of length \(10 \, m\) and resistance \(30 \, \Omega\).
- A cell with an EMF of \(1.5 \, V\) and an internal resistance of \(1 \, \Omega\) connected to one end of the wire, with the other terminal connected to a galvanometer and a slider on the wire.
### Step 2: Finding the Total Resistance in the Circuit
The total resistance in the circuit when the cell is connected is the sum of the resistance of the wire and the internal resistance of the cell:
\[
R_{\text{total}} = R_{\text{wire}} + R_{\text{internal}} = 30 \, \Omega + 1 \, \Omega = 31 \, \Omega
\]
### Step 3: Calculating the Current in the Circuit
Using Ohm's law, the current \(I\) flowing through the circuit can be calculated as:
\[
I = \frac{E_{\text{accumulator}}}{R_{\text{total}}} = \frac{2 \, V}{31 \, \Omega} \approx 0.0645 \, A
\]
### Step 4: Finding the Potential Drop Across the Wire
The potential drop across the wire can be calculated using Ohm's law:
\[
V_{\text{wire}} = I \times R_{\text{wire}} = 0.0645 \, A \times 30 \, \Omega \approx 1.935 \, V
\]
### Step 5: Setting Up the Balancing Condition
At the balancing length \(x\), the potential difference across the length of the wire is equal to the EMF of the cell:
\[
\frac{V_{\text{wire}}}{L} \cdot x = E_{\text{cell}} \implies \frac{1.935 \, V}{10 \, m} \cdot x = 1.5 \, V
\]
Solving for \(x\):
\[
x = \frac{1.5 \, V \cdot 10 \, m}{1.935 \, V} \approx 7.75 \, m
\]
### Step 6: Analyzing Changes in Balancing Length
#### (a) When a coil of resistance \(5 \, \Omega\) is placed in series with the accumulator:
The new total resistance becomes:
\[
R_{\text{new total}} = 30 \, \Omega + 5 \, \Omega + 1 \, \Omega = 36 \, \Omega
\]
Calculating the new current:
\[
I_{\text{new}} = \frac{2 \, V}{36 \, \Omega} \approx 0.0556 \, A
\]
Calculating the new potential drop across the wire:
\[
V_{\text{wire new}} = 0.0556 \, A \times 30 \, \Omega \approx 1.668 \, V
\]
Setting up the new balancing condition:
\[
\frac{1.668 \, V}{10 \, m} \cdot x = 1.5 \, V \implies x = \frac{1.5 \, V \cdot 10 \, m}{1.668 \, V} \approx 8.98 \, m
\]
#### (b) When the cell of \(1.5 \, V\) is shunted with a \(5 \, \Omega\) resistor:
The effective EMF of the shunted cell can be calculated using the formula for parallel resistors. The total resistance of the shunt is:
\[
R_{\text{shunt}} = \frac{1 \, \Omega \cdot 5 \, \Omega}{1 \, \Omega + 5 \, \Omega} = \frac{5}{6} \, \Omega
\]
The new current through the circuit becomes:
\[
I_{\text{shunt}} = \frac{2 \, V}{30 \, \Omega + \frac{5}{6} \, \Omega} \approx 0.0667 \, A
\]
Calculating the new potential drop across the wire:
\[
V_{\text{wire shunt}} = 0.0667 \, A \times 30 \, \Omega \approx 2.001 \, V
\]
Setting up the new balancing condition:
\[
\frac{2.001 \, V}{10 \, m} \cdot x = 1.5 \, V \implies x = \frac{1.5 \, V \cdot 10 \, m}{2.001 \, V} \approx 7.49 \, m
\]
### Summary of Results
1. The length of the wire required for zero deflection of the galvanometer is approximately \(7.75 \, m\).
2. When a \(5 \, \Omega\) coil is added in series, the balancing length changes to approximately \(8.98 \, m\).
3. When the cell is shunted with a \(5 \, \Omega\) resistor, the balancing length changes to approximately \(7.49 \, m\).
