To solve the problem step-by-step, we will analyze the situation with the galvanometer and the resistances connected in parallel.
### Step 1: Understand the Initial Condition
The problem states that when a resistance of \(5 \, \Omega\) is connected in parallel with the galvanometer, the deflection falls to \( \frac{1}{10} \) of its original value. This means that the current through the galvanometer is now \( \frac{1}{10} I_0 \), where \( I_0 \) is the initial current through the galvanometer.
### Step 2: Set Up the Current Division
Let \( G \) be the resistance of the galvanometer. The total current \( I_0 \) splits into two parts:
- Current through the galvanometer: \( I_g = \frac{1}{10} I_0 \)
- Current through the \(5 \, \Omega\) resistor: \( I_{5\Omega} = \frac{9}{10} I_0 \)
### Step 3: Apply Ohm's Law
Since the voltage across the galvanometer and the \(5 \, \Omega\) resistor is the same, we can use Ohm's law to relate the voltages:
\[
V = I_g \cdot G = I_{5\Omega} \cdot 5
\]
Substituting the values:
\[
\frac{1}{10} I_0 \cdot G = \frac{9}{10} I_0 \cdot 5
\]
### Step 4: Simplify and Solve for G
Cancelling \( I_0 \) from both sides (assuming \( I_0 \neq 0 \)):
\[
\frac{1}{10} G = \frac{9}{10} \cdot 5
\]
Multiplying both sides by \(10\):
\[
G = 9 \cdot 5 = 45 \, \Omega
\]
### Step 5: Introduce the Additional Resistance
Now, we connect an additional \(2 \, \Omega\) resistor in parallel with the galvanometer. The new configuration has the galvanometer resistance \( G = 45 \, \Omega \) and the \(2 \, \Omega\) resistor in parallel.
### Step 6: Calculate the Equivalent Resistance
The equivalent resistance \( R_{eq} \) of the galvanometer and the \(2 \, \Omega\) resistor in parallel is given by:
\[
\frac{1}{R_{eq}} = \frac{1}{G} + \frac{1}{2}
\]
Substituting \( G = 45 \):
\[
\frac{1}{R_{eq}} = \frac{1}{45} + \frac{1}{2}
\]
Finding a common denominator (90):
\[
\frac{1}{R_{eq}} = \frac{2}{90} + \frac{45}{90} = \frac{47}{90}
\]
Thus,
\[
R_{eq} = \frac{90}{47} \, \Omega
\]
### Step 7: Determine the New Current Through the Galvanometer
Now, we need to find the new current through the galvanometer. The total current \( I_0 \) is still the same, and it will now split among \( R_{eq} \) and the \(5 \, \Omega\) resistor.
Using the current division rule:
\[
I_g' = I_0 \cdot \frac{R_{eq}}{R_{eq} + 5}
\]
Substituting \( R_{eq} = \frac{90}{47} \):
\[
I_g' = I_0 \cdot \frac{\frac{90}{47}}{\frac{90}{47} + 5}
\]
Calculating \( \frac{90}{47} + 5 = \frac{90 + 235}{47} = \frac{325}{47} \):
\[
I_g' = I_0 \cdot \frac{\frac{90}{47}}{\frac{325}{47}} = I_0 \cdot \frac{90}{325} = I_0 \cdot \frac{18}{65}
\]
### Step 8: Final Deflection Calculation
The new deflection of the galvanometer is now \( \frac{I_g'}{I_0} = \frac{18}{65} \).
### Conclusion
The final deflection of the galvanometer when an additional \(2 \, \Omega\) resistor is connected in parallel is \( \frac{18}{65} \) of the original deflection.
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